A Probability Puzzle

Many of you are familiar with the old Monty Hall problem. You might also be aware that it rose to prominence as a result of a column in Parade magazine by Marilyn vos Savant. After Savant’s initial discussion of the problem, she received a flood of angry letters, some from actual mathematicians, saying that she was wrong and that her answer was foolish. Actually, Savant got it right, and the problem is now a staple of courses in elementary probability theory.

I recently got it into my head to track down the actual correspondence she received on the issue. Courtesy of the local public library, I found an anthology of some of the letters she received during the time she wrote the column. The Monty Hall discussion was in there, as I hoped.

To my surprise, however, I discovered a second probability puzzle that she addressed that also led to some heated controversy in the replies. It’s a classic puzzle, so perhaps you have heard it before. But just in case you haven’t, here it is:

A shopkeeper says she has two new baby beagles to show you, but she doesn’t know whether they’re both male, both female, or one of each. You tell her that you want only a male, and she telephones the fellow who’s giving them a bath. “Is at least one a male?” she asks him. She receives a reply. “Yes!” she informs you with a smile. What is the probability that the other one is a male?

Feel free to hash that out in the comments.

Comments

  1. #1 David D.G.
    December 26, 2006

    Well, I am no mathematician at all, but my intuitive reaction is to say that the odds for the other puppy being a male are exactly 50%. It has to be either male or female, one or the other, but the sex of the first puppy has no bearing whatsoever on the sex of the second one, so the chance is a straightforward 1 in 2.

    ~David D.G.

  2. #2 steve s
    December 26, 2006

    If you go about creating random puppy pairs, you’ll get, with equal probability, boy/boy, boy/girl, girl/boy, and girl/girl. Only the first three cases apply in this problem, and there are twice as many mixed pairs as boy/boy. So it’s 33% to 66%.

  3. #3 steve s
    December 26, 2006

    I was very confused by Marilyn’s answer when it came out, for a simple reason. Either she didn’t make it clear, or I missed, that Monty Hall knew what was behind which doors, and was only opening doors he knew didn’t contain the prize. It seemed, the way I read it, that we were just looking at the subset of cases where Hall had happened to open doors without the prize, and if you look at those cases, the odds were neutral on a switch.

  4. #4 steve s
    December 26, 2006

    I see the poor wording from that Wikipedia link:

    Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?

    This doesn’t say anything about him deliberately not opening a prize door, it just says he knows. This is mentioned by the Wikipedians:

    While this is a common presentation of the problem, it is problematic as it leaves important conditions of the problem unstated. In an unambiguous statement of the problem by Mueser and Granberg, the host is constrained to always open a door revealing a goat and to always make the offer to switch.

  5. #5 Fred
    December 26, 2006

    Seems to me like it’s a 50/50 chance that it’s a male. Nothing else has a bearing on it– there are only two choices.

  6. #6 Scott Stratton
    December 26, 2006

    I’m definately not a mathmetician. It seems like you could look at it two ways. The first Steve has already explained. To recap:

    There are four initial possibilities:

    M M
    M F
    F M
    F F

    You KNOW it has to be one of the first three cases, since you know that, the “frame” shifts. Now 100% of the cases is equal to the first three cases. So there is now a 1/3 (33%) chance of either M/M, M/F, or F/M. Since the question is about having TWO males, there is only a 33% chance.

    But I’m not sure why you can’t just say, since there are four possibilities, and based on the information you have, you know the case being asked about is the M/M case, and that happens 25% of the time. So the odds of the other eaglet being male is 25%?

    Consider me confused! Naturally, I didn’t get the Monty Hall problem right the first time either.

  7. #7 Karl
    December 26, 2006

    I have seen this before and I’ve seen the analysis given by steve s. I had a lot of trouble with that for a while. After all the two events are independent, so the fact that one is male doesn’t affect what the other one is, so it should be 1/2. But, then I thought of it this way: suppose that you have 100 pairs of puppies. Think of them as distinguishable by age. Of the 100 older ones 50 will be m and 50 f. Then, of these 50 older m’s, 25 of the second borns will be m and 25 f. That gives 25 mm and 25 mf (in order by age). Same for f being the older one. There will be 25 ff and 25 fm. Notice that mf and fm are two separate sets when examined by age. But when examined only by gender there are 50 that are one f and one m. So, the probability is that there will be 25 pairs that are ff, 25 pairs that are mm, 50 pairs that are either fm or mf. Then, once you know that one of them is m, you only have 75 pairs to examine. And, of those, only 25 are mm. So the odds that the second one is m given that the other one is m. is 25/75 – which is 1/3 or 33%. I think that’s the same thing steve s said but with a little elaboration. The key is that this a conditional probabilty rather than an absolute, or undonditional situation.

  8. #8 Jason Rosenhouse
    December 26, 2006

    steve s-

    The poor wording of the problem was due to the person who originally wrote the letter. You’re right that the original wording does not make it clear that the host always opens an empty door. However, vos Savant’s answer made it clear that that assumption was essential. In her initial reply she wrote (among other things):

    Suppose there are a million doors, and you pick door number one. Then the host, who knows what’s behind the doors and will always avoid the one with the prize, opens them all except door number 777,777. You’d switch to that door pretty fast, wouldn’t you?

    Her interlocutors did not harp on this point in their replies. Instead they all gave some version of the argument that after Monty Hall opens a door, there are only two equally likely doors remaining, and that’s that.

    Vos Savant addressed the question two more times, and in one of those cases she made it absolutely explicit, putting it in italics even, that it was crucial to assume that Monty Hall always opens an empty door.

    As for the beagles (not eaglets, note) I’ll give more people a chance to weigh in before giving my thoughts on the subject!

  9. #9 Corkscrew
    December 26, 2006

    define A: one or more is male
    define B: both are male

    Then what we’re after is P(B|A) = P(BnA)/P(A)

    P(BnA) = P(B) as B=>A
    P(B) = 0.5^2 = 0.25
    P(A) = 1 – P(A’) = 1 – P(B) as P(both female) = P(both male)

    So P(B|A) = P(BnA)/P(A) = P(B)/(1-P(B)) = 0.25/0.75 = 1/3

  10. #10 Kevin
    December 26, 2006

    Seems to me to be 50%. We start with four possibilities:
    MM
    MF
    FM
    FF

    But since we haven’t declared specifically that we care if beagle A is male, really there are only 3 possibilities:
    MM
    MF
    FF

    So there’s a 66% chance that at least one beagle is male. There is a 33% chance that neither beagle is male. Since the phone call assures us that at least one is male, we know that FF is not a possibility. So we’re left with
    MM
    MF

    Looks like a 50% chance to me. 8/ It hasn’t asked us to choose or commit to any beagle, unlike the monty hall problem, so the knowledge isn’t altering anything. If it’s ordered, then it’s different though. If we care about which beagle is which, it becomes more like the monty hall problem. After we know that there’s at least one male, our choices are left with
    MM
    MF
    FM

    and that means there are 2 permutations possible where the other beagle is a female. So the chance of the other beagle being male is 33%. The question wording is rather ambiguous.

  11. #11 doug bennion
    December 26, 2006

    You could refine your answer if only you knew how long it took the guy taking the bath, to come up with the answer.

    There are three remaining cases, MF, MM and FM. If he was away for a short time, and he had only time to examine one puppy, the answer is 1/2, because he ran into either MF or MM. If he was away for long enough to examine two puppies, the answer is 0, because he ran into FM.

