Dave Bacon is a theoretical ski bum who is also an assistant research 
Last quarter I taught discrete math. One component of the class was to cover some basic probability theory. On one of the homeworks we asked the following two questions about random five card poker hands:
- Given that the hand contains an ace, what is the probability that the hand contains another ace?
- Given that the hand contains the ace of diamonds, what is the probability that the hand contains another ace?
I find this problem interesting because while I can do the calculation and understand why it comes out the way it does, I was never able to give myself a good intuitive explanation beyond comparing this problem to a much simpler case with a similar behavior. Anyone have a good inuitive explanation for why one of the above probabilities is bigger than the other? (And yes I'm avoiding telling you the answer of which one is bigger: work it out for yourself! Or see the comments were I'm sure someone will leak the correct answer.)
Comments
Without calc: The first hand.
Posted by: Who Cares | January 2, 2008 2:13 PM
Am I missing something?
Case 1 will be larger. The first ace *could* be the Ace of diamonds, in which case you have case 2. Or it *could* be one of the other aces, which means there are more ways of getting "yes" on case 1 than there are on case 2.
Of course, I can't say how much larger as that might need calculation ;)
Posted by: Rupert | January 2, 2008 2:21 PM
Suppose we play the following game. I deal you a poker hand, you check whether it contains an ace. If not, we start over.
If the hand does contain an ace, I can decide to guess that there are are multiple aces in your hand. If I'm correct, you give me some amount of money, if I'm wrong you win some smaller amount. (So, I don't *have to* make a guess!)
However, suppose you are forced to tell me whenever there is the ace of diamonds in your poker hand. That certainly gives me more information: namely, I know you do not have a hand which contains only 1 ace that is not a diamond.
Since there is only 1 in 4 chance a 1-ace hand contains the ace of diamond, but an at-least 50% chance a multiple-ace hand contains the ace of diamond, I would decide to make my guess only when you give me that extra information.
Posted by: Steven | January 2, 2008 2:25 PM
To be honest my first guess was both have the same probability until I read the anyone has a good explanation why one of these is bigger.
Posted by: Who Cares | January 2, 2008 2:27 PM
I agree with "Who Cares". The suit of the ace doesn't change the number of aces still available or the pool from which they come. Purely on intuition, I wager Case one and Case two have equal probability.
Posted by: Joe Gavin | January 2, 2008 2:40 PM
Apparently, my intuition disagrees with everyone else's (my comment above implies I vote for case 2).
Posted by: Steven | January 2, 2008 2:42 PM
Rupert: I think what you just said is that the probability of having a hand with the ace of diamonds is lower than the probability of having a hand with an ace of any suit. What I'm trying to ask about is the probability that, given one of those two events is true, that you have at least one other ace in the hand.
Who Cares and Joe: The fact that the probabilities aren't the same really is surprising, isn't it!
Posted by: Dave Bacon | January 2, 2008 2:45 PM
Hmm, except that "my comment above" doesn't appear :-(
Posted by: Steven | January 2, 2008 2:46 PM
Because the second set has a more specific "given", there are a large number of one-ace hands that are a part of the first set, but not a part of the second set. There are also two-ace and three-ace hands that are excluded from the second set, but since these should be rarer than the one ace hands, I don't expect them to tip the balance.
Therefore the second set (with the ace of diamonds) should offer higher probability.
Still, subtleties of phrasing make a big difference in problems like this. For example, if you were to say "the top card in a deck is an ace. What is the chance that the next four cards contain an ace?", then it would offer the same probabilities as "the top card in a deck is an ace of diamonds. What is the chance that the next four cards contain an ace?"
The difference between my examples and yours is that mine analyzes a single card for the first ace, while yours allows any of the five to be the first ace. Tricky stuff!
Posted by: Spaulding | January 2, 2008 2:48 PM
Steven: sorry somehow it got stuck in the junk comment folder. Not sure why. Must be biased against poker players.
Posted by: Dave Bacon | January 2, 2008 2:49 PM
Dave: You can delete my later duplicate comment, it seems fixed now!
Posted by: Steven | January 2, 2008 2:52 PM
My intuition tells me the second hand is more likely. My reasoning is that the ace of diamonds specifically would be more common in hands with multiple aces. Hmm... I guess that reasoning isn't particularly coherent. Now I am curious.
Posted by: miller | January 2, 2008 2:54 PM
The ace of diamonds is contained in 1/4 of the one-ace hands, 1/2 of the two-ace hands, 3/4 of the three-ace hands, and all of the four-ace hands. So once we know that the hand contains the ace of diamonds, we have excluded 3/4 of the one-ace hands and we have excluded a smaller fraction of the multiple-ace hands. That makes a multiple-ace hand more likely than if we knew only that the hand contains at least one ace.
Posted by: John Preskill | January 2, 2008 3:10 PM
I tried it with the math, and it turns out that it's not too hard if you're willing to approximate. (spoiler alert)
We can define P(n) to be the probability of having exactly n aces. The probability of having exactly n aces AND having the ace of diamonds is P(n)*n/4. The probability in the first case is [P(2)+P(3)+P(4)]/[P(1)+P(2)+P(3)+P(4)]. If we assume P(n-1) is much greater than P(n), then we can approximate this to P(2)/P(1). Similar approximation of the second case leads to 2*P(2)/P(1). Therefore, the second case is more likely by about a factor of 2. My numerical calculations (not shown) agree with this assessment.
Posted by: miller | January 2, 2008 3:24 PM
Dave: I like that example a lot. Just to be sure that I've got it right, could you use that to talk about quantum peeking or measurement? That is, once you know more information, you thn change the probability of what follows; a Texas hold 'em EPR.
Posted by: Justin | January 2, 2008 3:31 PM
Reminds me of the Monty Hall problem, although that one's easier.
Posted by: jeffk | January 2, 2008 3:50 PM
Shouldn't the probability of case one just be P(2)+P(3)+P(4). Why are you dividing there? Also the assumption of P(n-1) being MUCH greater than P(n) raises some questions. Is difference between P(n) and P(n-1) great enough to allow the approximation, I think your in some murky water there.
Posted by: doodle | January 2, 2008 4:09 PM
can i take that last comment back? pls?
Posted by: doodle | January 2, 2008 4:18 PM
Intuition tells me that case 1 has greater probability simply since getting ANY ace is more likely than getting a SPECIFIC ace. But often in probability, intuition is flat out wrong.
I believe conditional probability applies here. P(B|A) = P(A intersect B)/P(A).
P(A) in the first case is 4/52 and 1/52 in the second.
P(A intersect B) is 4/52 * 3/51 in the first case and 1/52 * 3/51 in the second case.
So in case 1 P(B|A) = (4/52*3/51)/(4/52) = 3/51
And in case 2 P(B|A) = (1/52*3/51)/(1/52) = 3/51
If I'm remembering my probability correctly the odds of having the second ace are identical. However the odds of a 2 ace hand and the odds of an ace of diamonds plus another ace hand are not the same.
Posted by: gex | January 2, 2008 4:37 PM
Reminds me of the Monty Hall problem, too...
OK, so we agree that it's more likely to get a second ace if you're told the hand contains the ace of diamonds specifically, yes?
I'm good on the simpler explanation. Say you're pulling two cards in a four card deck that consists only of the Ace of diamonds, the Ace of clubs, and two jokers. You can pull any of these combinations:
a) Ad Ac
b) Ad J1
c) Ad J2
d) Ac J1
e) Ac J2
f) J1 J2
If you're only told you have one ace, any of cases a through e might apply, and you have a 1/5 chance of having a second ace. If you're told you have the ace of diamonds, only cases a through c might apply, and you have a 1/3 chance of having a second ace.