    Combined they give the correct answer of 1/3 (2/3*1/2 + 1/3*0).

  12. #12 John
    December 26, 2006

    All the fancy math aside, I’m going to say that these are independent events and the probability of the second pup being male is 1/2.

  13. #13 steve s
    December 26, 2006

    Yeah, I wasn’t confused after I saw one of the followups where she was emphatic about that point. It was just a case of my reading too fast.

  14. #14 Jeff Chamberlain
    December 26, 2006

    Given the problem as stated, isn’t option MF the same as option FM?

  15. #15 Koray
    December 26, 2006

    Scott S, your first answer is correct because it’s a conditional probability problem. (Bayes’ theorem)

  16. #16 Roy
    December 26, 2006

    The Monty Hall problem, solution, and arguments against, are the result of mathematicians knowing their math but not knowing game shows.

    If the problem is cast as a lone player dealing with a mechanism, where the innards could be examined to see that they abide by the rules, then the problem is straightforward, as is its solution.

    When cast as a game show, the mathematician is in over his head because he cannot say how many players there are. We have the contestant for sure, but the emcee may also be a player, and the floor manager may be a player, and the production manager may be a player, and there may be a test audience as players. As an illustration, suppose that the test audience doesn’t like the egghead contestant, so the managers have the crew behind the scenes make sure he loses: whatever choice he makes, he ends up with the bad prize.

    How would a mathematician figure the probabilities in a real game show? If he thinks that he can, he’s a fool.

    I watched “1 vs 100″ just once to see how it went. I conclude that I cannot guess what the rules actually are. Some questions are so simple, I am surprised anyone gets them wrong. Then comes a question almost no one on the planet will know the answer to. And with 100 initial players in the ‘mob’, who knows what directions they got before the start of play and what they were given after play began? This game show cannot be analyzed because nobody can find out how it’s played.

  17. #17 doctorgoo
    December 26, 2006

    The trick to this problem, that most have recognized, is that you start with two puppies, and therefore four permutations (MM,MF,FM,FF). Of the 4 possibilities, one (FF) is eliminated right away, leaving 3 left. Of these, only one allows for a second male.

    But it’s easy to see why people get confused by the solution to this problem. So here’s an even more basic math problem for you to solve:

    What does 1+2×3 equal?

    Of course, the answer is 7 (because multiplication/division always occurs before addition/subtraction). But many people will claim 1+2×3=9 just because they’ll do the math from left to right.

    Go on, try it with your friends and family… see how many answer 9. And when you explain why it’s 7, get prepared for an argument! :-)

  18. #18 doctorgoo
    December 26, 2006

    Of course, there is an easy way to test this. Instead of using beagle puppies, use a flip of 2 coins:

    Step 1. Flip 2 coins at the same time. Let heads=male & tails=female.
    Step 2. If both are tails (FF), then disregard and flip again.
    Step 3. Now that at least one coin is heads, pull one of the heads aside, and record the condition (heads or tails) of the other coin.

    Repeat these steps 100 times (or however many times is required to be statistically valid), and if you end up with approximately 33 heads instead of 50, then you proved it empirically.

    Also, note that it is the second step that changes the odds from 50:50 to 33:66.

  19. #19 Kevin
    December 26, 2006

    Only problem, Dr Goo, is when you end up with 41 heads. ;)

    It all comes down to whether we care – are we considering MF to be a different, or the same, outcome as FM.

  20. #20 Dave
    December 26, 2006

    Kevin,

    You have to be careful if you decide to list only 3 outcome
    MM
    FF
    MF
    because the MF outcome has a higher probability (50%) than either MM (25%) or FF (25%). If that doesn’t seem right, try repeatedly flipping two nickels and count the number of HH, TT, and HT results (heads H, tails T) You’ll get 50% HT, 25% FF, and 25% TT.

  21. #21 Russell Blackford
    December 26, 2006

    I’m going to be perverse and go with the minority saying it’s 50:50.

    There are two dogs which can be distinguished as, say, the older dog and the younger dog or the dog on the left and the dog on the right.

    There are four possibilities: MM, MF, FF, and FM, where the first letter refers to the dog on the left and the second to the dog on the right. All possibilities are equal before we are given any information. Thus, before we have the information there is a 25 per cent chance that both dogs are male.

    We are then told that one dog is male. The person telling us this either had in mind the dog on the left or the dog on the right. We now branch the analysis:

    If it was the dog on the left, we have eliminated both FF and FM. There are now only two possibilities: MF and MM. There is no reason for the probabilities for these to be other than 50:50.

    If it was the dog on the right, then the only possibilities left are FM and MM. Again, there is no reason for the probabilities to be other than 50:50.

    Whichever dog the person had in mind, then, the probability that the other dog is male has become 50 per cent.

  22. #22 Kevin
    December 26, 2006

    Dave-
    I would need to be careful if the question was about the probability of a given combination/permutation. But it’s not. The fact that one beagle is male is a given. So what matters is the ambiguous part of the question – do we care which beagle we have been told is male? Is MF the same as FM? If we don’t care, there’s only two outcomes: male or female, 50% chance. If we do care, its more complicated. If Beagle A is male, Beagle B could be M or F. If Beagle B is Male, then Beagle A could be M or F. Now, MF and FM are different outcomes we must take into account, so its a 33% chance, because in both cases MM is an identical outcome – but not one that has a greater chance of occurring.

  23. #23 Kevin
    December 26, 2006

    Russell-
    Your branch is flawed, since MM is an identical outcome regardless of branch. Your breakdown would give the following outcomes:
    MM
    MF
    FM
    MM

    If those were all unique, then the chance of the second dog being a male would be 50%. But MM is an identical outcome. You’re left with
    MM
    MF
    FM

    and back to the 33% chance.

  24. #24 Fred
    December 26, 2006

    Guys, you’re letting all of the surrounding stuff fool you. It’s a 50/50 chance.

    If I told you to flip a coin, surely you’d say that the chances of one outcome or the other is 50%. Now what if I said “There are three doors, pick one, then go through. Now you will be presented with two more doors. Pick one. Now flip the coin. What are the chances the coin will be heads?” The answer is 50/50. The nonsense with the doors doesn’t matter. Likewise, how many other dogs in a litter are male or female doesn’t matter. Maybe for an animal that can’t have two babies at the same time that are the same sex it would be different, but that’s not the case with dogs.

  25. #25 Kevin
    December 26, 2006

    Fred-
    it absolutely does matter. Your examples are not representative of the problems.

    Monty Hall-
    Choose one of 3 doors. You pick a door, it has a 33% chance of being a prize. This is set – you can now never change this.
    Of the remaining 2 doors, there is a 66% chance that one of them has a prize. Your door has a 33% chance, the remaining two doors add up to a 66% chance. Now, someone eliminates one door that they 100% know does NOT have a prize. Your door still has a 33% chance of having a prize – you can’t change that. The other door has to have a 66% chance of having a prize – it’s the only other door.

    As for the puppies, I don’t think you understand the question. Sorry.

  26. #26 doctorgoo
    December 26, 2006

    Wrong Fred.

    Go back to my 3 steps of the coin flips. When you flip two coins and disregard any time it comes up tails-tails (FF), then it is twice as likely to get one heads and one tails as it is to get head-heads.

    Therefore, it’s a 33% chance the second coin is heads, and similarly, a 33% chance the second beagle is is male.