This scales. When you say someone has one ace, of any suit, instead of just the ace of diamonds, you're allowing for proportionately more one-ace situations than you are for two-ace situations. As in the case above: AA can only happen one way, but AJ1 can happen two ways. By limiting yourself to Ad cases, you cut out more one-ace cases than two-ace cases.
My God, that's unclear...
Posted by: ThePolynomial | January 2, 2008 4:40 PM
Let p be the probability of another ace in case 1 and q be the probability of another ace in case 2. q>p.
If I am told the hand contains an ace, the probability there is another ace is p. Now I ask the suit of the ace, knowing the answer will be one of four things. Regardless of the answer I get, I will raise the probability from p to q. The strange thing is that the probability of their being an ace does not get raised until I get a response, even though it doesn't matter which response I get.
Posted by: Matt Elliott | January 2, 2008 5:36 PM
The first question is equivalent to: I hold an ace from a 52-card American deck, and I will fill a 5-card hand by drawing 4 more cards from the deck. What is the probability that I will draw at least one more ace?
If you accept that as valid, then the second question is equivalent to: I hold an ace of a certain suit from a 52-card American deck, and I will fill a 5-card hand by drawing 4 more cards from the deck. What is the probability that I will draw at least one more ace?
It doesn't matter if I tell you the suit or not. It doesn't matter if I know the suit. Whichever ace I hold, it is of a certain suit. This is implied in the first question and explicit in the second question. The card's rank and suit are given: they are both certainties, of unit probability. Therefore the sole card of that rank and suit no longer plays any part in the probabilities befalling the remaining 51 cards.
The two questions are equivalent and their answers must be identical.
Posted by: CRM-114 | January 2, 2008 5:41 PM
One point that occurs to me: The "given" Ace of diamonds may not be the first ace -- that multiplies the probabilities of the multiple-ace cases.
Posted by: David Harmon | January 2, 2008 5:44 PM
I've been thinking about this all day and I agree with CRM-114. I think this is a mistake that's being made: the "probability that a hand has two aces given that one is the ace of diamonds" and the "probability that a hand has two aces and one is the ace of diamonds" are two different questions. I think the probability is identical.
Posted by: jeffk | January 2, 2008 5:48 PM
There are six 2-ace hands with all suits allowed. There are three 2-ace hands with a diamond ace. My intuition is that 1. is more likely. I don't see how you can increase the probability by specifying a suit. Let's say I'm playing against you and you tell me you have at least one ace. I then have to bet on whether you have another one. If you then tell me you have "the ace of ____" where ____ is whatever you have, it seems to me like all you're doing is reducing the number of possible 2-ace hands you could have and thereby reducing the chance of you having a 2-ace hand.
Posted by: Geordie | January 2, 2008 5:50 PM
Ah, now this is becoming fun to watch! :)
Posted by: Dave Bacon | January 2, 2008 5:58 PM
Ok, here's my go at this:
The total pool of possible hands is defined by the first condition: either it has A) an ace (any ace) or it has B) the ace of diamonds. 3/4th of the hands that have a single ace do not have the ace of diamonds, so the total potential hands is smaller if you know that it has the ace of diamonds.
Multiple-ace hands are rarer, but the ace of diamonds is in a larger fraction of them: 1/2 of the 2-ace hands, 3/4th of the 3-ace hands, and all of the 4-ace hands.
So, if you know it has the ace of diamonds, you disproportionately eliminated from consideration hands that don't have more than one ace (relative to those that have 2 or more), and so are more likely to have a hand that has another one.
BTW, a nifty problem, as my first reaction was the same as the others: what difference does it make?
Posted by: divalent | January 2, 2008 6:03 PM
You're reducing the number of possible two-ace hands by half...but aren't you also reducing the number of one-ace hands by 3/4? Perhaps I am confused.
Posted by: ThePolynomial | January 2, 2008 6:06 PM
This seems like the case of many discrete probability problems. It depends how you generate the sets.
Deal 5 cards turn over the first one it "happens" to be an ace. Chance of any of the other 4 being an ace 1-48/51*47/50*46/49*45/48. (1-P(no aces))
and same thing; 5 cards first one show "happens" to be ace of diamonds then we are in the same case as above.
however if the sets are generated thus
1) generate all poker hands
2) remove all poker hands with no ace of diamonds
3) calc number of 2 ace hands / ace of diamonds hands = p1
1) generate all poker hands
2) remove all poker hands with no ace
3) calc number of 2 ace hands / number of ace hands = p2
So we can now see that p1 Prob(2 aces given ace of diamonds) and p2(prob of 2 aces given one ace) are not necessarily the same.
Then the logic of ThePolynomial and miller hold
There is a host of similar questions that rely on how you form the sets. I feel in most cases the intuition is generally to pick the first kind of assumption because that is how you would come across these problems in real life. Then when some math shows the other is correct you feel you are missing something. In reality both are correct depending on the set generation assumption. More trained mathematicians may think more like the 2nd set as that is either more interesting or just something they have become used to
Posted by: KJC | January 2, 2008 6:07 PM
KJC: I agree that they yield different answers, but how would you ever interpret "Given that the hand contains an ace" as "Given that the first card dealt to you is an ace?" (and similar for the ace of diamonds) I think the english language is strong enough in this case to exclude the later setup. Or am I wrong?
Posted by: Dave Bacon | January 2, 2008 6:14 PM
Fair enough, but I think it is how most people see the problem. The problem still holds even if it isn't the first card. If I get dealt 5 cards and there is the chance of 0 aces and then I turn over any of those cards and it is the ace of diamonds I don't gain any more info than it just being an ace (info about the chance of more aces anyway). So I don't think the sentences as written contain no ambiguity. It is "contains ace" but still "generation forces the ace" or "chance happend to give you the ace". In practice the hand I just picked up off the table, could have contained no aces, but one of the cards I glimpsed happens to be an ace, I think is how people see this played out in there heads. The language is strong enough. If you asked. how many of all poker hands containing 1 aces contain at least 2 aces, and how many of all poker hands containing the ace of diamons how many contain 2 aces it is unlikely peoples intuition will go wrong. "given it contains an ace" doesn't tell me how it was generated IMHO. I think most trained mathmeticians would understand the language in the way you describe, for us lesser trained (or perhaps people who play too much poker) I am not so sure
Posted by: KJC | January 2, 2008 6:30 PM
However if you original question was: Given we are creating sets to ensure the situation described then how can you intuitively see which is the bigger probability (i.e. you were not trying to get a way of understanding / explaining why it wasn't the same prob in either case) then perhaps my insights are pointless and (slightly) off topic.
Posted by: KJC | January 2, 2008 6:39 PM
Sorry for the comment spam. But is my longer comemnt to your question in moderation or did I mess up posting (just don't want to re-type it was a bit long (over long?)
Posted by: KJC | January 2, 2008 6:40 PM
KJC: Sorry the spam filter seems to be over eager. Probably the word "poker!" It should be posted now.
So is there a better way to phrase it? Should it be "Given that an honest observor tells you that the hand has at least one ace..." Or even more verbose "Given that an honest observor tells you that the hand has at least one ace, but doesn't tell you which of the cards are aces..."?
Posted by: Dave Bacon | January 2, 2008 6:49 PM
I believe the question is asking what the probability is of having a second ace, not the probability of a 2 ace hand versus a 2 ace hand with one having a specified suit.
Posted by: gex | January 2, 2008 6:57 PM
I still don't think this: There are six 2-ace hands with all suits allowed. There are three 2-ace hands with a diamond ace. is what the problem is asking. If that's what it intends to ask, then it's worded wrong. Why should the probability be any different if you're told the suit of the ace? In other words, I think it only appears to be like the Monty Hall problem, where the seemingly useless information actually does matter.
Anyways, I assume the blogger knows the answer. When can we know?
Posted by: jeffk | January 2, 2008 7:13 PM
At first glance, I thought they were the same, also. After being told one is larger, I figured the top one would have better odds... and I worked out a reduced-deck example.