  27. #27 R
    December 26, 2006

    Kevin, I can’t quite follow your rebuttal but I must admit that I have this instinctive feeling that you are right and my analysis is flawed in some way that I can’t quite see … and the answer is “p = 1/3″. After all, all we’ve actually been told is that there are two dogs distributed as either MM, MF or FM (but not FF).

    I’ll await further illumination from the real mathematicians here.

  28. #28 Russell Blackford
    December 26, 2006

    ^whoops, that was me

  29. #29 doctorgoo
    December 26, 2006

    Kevin said:

    Only problem, Dr Goo, is when you end up with 41 heads. ;)

    Well, this just means that you haven’t collected enough data points. So keep going until you get the result you want.**

    **This is called “Testing to Compliance”. Not recommended for FDA/GxP work, but probably good enough here. ;-)

  30. #30 Dave
    December 26, 2006

    Kevin,

    Sorry, the eggnog may have slowed me down, but I don’t follow your reasoning about what we care about. Whether we care doesn’t change the probability. Only what information we have changes he probability. (I’m probably missing something.)

    Hey Fred,

    I disagree. Suppose you flip a nickel and a penny. There are four possible outcomes:
    A penny heads nickel heads
    B penny heads nickel tails
    C penny tails nickel heads
    D penny tails nickel tails

    If you tell me that “the penny is heads”, then I can narrow the possibilities to A and B. If you ask me if the other coin is heads, given the info I have it is 1 chance (A) out of 2 (A or B). 50% Nothing wierd there.

    But if you instead tell me that “at least one coin is heads”, then I can narrow the possibilities to A, B, and C. If you now ask me if the other coin is heads, given the info I have it is 1 chance (A) out of 3 (A, B, or C). 33%

  31. #31 Kevin
    December 26, 2006

    Dave-
    Here’s why it matters. Russell may have said it most clearly: “After all, all we’ve actually been told is that there are two dogs distributed as either MM, MF or FM (but not FF).”

    1 in 3 chance, because the options are MM, MF, or FM. However, do we consider MF and FM to be different outcomes? Or do we not care which dog is male, just that one is? In that case, MF and FM are the same outcome, and we’re left with just MM and MF as possibilities. That would make it a 50% chance.

  32. #32 Dave
    December 26, 2006

    Kevin,

    I disagree. I see no ambiguity.

    You get the same answer either way: 1/3.

    You can call it three outcomes. Then the outcomes have probabilities MM (1/3) MF (1/3) FM (1/3).

    If you want to call it two outcomes MM or ONEofEACH, then the probabilities are MM (1/3) and ONEofEACH (2/3), because there are two ways to get ONEofEACH (MF and FM). So you can’t just say “two possibilities MM and ONEofEACH so it is 50:50.” No. The two are not equally likely.

  33. #33 doctorgoo
    December 26, 2006

    Kevin said:

    Or do we not care which dog is male, just that one is? In that case, MF and FM are the same outcome, and we’re left with just MM and MF as possibilities. That would make it a 50% chance.

    Whie it’s true that you just have MM and MF as possibilities, the MF (which is equivalent to FM) is twice as likely to occur. Therefore, it’s a 33% chance.

  34. #34 Amos
    December 26, 2006

    Though I admit I’m no expert on probability, it seems to me that those who say 50% are reading “What is the probability that the other one is a male?” at face value, while those who say 1/3 are implicitly converting it to the alternate question “What is the probability that both are male, given that one is male?”

    My naive take is that if two readings give different answers and one requires conversion to a different question, then that conversion to a different question must be a mistake. My vote is for 50%.

  35. #35 Dave
    December 26, 2006

    Amos,

    I disagree. In my reading, that is the question: “What is the probability that both are male given that one is?” Otherwise, what is the point of the rest of the statement of the problem?

    I have flipped a coin. It came up heads. What is the probability that it is tails?
    The answer is 0%.

    If you answer, “the probability of flipping a coin and getting tails is 50%”, then I would say you haven’t answered the question I posed.

  36. #36 Tom Renbarger
    December 26, 2006

    Amos, you aren’t told which one is male, just that at least one is. In effect, it’s the conversion to the question that has the 50-50 answer that is the incorrect conversion.

  37. #37 Amos
    December 26, 2006

    If you were to ask me “I have flipped a coin. It came up heads. What is the probability that it is tails?” I would also answer 0%. You’ve done a good job of posing an unambiguous question.

    What if we were asked this:

    “I’m going to flip two coins. I want to know if both are heads, both are tails, or they are one of each. The first is heads. Given that, what is the probability both are heads?”

    or this:

    “I’m going to flip two coins. What is the probability that one is heads? What is the probability the other is heads?”

    For the first question, I’d say 1/3 and for the second, 1/2. The question we’ve been given has the beginning of my first question, but the end of my second. But the end of the question is where the actual question is posed. Since that’s where the real question being asked is, we have to answer my second question, giving 1/2.

  38. #38 Baratos
    December 26, 2006

    The answer is red.

  39. #39 Kevin
    December 26, 2006

    dr goo-
    Hmm, I have to disagree still with “Whie it’s true that you just have MM and MF as possibilities, the MF (which is equivalent to FM) is twice as likely to occur. Therefore, it’s a 33% chance.”

    You’re still treating MF and FM as different outcomes. Once you stop caring which dog is male, it becomes an independent event sequential event. Take a coin example. I’m going to flip a coin twice. The first time it’s heads. What are the odds that the next flip will be heads? It’s 50%. To get it to a 33% chance, you would have to flip the two coins simultaneously and instantly cover them up. While still covered, someone tells you they saw one is heads, but they don’t know which one. The question now is, what is the chance they are both heads? It’s 33%.

  40. #40 Tom Renbarger
    December 26, 2006

    “The other one” doesn’t refer to a specific puppy, just to the as-yet-to-be-identified one that we aren’t sure is male. It’s a probability question, not a grammar question, though I can see where one might find the wording to be a little shaky.

    Think of it this way — I’m going to flip two coins out of your sight, and I’m going to show you a heads so long as at least one of the coins lands that way. What’s the probability that the other one (that is, the one still hidden) is also heads?

  41. #41 Dave
    December 26, 2006

    Amos,

    You agreed that my silly problem with 0% answer is unambiguous. Yet I didn’t use the words “Given it is heads,” when I asked “What is the probability that it is tails?” So when you read the sentence “What is the probability that it is tails?”, you didn’t have a problem with inferring that the information in the previous sentences was to be taken into account, though this connection was not stated explicitly.

    So what’s the difference with the beagle problem? The puzzle is stated in 5 or so sentences. When the final sentence reads “What is the probability that the other one is a male?” Surely it is logical to infer that this means, “given what was just said in the previous 4 sentences”.

  42. #42 doctorgoo
    December 26, 2006

    Once you stop caring which dog is male, it becomes an independent event sequential event.

    Why do you think it could become an independent sequential event? The premise of the original question sets up a two-coin simultaneous flip (which is then simplified to eliminate 25% of the possibilities.) It does NOT set up two independent sequential events as you (and maybe Amos) suggest.

    The original question asked “what is the probability that the other one is a male”. And the odds of BOTH being male, once you know that one MUST be, is 33%.

    Sorry, I’m not sure I can explain it better than this without looking it up some old college textbooks. But don’t worry, this is why Jason said there was lots of “heated controversy”.

    :-)

  43. #43 Amos
    December 26, 2006

    Dave,

    I don’t have a problem with the use of information from previous sentences when that information is relevant to the question; and in your 0% example the connection was clear. In the beagle problem, as I showed in my previous post, I read the last sentence to ask a question about an independent event, making all the given information irrelevant. I see it all hinging on how that last sentence is interpreted.

  44. #44 Amos
    December 26, 2006

    doctorgoo,

    I do think it starts out setting up a simultaneous flip situation. But I believe it then asks us to solve a different problem.

  45. #45 Kevin
    December 26, 2006

    Dr Goo-
    I think the question is ambiguous. The answer they are likely looking for is 33%. I have no trouble with that. But here’s the issue, let’s give a little more information:

    2 dogs, one black, one white. At least one is male. What is the probability they are both male? 33%
    (Here you need to take into account each possibility – BM-WF, BF-WM, BM-WM)

    2 dogs, one black, one white. The black one is male. What is the probability they are both male? 50%
    (only options are now BM-WF and BM-WM)

    I realize example 2 is not what the question is really asking, but I also think people could easily be interpreting it that way – that’s the source of all the debate.

  46. #46 Kevin
    December 26, 2006

    basically, people read it and put themselves (mistakenly) in the shoes of the dog washer, not on the other end of the telephone. Dog washer picks up a dog, looks, sees it’s male. The probability that the other dog is male is 50%.

  47. #47 JYB
    December 27, 2006

    I think it should be 33% as well. Since there’s four possibilities:

    M M
    M F
    F M
    F F

    If you eliminate FF since you know one is a male then that’s 3 possibilities. In only 1/3 (MM) is the other dog a male, thus 33%.

  48. #48 Mark
    December 27, 2006

    Put me down for 1/3. Like corkscrew said

    define A: one or more is male
    define B: both are male

    Then what we’re after is P(B|A) = P(BnA)/P(A)

    P(BnA) = P(B) as B=>A
    P(B) = 0.5^2 = 0.25
    P(A) = 1 – P(A’) = 1 – P(B) as P(both female) = P(both male)

    So P(B|A) = P(BnA)/P(A) = P(B)/(1-P(B)) = 0.25/0.75 = 1/3

    Basically, its one of those arguments where some people come to an answer based on what seems right, while others use mathematics and proper definitions of words. Ultimately situations like this can only be resolved to the satisfaction of both parties by a brute force computer simulation.

  49. #49 Rodrigo Mendoza
    December 27, 2006

    It’s easier if you see it this way.
    There can only be three possible cases (or events) :

    1. Both Male
    2. Both Female
    3. One Male, One Female

    Now, each case is an event that may happen. There are three events in this problem. If we know that at least one is male, then we must eliminate the second and third events. So only one event is probable from a list of three (because only the first event would satisfy the given condition). Hence, the probability that the other one will be male is 1/3.

    Rod.

  50. #50 Karl
    December 27, 2006

    Well,,,look at it this way: Two dogs, cover one up, do not determine its gender. The UNIVERSE for this probability is ALL pairs of dogs (MM, MF, FM, and FF). And since we know that male and female occur in (approximately) equal numbers we know that the four groupings occur the same number of times. For any random collection of four pairs, there will be 4 F’s and 4 M’s which gives a probability of 1/2 for M in THIS UNIVERSE.
    Now uncover the dog, look. it is male. Now you have CHANGED THE UNIVERSE that is under consideration. This is called a CONDITIONAL probability, The condition being, obviously, that the first one is male. Now you are only thinking about pairs of dogs where one is male. This universe consists of pairs MF. FM (which are two different groupings), and MM, with each pairing occuring in equal numbers. In THIS universe only 1/3 of the time will the second dog also be a male. So, for the question being asked, with the conditions as given, the answer is 1/3.
    It may seem weird and counter-intuitive, but if you want to bet real money on it, in this second situation you will lose if you give even odds that the second dog is M.

  51. #51 Peter
    December 27, 2006

    Consider this: you’re bathing two puppies, and someone calls and asks if at least one of them is male. You check one puppy; if it’s male, you answer “Yes!” and hang up; if it’s female, you check the other puppy and answer yes or no accordingly. The “other” puppy is therefore either female or unexamined. If unexamined, it’s got a 50/50 chance of being male or female. The odds of the “other” puppy being male are therefore 25%.

  52. #52 SmellyTerror
    December 27, 2006

    Regarding the coin toss/analogy problem, for those who think the answer is 50%: keep in mind that there are two completely different problems and answers here:

    1: I toss a coin. It is heads. Now I’m going to toss a second coin. What are the chances of it being heads, too? 50%. It’s not dependat on the first toss.

    2 (which is really another verions of 1): I toss two coins at the same time. The big one is heads. What are the chances of the small one being heads? Answer, again, is 50%.

    3 (the real question here): I toss two coins. ***At least one is heads****, but we don’t know or care which. Now what are the chances the other is heads? This time the answer is 1/3.

    The difference is that problem 3 actually looks at both coins, and selects one that is heads. It’s not just limited to the first coin or whatever – it will pick either, as long as it’s heads.

    So the only way the second could be heads is for *both* to be heads.

    To look at the chances more closely, let’s use two different coins – a Big coin, and a Small coin. The possible outcomes are:

    Big: Heads – Small: Heads
    Big: Heads – Small: Tails
    Big: Tails – Small: Heads
    Big: Tails – Small: Tails

    Note that one-coin-heads-one-coin-tails is actually two distinct results. Sure they’re both “one of each”, but two separate results give you that answer.

    so the chances of two heads is one-in-four. But the question at least tells us that it’s not tails-tails. So that leaves one result in three being heads-heads.

    To put it another way:

    Now, you could say “but if he says that the Big coin is heads, that only leaves two answers: Small has 1/2 chance of being heads or tails”. But remember that the problem wasn’t just going to show Big, no matter what. It was just going to show one coin that got heads, if at least one did. It’ll show Small as heads just as often. So when it shows you that Big is heads, you’ve got to ask yourself why it didn’t show Small as heads? If both were heads, it’d have a 1/2 chance of showing Small, right?

    So you’re not just looking at a blind 1/2 chance of Small being heads – you’re also looking at the possibility that Small would have been shown if it had had heads, but that it didn’t have heads. By showing Big as heads, there is now a smaller chance that Small is heads.

  53. #53 Russell Blackford
    December 27, 2006

    Peter, I don’t think that’s right. Doing it your way, one gets p = 1/2. Thus:

    One dog has been examined – it is male. The other dog is not “either female or unexamined”. It is simply “unexamined”. On the face of it, there is still a 50:50 chance that when it does get examined it will turn out to be male.

    This is not meant to defend the 50% probability answer, in which I have no real confidence as I said above, just to query your line of reasoning.

  54. #54 Russell Blackford
    December 27, 2006

    Hmmm, forget that. I’m wrong, now I think about it. Damn!

  55. #55 Russell Blackford
    December 27, 2006

    Let’s try again. A asks B to ask C if at least one puppy is male. A gets back the advice, “Yes.” However, A does not know whether C looked at both puppies or just the closest puppy. If C only looked at one puppy then there are two possibilities: MM and MF. There is still a 50% chance that the other puppy is male. There is a 2/3 chance that this is what happened. If C had to look at both puppies than there is a 0% per cent chance that there is another male puppy – the closest puppy was assessed as female. However, there is only a 1/3 chance it happened like this.