Posted by: ocmpoma | January 2, 2008 7:26 PM
It seems that the sequence in which the cards are dealt, and the knowledge of the cards that have actually been dealt are important for this question.
For the first situation, the sequence of the discrete probabilities for two aces (dealt one after another) might look like this:
Situation 1: 1/52 --> 1/51
You're told that there's an ace among the five cards dealt, but you aren't told that when the cards were dealt, it was the first one to show up among these cards. So for all intents and purposes, the chances of getting an ace of diamonds for the first card dealt would be 1/52; the chances of getting another ace would be 1/51.
The sequence of discrete probabilities for the second situation would look like this:
Situation 2: 1/1 --> 3/51
Here, you're basically told there's already an ace among the five cards that are dealt, so for all intents and purposes, you could say that the probability of getting an ace as the first card *was* 1/1. This would mean that there's a 3/51 chance that the second card dealt will be an ace. So if we just go by the probabilities of the 2nd sequence probabilities for aces dealt, it looks like there's a higher probability for the second situation.
Posted by: Tony Jeremiah | January 2, 2008 7:53 PM
The reason this breaks intuition is in a subtlety that a few people are pointing out. The following cases are different.
(1) Before the deal, we decide that we will only accept hands with the Ace of Diamonds (others will be redealt). What is the chance that the hand has two Aces?
(2) Before the deal, we decide that we will only accept hands with one ace, and we will be told what the suit of that Ace is (e.g. Diamonds). What is the chance that the hand has two Aces?
The English phrasing in the original problem setup could be parsed either way.
Personally, Aces are special to me, and I assume that the hand was redealt until as Ace arrived. But the Ace of Diamonds isn't really a special Ace, so I figured that we just took whichever one came along.
Posted by: Mory | January 2, 2008 8:21 PM
Mory I'm confused: "two Aces" should be "two or more aces"? And I don't get your (2) at all... If the hand has "one ace" it cannot have "two Aces?" You mean "at least one ace"? And if there is more than one ace, I'm telling you the suit of which one exactly?
Posted by: Dave Bacon | January 2, 2008 8:28 PM
It seems to me that the question at least partially depends upon whether one takes a Bayesian approach or not, i.e. does knowledge of the ace of diamonds really condition our knowledge of the other aces? Quick, someone get Chris Fuchs (who is not only a Bayesian but also a Texan...)
Posted by: Ian Durham | January 2, 2008 8:29 PM
That phrasing can also be misinterpreted. Did the observer pick a random card in the hand, which happened to be an ace, and tell you about it? Or did you ask the observer if there was at least one ace and he answered "yes"? These two cases give two different answers. I assume you want the latter case.
Posted by: miller | January 2, 2008 8:33 PM
I think it is important to include some clarity of how the hand was generated. Like Mory says we have to say up front before we deal which hands we are dealing with. It all stems on whether, before the deal, there was a chance that I get 0 aces. I think that is a very common assumption. If you eliminate that possibility clearly then you have the right language (IMO). I don't think it is easy to clarify it though without getting too verbose. In normal language "Given..." clearly means to some people "It happened that..." and to other "We assure that...". Obviously there are lots of ways to specify it clearly, but they are to verbose. I cannot come up with a pithy way that I think is 100% clear.
Posted by: KJC | January 2, 2008 8:34 PM
take a look at the "principle of restricted choice" in card play, specifically in bridge.
Posted by: matt | January 2, 2008 8:44 PM
Stupid English. Let me try and restate
"Two Aces" should be "Two or more Aces".
Case (2) restated:
(2) Before the deal, we decide that we will only accept hands with at least one Ace, and we will be told what the suit of the Ace on the left (e.g. Diamonds). What is the chance that the hand has two or more Aces?
Posted by: Mory | January 2, 2008 8:46 PM
OK, I thought of an analogy that may help explain intuitively why there is a difference in the two cases. Instead of cards, imagine a pair of dice where one is red and the other is blue.