    It looks like the odds are 2/3 of 50%, i.e. p = 1/3 when we do it Peter’s way.

    To say the odds are 50/50 we would have to know that C was able to answer the question “yes” after looking at only one puppy. I think my original analysis was covertly making that assumption, which it was not entitled to make. So I’m formally changing my answer to “P = 1/3″ and will slash my wrists (or at least be unhappy for 30 seconds) if it turns out I’m still making a mistake.

    Enough!

  56. #56 SmellyTerror
    December 27, 2006

    Russel: “To say the odds are 50/50 we would have to know that C was able to answer the question “yes” after looking at only one puppy.”

    Good line, well summed up.

  57. #57 Fred
    December 27, 2006

    If a woman is in the delivery room giving birth to twins, and a nurse comes out and tells you “the first one is a boy,” aren’t the chances that the second one will also be a boy 50/50? I mean, what are the alternatives? (Yeah, I know that with humans the odds are slightly in favor of girls, but for the sake of argument I’m rounding off.)

    This whole thing strikes me like the foolish thought that if a slot machine hasn’t paid out in a while, it is “due.” Each turn on the machine resets the odds, in that it doesn’t take previous turns into account.

  58. #58 SmellyTerror
    December 27, 2006

    Fred: again, there are two completely different questions. The twins are the same as the coin analogy, but I’ll repeat it with the twins (and we’ll assume they’re non-identical :p ).

    1. The first is a boy. The chances of the second being a boy is 1/2. You are correct.

    2. Both are born, someone looks at both and tells you that *at least one* is a boy. In this case, the chance of the *other* being a boy is 1/3.

    You see the difference? In the first case the announcement that the first is a boy can have no bearing on sex of the second. But in the second case the announcement took both babies into account. The nurse looked at both, and came back with the answer that at least one was a boy.

    The nurse was able to *select* which child he was reporting on, rather than being forced to report on the sex of one particular child.

    So although the chances that any particular kid is a boy is 1/2, the chances both kids in any random set of two being boys is 1/4. So, say, if the nurse was sent to report if at least one is a boy, and comes back and says the second kid is a boy. What about the first? Why did the nurse not use him as the example of “at least one boy”? We assume that, if both were boys, there would be a 1/2 chance that the nurse would have picked the first. So we have to admit the possibility that the nurse *would have* picked the first to announce as a boy, but that it *wasn’t* a boy. This chance reduces the 1/2 chance of it being a boy by a little bit.

    Or, to put it the other way, consider the four possibilities:

    First kid: boy; Second kid: boy
    First kid: boy; Second kid: girl
    First kid: girl; Second kid: boy
    First kid: girl; Second kid: girl

    As you can see, there is only 1/4 chance of both being boys.

    Now, if you say the *first* is a boy, you are right: you’re left with two possibilities for the second. But if you say *one of them* is a boy, you’re left with *three* possibilities, only one of which involves both boys.

    Hence, a 1/3 chance.

  59. #59 SmellyTerror
    December 27, 2006

    Err, that last ‘*one of them*’ should be ‘*at least one of them*’.

    My fingers are possessed by the souls of a thousand poets, so I don’t have time to proof-read.
    :p

  60. #60 Richard Wein
    December 27, 2006

    The question is subtly misleading, when it asks “What is the probability that the other one is a male?” Strictly speaking, the concept of “the other one” is undefined here, because the conversation up to that point has not referred to a specific one of them, only to “at least one of them”. So I would say there are two possible valid responses to the question:

    1. “I take your question to mean ‘What is the probability that both of them are male?’ The answer to that question is 1/3.”

    2. “Your question is meaningless, since it relies on an term (‘the other one’) which cannot be evaluated.

  61. #61 Edmund
    December 27, 2006

    The question would be so much more fun if the possibility of hermaphrodite puppies was included.

  62. #62 Dave S.
    December 27, 2006

    Jason puzzels:

    A shopkeeper says she has two new baby beagles to show you, but she doesn’t know whether they’re both male, both female, or one of each. You tell her that you want only a male, and she telephones the fellow who’s giving them a bath. “Is at least one a male?” she asks him. She receives a reply. “Yes!” she informs you with a smile. What is the probability that the other one is a male?

    To me, the problem is the final question is worded “What is the probability that the other one is a male?” and not “What is the probability that the other one is also a male?”

    Let’s try it this way. We know there are 2 pups (Dog1 and Dog2), either both male, both female, or one of each.

    Dog1(M)-Dog2(M): The dog washer is asked if at least one dog is male. YES is the answer. Is the other one male? If the washer considers Dog1 as “this one”, then Dog2 is the “other one” and the answer is YES. However, if the washer considers Dog2 as “this one”, then Dog1 is the “other one” and the answer is still YES.

    Dog1(F)-Dog2(M): The dog washer is asked if at least one dog is male. YES is the answer. Is the other dog male? If you consider Dog1 as “this one”, then Dog2 is the “other one” and the answer is YES. However, if the washer consideres Dog2 as “this one”, then Dog1 is the “other one” and the answer is NO.

    Dog1(F)-Dog2(F): The dog washer is asked if at least one dog is male. NO is the answer. Is the other dog male? The answer obviously must be NO regardless of which dog the washer considers “this one” and which one he considers the “other one”. Since we know the answer to question one was YES, this scenario is irrelevant to the problem and can be ignored.

    In summary, if the dog washer answers YES to the question “Is at least one a male?”, then there are 3 ways to respond YES to the question of whether the “other dog” is male and only 1 way to respond NO. This means the probability ‘the other one is a male’ will be 0.75.

    Which I suspect is way wrong.

  63. #63 Dave S.
    December 27, 2006

    Note:

    I don’t think it makes a difference in the second case if you call Dog1 the male instead of female and vice versa. What’s relevant here is that there are 3 configurations – 2 males, a male and a female, and two females. The latter two answers will change from YES and NO, to NO and YES, and so the final probability will be the same.

  64. #64 Brian
    December 27, 2006

    I’m at work, so I don’t really have much time to read through all the comments, so I apologize at once if I state anything that’s been previously stated.

    I’ve seen many folks comment that the probability of both puppies being M is 1/3, observing that the possible outcomes are:

    M/M
    M/F
    F/M
    F/F

    And furthermore that the F/F choice has been eliminated.

    But wouldn’t M/F and F/M be equivalent? Is the probability that the second puppy is F actually 2/3? If so, why?
    The probability is 1/2, so far as I can tell, and not 1/3.

  65. #65 Brian
    December 27, 2006

    Okay, wait a minute. I’m thinking there’s a flaw with assuming that the probability of getting two males is 1/3. The probability of one puppy being M is 1/2. The probability of another puppy being M is 1/2. Therefore, the probability that both are M should be 1/2 x 1/2 = 1/4.

    We are told that one is M, but its sex has no bearing on the probability that “the other one” is M. Therefore, even though the probability that BOTH are M is 1/4, the probability that the OTHER one is M is still 1/2.

  66. #66 Jesse
    December 27, 2006

    There are no prior probabilities assigned to the pair of puppies. Everyone is arguing about whether the support set contains three uniform possibilities or four, and there’s no reason to believe they are uniform. The answer is: THERE IS NO PROBABILITY until you know the priors.

    We can still do some analysis.

    Define:

    P{MM}: both are male
    P{FF}: both are female
    P{MF}: one of each (we can distinguish MF from FM later if need be)

    There is a second point of contention: Did the fellow bathing the puppies answer “yes” based on a random sample of one of the puppies (the alternative answer thus being “i don’t know”), or did he check both puppies before answering. We’ll consider each of these possibilities:

    P{M}: probability that at least one is male
    P{M’}: probability that a randomly sampled puppy is male

    In particular, P{M | MF} = 1, but P{M’ | MF} = 1/2.

    We may be interested in two things,
    1. P{MM | M}, and
    2. P{MM | M’),
    depending on whether we think M or M’ describes what the puppy bather did.

    For case 1:
    P{MM | M}
    = P{M | MM} P{MM} / P{M} (by Bayes’s Rule)
    = P{M | MM} P{MM} / (P{M | MM} P{MM} + P{M | FF} P{FF} + P{M | MF} P{MF}) (by Chain Rule)
    = P{MM} / (P{MM} + P{MF}) (because P{M | MM} = 1, P{M | FF} = 0, and P{M | MF} = 1)

    For case 2:
    P{MM | M’}
    = P{M’ | MM} P{MM} / P{M’} (by Bayes’s Rule)
    = P{M’ | MM} P{MM} / (P{M’ | MM} P{MM} + P{M’ | FF} P{FF} + P{M’ | MF} P{MF}) (by Chain Rule)
    = P{MM} / (P{MM} + 1/2 P{MF}) (because P{M’ | MM} = 1, P{M’ | FF} = 0, and P{M’ | MF} = 1/2)

    Now we have a choice of priors. Some have contended that P{MM}, P{FF}, and P{MF} are the three possibilities, uniformly distributed at 1/3 each. Others have contended that each puppy is an independent “coin flip”, in which P{MM} = P{FF} = 1/4 and P{MF} = 1/2. We have no information to prefer either of these priors over each other (or any other), but we can still consider what each gives. So we get four answer in the end:

    – If P{MM} = P{FF} = P{MF} = 1/3, then
    1. P{MM | M} = 1/2
    2. P{MM | M’} = 2/3

    – If P{MM} = P{FF} = 1/4 and P{MF} = 1/2, then
    1. P{MM | M} = 1/3
    2. P{MM | M’} = 1/2

  67. #67 Dave M
    December 27, 2006

    This problem, or one similar to it, depending on how you read it, was one of several in a brainteaser quiz at 3quarksdaily last month. There was much discussion after the answers were revealed, so if you’re not satisfied then check it out. (I think that after consulting with professionals they finally settled on 50-50.)

    Remember, there are other questions, so go look at the quiz first if you don’t want to spoil it. In fact why don’t I leave off the direct link to the answers (the quiz links to them).

  68. #68 Brian
    December 27, 2006

    Back again, with yet another take on it. I find myself confused by much here.
    I’m specifically concerned with this:
    There’s a 1/4 (sight unseen) chance of M/M, as we’ve established. We are told one is M, meaning that the choice of F/F is eliminated.
    I assume that means we’ve bettered our chances of arriving at M/M, so there is now a 1/3 chance of a M/M pair.
    NOW we’ve essentially got a Monty Hall problem, haven’t we?
    One is M, we’ve established that, so set him aside, and now we’re left with an as-yet unsexed beagle puppy. There are 2 possible combinations in which that puppy would be F, and only one in which it would be M [assuming pair order is meaningless (i.e., MF is equivalent to FM)]. 1/3 of the time it’s M, 2/3 it’s F.

    Therefore, there is only a 1/3 chance that the second one is M, no?

    So many considerations. My intuition has been lying to me all this time…

  69. #69 Brian
    December 27, 2006

    Last time, I promise. I think I’ve finally figured it out.

    The probability that the other one is M is 1/4.

    Think of it like this. The problem is similar to the game show. You have four doors, and you’re to pick the one behind which stands a pair of males. You have a 1/4 chance of being right, and 3/4 chance of being wrong.

    When you’re told one is a M, you are effectively being shown the door with F/F, meaning you have eliminated one door choice. You STILL only have a 1/4 chance that you guessed right. This means there’s still a 75% chance you guessed incorrectly.

    In other words, there’s a 1/4 chance the other one is M, while there is a 3/4 chance that it is F.

  70. #70 Jeff
    December 27, 2006

    My vote’s for 50%.

    We know one is male. We can arbitrarily say “Dog #1 is Male.”

    That leaves us with Dog #2. It can be male or female.

    Thus, 50% chance of both dogs being male (ie, 50% chance of Dog #2 being male)

  71. #71 Kevin
    December 27, 2006

    Brian-
    No, you’ve eliminated a door choice. Now, if you’ve chosen a door beforehand, your odds were 1/4. When the FF door is eliminated, there is the possibility that YOUR DOOR is eliminated. There is no guarantee (unlike the monty hall problem) that your door will not be eliminated. If the FF door is eliminated before you pick, then your odds are 1/3.

    Jeff-
    You can’t arbitrarily say that, because you can’t identify one dog for sure. You only know that at least one of the two is male. If I put two dogs in front of you, and say “at least one is male” – which dog do you declare the male? You don’t know which one it is, you can’t exclude the possibility that either of them might be female. You have to make the choice – is dog 1 male and 2 female? is dog 1 female and dog 2 male? or are both male? you have a 1/3 chance of guessing right.

  72. #72 Jeff
    December 27, 2006

    It’s a matter of perspective. All we need to know is that one IS male, not which.

  73. #73 Kevin
    December 27, 2006

    Dave M-
    I checked that page – the correct answer in that case is indeed 1/2, but that is not the same question as we have here with the beagles. in the sister/brother question, since the reader has seen and positively identified one child as a girl, you can exclude her from the equation. All that’s left is one more child, which is 50% chance of being male or female.

    In our beagles question, we cannot positively identify either beagle as male to remove it from the equation. We must take both beagles as a single set, knowing that at least one (but not knowing which one) is male. probability of labeling each dog correctly, sight unseen, is 1/3.

  74. #74 Kevin
    December 27, 2006

    Jeff-
    it’s not just perspective.

    I put two dogs in front of you. I give you 3 labels – 2 say male, 1 says female.

    I now ask you to place labels on the dogs. Your odds of being correct are 1/3.

  75. #75 Brian
    December 27, 2006

    Kevin,

    Good point. I had considered that myself, but thought “Nah, forget it, ‘Yo Holmes, to Bel Air!” and went ahead with 1/4.

  76. #76 Amos
    December 27, 2006

    Richard Wein,

    I’ve been convinced that your analysis is right, and so I retract my answer of 50%. I had mistakingly given “other one” a referent when it had none. But until someone makes a convincing argument on the semantics, I continue to maintain that a strict reading of the question doesn’t justify going to your first prong (both being male) and so I change my opinion to your second prong, that the question is meaningless.

  77. #77 Kurt
    December 27, 2006

    It’s kind of curious that, even knowing that this is supposed to be a “trick problem”, about half the comments seem to be getting it wrong. This includes some incorrect statements about the Monty Hall problem, whose solution wasn’t in dispute.

    Looking at this as a cognitive issue, it would seem that people’s intuitions about conditional probabilities are often simply wrong (and assuming that two events are independent when in fact they are not seems to be the most common error). This might explain some of the traction that ID belief gets among the general population.

  78. #78 Kevin
    December 27, 2006

    Part of it is that many trick questions are worded poorly to be tricky. This is one of them. It’s too easy to literally infer that since we know one is male, we can exclude one dog and the chance the other is male is 50%. I don’t believe that’s the spirit of the question though. I think what they’re trying to ask can be boiled down to “you have two dogs. I assure you one is male. what are the odds that you will correctly guess the gender of both dogs?”

  79. #79 Kevin
    December 27, 2006

    “I assure that at least one is male” is what that should read, otherwise we’re back to ambiguity again. ;)

  80. #80 Jason Rosenhouse
    December 27, 2006

    Goodness! I never expected so enthusiastic a reply to this problem.

    I’ll devote a separate post to trying to sort out some of the issues raised by this problem. Stay tuned!

  81. #81 David D.G.
    December 27, 2006

    “Goodness! I never expected so enthusiastic a reply to this problem.

    I’ll devote a separate post to trying to sort out some of the issues raised by this problem. Stay tuned!”

    NEVER MIND THAT! WHAT’S THE FLIPPIN’ ANSWER? Come on, man; when you ask a riddle and people have given it their best shot, you have to drop the other shoe and reveal the solution!

    ~David D.G.

  82. #82 MarkP
    December 27, 2006

    The answer is 1/3. If you find the previous explanations given unsatisfying, consider a more generalized version of the problem. Ask it this way:

    Given n dogs, and that at least one is male, what is the probability that all are male?

    Well, given n dogs, there are 2^n possible outcomes. Being told “at least one is male” is the same as saying “It is not the case that all are female”. Since there is only one scenario that is all female, elminating that possiblity from the total leaves 2^n-1 possibilities. And since only one of these possiblities is the desired solution (all male), the probablity is 1/(2^n-1).

    If you accept that, then just assume n=2 for the specific problem posed here, and you get the answer: 1/3.

  83. #83 peter
    December 27, 2006

    curious question about the monty hall problem.

    given the nature of the setup, choice of 1 of 3, expose one, give new choice 1 of 2.

    initially each door has an even chance of being the car, 1/3
    odds that the car is behind a door you didn’t choose, 2/3

    one door is opened, odds that you have the correct door still haven’t changed, (1 door in 3) but then you are given a new choice. car is behind 1 of 2 doors. guaranteed. since that no matter what your original choice might be, monty can open an empty door, (or goat) and then allow you a second choice. do not the odds reset? (similar to the lottery problem having no memory?)

    as I drew it out for myself, if you go by the assumption that monty will open a goat door after you choose, there are the same number of possible positions whether you have chosen the correct door initially or not. (i.e. if you choose the door with the car, monty could open either other door, (2 possiblities) and the third door is a goat, or if you choose one of the goat doors, monty can only open one of the other doors, as the other has the car. (2 other doors, 2 other possibilities.) an example would be (assuming car in first door,):
    x,O,_
    x,_,O
    _,x,O
    _,O,x
    key:
    x is your choice
    O is the door monty opens
    _ is the mystery.
    (you can expand this to the other 8 possibilities based on the car being initially being behind doors 2 and 3…) after your choice and monty’s exposure, there are 12 states to be in. 6 states where you have chosen the correct door, and 6 where you have not. (as opposed to the state before monty exposes the goat, where there are 9 states. 3 where you have chosen the car vs. 6 where you have not.)

    At this point, Monty gives you a new choice. 2 doors, one car, one goat. odds are 50/50.

    If monty did not give you the new choice, the odds of your original choice would still be 1/3, but because you are given the new choice, your odds would seem to be even. somebody please correct me, I remember the original parade article, and I seem to remember having trouble with it then. but I can’t remember whether it was because I thought it was poorly worded or not.

  84. #84 mark
    December 27, 2006

    Maybe I missed it, but did anyone note that it was not established whether the dogwasher had seen more than the first dog? Thus, if he had only washed the first puppy, and it were female, he could not answer in the affirmative. We know he could not have sexed both and found 2 females. If he only sexed the first one he was washing, the combination could be MM or MF; if he had sexed both, the combination could be MM MF or FM. It seems this would affect the probability of the second dog being male.

  85. #85 Kevin
    December 27, 2006

    Peter-

    When monty gives you the new choice, the odds don’t reset. He opens the door 100% sure that there is no prize behind it. The fact that the door doesn’t have a prize does not change the odds of your door having a prize. Once you choose, the doors are divided into 2 groups – one you chose, one you didn’t. the one you chose has a 33% chance of being right. the other group has a 66% chance of being right. Basically, monty is asking you – do you want to stick with your door, or do you want to go with the other group, which has 2 doors? – you get to see both doors in the other group, since he opens one for you. Of course you switch – it’s the same as him saying from the beginning “choose 2 doors, we’ll open both, and if the car is behind either one, you win”

  86. #86 kevin
    December 27, 2006

    Mark-
    “A shopkeeper says she has two new baby beagles to show you, but she doesn’t know whether they’re both male, both female, or one of each. You tell her that you want only a male, and she telephones the fellow who’s giving them a bath. Is at least one a male? she asks him. She receives a reply. Yes! she informs you with a smile. What is the probability that the other one is a male?”

    That’s all the information we get. Your question is not answered, we don’t know whether he checked one, or both, or already knew.

    I do know is that I kinda want a beagle puppy now.

  87. #87 peter
    December 27, 2006

    Kevin-
    but hang on, does the fact that you have already made a choice really affect the odds in the second choice? when monty opens the goat door, it can also be seen as reaffirming the original choice. “at least you didn’t choose this one.” sort of thing? if the odds don’t reset, the third choice also still only has a 33% chance of being the one with the prize. all you really know is that he has eliminated 33% of the possiblities with the exposure. if, as you say that the choice is between this one door and one of the other two, then the odds start out at 50% and stay there, no? you still don’t know what’s behind the third door unless you choose it, and there the odds are 50% again.

    this is why I get a bit confused about this, as there seem to be those 4 possible states for the doors to be in for each position of the car.

    similar to the puppy example where there are 2 seperate states for one of each, MF and FM, monty has two ways to fool you if you have already selected properly and only one each if you haven’t.

    btw, thank you for taking the time to discuss this with me.

  88. #88 Kevin
    December 27, 2006

    Peter-

    I think you’re actually overthinking this!

    From the beginning:
    there are 3 doors. one door has a prize, the other two have nothing. You get to choose one. What is the chance you get a car? 33% What is the chance you get no car? 66% – right?

    Now, choose two doors. What is the chance you get a car? 66% What is the chance you get no car? 33% – right?

    So now, when you choose your door, you have a 33% chance of being right, and the remaining doors combined have a 66% chance of having the car. Monty simplifies the 66% chance group. You get two doors for the price of one. You don’t get to choose which door monty opens – he opens a bad door that isn’t yours every single time.

    The two groups are not 50% for one door and 50% for the two doors because you didn’t choose as a group. You had to pick one. Once you did that, you locked the groups into a 33% group and a 66% group.

    Think of this – you choose your door, you have a 33% chance of being right. Now, Monty, without opening a door, offers you BOTH the other doors. Of course you take it, there’s a 66% chance of 2 doors having a car.

    There is a way to make it a 50% chance – but that means you have no extra information. For example, you choose a door, monty opens a bad door, and then brings someone in from backstage who doesn’t know what door you chose. That person has a 50% chance of picking the right door – because he has no extra knowledge, and never had a chance to pick the bad door monty threw out.

  89. #89 peter
    December 27, 2006

    Kevin-

    ah, that’s the analogy that works. I get the rationale now.

    It makes much more sense when you ask the question “Do you want your door or these other two…” as it demonstrates the exposure as the red herring that I was having trouble getting past. the trouble in my head is that if you go into the challenge knowing ahead of time that no matter which one you choose, one of the others will be eliminated, the second choice really looks 50/50. it’s a bit hard to give that up.

    thank you!

  90. #90 peter
    December 28, 2006

    Kevin-
    took me a bit longer, and I realized that the answer that I needed to figure out is that no matter which door you choose first, switching will give you the 100% chance of reversing your result. so if you have a 66% chance of getting the first selection wrong, switching doors means you have a 100% chance of getting the prize. hence the advantage to switching. thanks again for helping me air this out…

  91. #91 Christopher Heard
    December 28, 2006

    I think the answer is either 25% or 50%, but I’m having a hard time reasoning through to the solution. (I’m not a mathematician, statistician, logician, magician, or any other -ician.)

    Here’s the key question: does our knowledge that one puppy is male affect the probability of two puppies being male?

    It seems so simple: if you have a pair of puppies and you know that one is male, then the possible pairings are MM and MF (discounting the possibility of hermaphroditic puppies), so the probability that both puppies are male is 50%.

    But is it really that simple? If we have a pair of puppies and we don’t know either one’s gender, we would say that the probability of both being male would be 25%. If we later learn that one of the puppies is indeed male, that doesn’t mean that the probability of that puppy having been male was really 100% after all, but rather that this puppy “beat the odds” (even odds, but you know what I mean) and fell on the male side of the fence.

    To use a different analogy: if you and I each flip one coin, the probability that both coins will turn up heads (discounting the possiblity of the old Batman canard of Two-Face’s coin landing on edge) is 25%. When I uncover my coin and see that it’s tails, I now know for certain that both coins are not tails. But this does not mean that the probability of their having been both heads was really 0% all along, but rather that reality fell in a certain way relative to the odds.

    I think that the probability that the second puppy is male was set before we ever knew the first puppy was male, and the probability that the second puppy is male is 25% because the probability of both puppies being male was 25% all along.

    Of course, I’ve been wrong before–but does that affect the odds of my being wrong now? ;-)

  92. #92 SmellyTerror
    December 28, 2006

    Christopher: The crucial difference is that you’re not simply uncovering one coin to see what it is. You are uncovering *both*, then hiding one of the answers.

    It is *not*: “the first coin is heads, what are the chances fo the second being heads?” That answer is 50%.

    The question is: “at least one of the coins is heads. What are the chances the other is, too?” The answer here is 33%.

    The question *looks at both coins* and tells you that one or the other is heads. You are left with three possibilities:

    1. The first is heads while the second is tails
    2. The first is tails while the second is heads
    3. Both are heads

    The answer “at least one is heads” satisfies all three of these equally likely possibilities. Only one of these possibilities, however, means that both are heads.

    So the chance of both being heads is 1 in 3.

  93. #93 Kevin
    December 28, 2006

    Christopher Heard-

    You’re totally right about the 25% – except for one little detail – it’s 25% because its 1/4 of the following choices:
    MM
    MF
    FM
    FF

    Since we are told that FF is not a possibility, it reduces down to:
    MM
    MF
    FM

    The coin analogy doesn’t work because there is still the possibility that both can be tails – there is nothing stopping it from happening. In the case of the puppies, we know that FF is not a possibility. It would be as if somehow, when one coin lands tails some cosmic force makes it impossible for the other to do so.

    So if both of us flip coins, whats to stop us from both getting tails? What would have to happen is that as soon as you see tails on yours, we throw my coin away and call it heads. if you see heads, then i have to check mine to see if it’s tails or not. The possible outcomes here are HH, HT, TH, and the odds of HH turning up are 1/3.

  94. #94 D
    December 28, 2006

    Consider two bits of code:

    while() {
    a = rand()%2;
    b = rand()%2;
    if (a==0) continue; // first dog is male

    if (b==0) nm+; // second dog male
    else nf++; // second dog female
    }

    run it. you do in fact get about equal numbers for nm and nf. This makes sense – the moment you say a and b are independent distributions, throwing away half the data makes no difference.

    while() {
    a = rand()%2;
    b = rand()%2;
    if (a==0 || b==0) continue; // at least one male dog

    if (b==0) nm++; // second dog male
    else nboy++; // second dog female
    }

    Now observe the entirely different results. Point is, you’re making cuts on the variable you’re observing.

  95. #95 Jettlogic
    December 28, 2006

    Yikes look at all these comments! steve s [1] got the right answer, 33%, right away. At least, if you make the natural assumption that the sex of a beagle pups are independent Bernoulli trials [2] with p=0.5, then the answer of 33% follows from freshman year probability theory. Dave’s post [3] explains it easily without equations.

    Unless of course the problem is even more sly than Monty Hall.

    Please no more 50%-ers, hopefully Fred/Scott/Kevin have worked it out and found their calling to take up a statistics major (or veterinary science specialising in Beagles).

    [1] http://scienceblogs.com/evolutionblog/2006/12/a_probability_puzzle.php#comment-299224

    [2] http://mathworld.wolfram.com/BernoulliTrial.html

    [3] http://scienceblogs.com/evolutionblog/2006/12/a_probability_puzzle.php#comment-299593

  96. #96 The Decidenator
    December 28, 2006

    Kevin said:

    Since we are told that FF is not a possibility, it reduces down to:
    MM
    MF
    FM

    Look at it this way: The first one you learn about is either male or female, and the second one you learn about is either male or female. We’ve already learned that the first one we learn about is male, so there’s no need to guess that the first one might be female, which is Kevin’s third choice. The only 2 choices are:

    MM
    MF

    50/50.

  97. #97 Tom Renbarger
    December 28, 2006

    There is no “first one” in this problem, just an “at least one.”

  98. #98 Russell Blackford
    December 28, 2006

    I’m still hanging in with the “p = 1/3″ camp, but, dog dammit, when is Jason going to give us his answer?

  99. #99 Kevin
    December 29, 2006

    Decidinator-

    It’s only 1/2 if you cheat and always segregate the male dog as a male – but then you have to know beforehand which dog is male. Since we don’t, we have to take into account all possibilities, which are
    MM
    MF
    FM

    That does not equal MM and MF, it equals MM and 2(MF). So still 1/3.

  100. #100 AJS
    December 29, 2006

    It’s tempting to view the sexes of the puppies as independent events, but there is a catch. The events are not independent because we don’t know which puppy is confirmed as male.

    To consider the events as independent, we need to be able temporarily to distinguish the puppies from one another, so let’s say one of them has a spot. Now our sex distributions are: Both male; spotted one male, plain one female; spotted one female, plain one male; or both female. We’ll suppose for the sake of argument that all outcomes are equally-probable.

    What we know is that at least one of the puppies is male. We don’t know which one, and this is why the events are not independent: what we have is a report on the combined outcome of the two events, not a report on a single event. It could be the spotted puppy who is male and the plain one female; the plain one who is male and the spotted one female; or they could both be male. Any of these outcomes is equally likely. Therefore, the probability that both puppies are male is 33.3…%.

    If we had asked whether the puppy with the spot is male and been told yes, then the probability of the puppy without the spot being male (i.e. both male) would be 50%. But that wasn’t what we asked.

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