Case 1: Without looking at the dice what is the probability of rolling snake eyes (two 1s)? As any craps player knows, it is 1/36.
Case 2:We sneak a peak and notice that the red die is a 1. Now what is the chance that the blue die is a 1? Well, now only the blue die is unknown and so the probability is now calculated on a single die! The answer here, of course, is 1/6.
Thus, conditioning really matters and Pr(Case 2) ≥ Pr(Case 1) where the inequality holds if both cases have zero probability (if not, they asymptotically approach one another as the number of configurations increases).
Posted by: Ian Durham | January 2, 2008 8:47 PM
Ian, I read the analogy more like this:
Case 1: Without looking at the dice what is the probability of rolling snake eyes (two 1s), given that one die is a 1? As any craps player knows, it is 1/6.
Case 2:We sneak a peak and notice that the red die is a 1. Now what is the chance that we rolled snake eyes? Still 1/6.
Posted by: jeffk | January 2, 2008 9:39 PM
I think the non-intuitive part of this problem lies in whether we treat all of the aces as indistinguishable or not, and whether their order in our hand is relevant.
Posted by: Caledonian | January 2, 2008 9:58 PM
I should clarify, that should read "we know that it's the red die that's a 1, and we don't know the other die" vs. "we know that one of the dice is a 1, and we don't know the other die".
In other words, it doesn't matter what color the dice are because they're not different. In the problem in this post, it doesn't matter the suit of the aces. If it matters, I think the problem is not clearly stated.
Posted by: jeffk | January 2, 2008 10:16 PM
jeffk, you have an interesting point. That is an extremely subtle difference and is dependent upon when the knowledge is revealed. If we're revealing both dice simultaneously, it is 1/36 but if we reveal them one at a time and know the first is a 1, then the probability of the second being a 1 is 1/6. In that sense there is no difference between the cases and I would agree that the problem, as stated, should have been more properly worded.
Posted by: Ian Durham | January 2, 2008 10:20 PM
Dave Bacon (#7) wrote:
Not really. I've worked quite a bit with math and the one thing I've learned is to never trust intuition if I have a chance. That said I've just dug out a discrete math book (university comp sci introductory level) to see if they have an explanation.Posted by: Who Cares | January 2, 2008 11:00 PM
First, the easiest way to NOT the same is to think about a coin toss. The red/blue distinction someone mentioned is useful and can make it more intuitive, so let's say one is red and one is blue. Using the following abbreviations (R = red, B = blue, H = heads, T = tails), the only possible outcomes after flipping both coins are, of course, RH&BH, RH&BT, RT&BH, RT&BT.
Now, two cases.
Case 1) I tell you that the red coin landed heads up. What are the odds that the blue coin also landed heads up? Obviously, the outcomes where the red coin landed tails up are eliminated, leaving us with RH&BH and RH&BT. So there's a 50% chance that the blue coin landed heads up.
Case 2) I flip the coins without you looking and tell you that AT LEAST ONE coin landed heads up, without specifying which. What are the odds they both landed heads up? Well, the only outcome eliminated by the informatoin I gave you is the one where neither lands heads up - RT&BT. So we still have RH&BH, RH&BT, and RT&BH. Each is equally likely (because each had a 1/4 chance to begin with). Thefore, there is only a 33% chance that both coins landed heads up.
This is extremely UNintuitive, I think, and it takes a really long to convince some people. It seems that if I tell you one coin landed heads up, it should be just like if I told you which coin (i.e., you can arbitrarily pick a coin to be the one that landed heads, because you know both had the same chance). But it doesn't work. Let's try it: I tell you one coin landed heads up, and you assume the red coin landed heads up, so there's a 50% chance as in case 1. But wait; what if the red coin landed heads down? You've unfairly excluded that equally probable outcome, so 50% can't be right.
Eliminating certain outcomes will probably have an effect (I assume there are cases where you could introduce additional information with the discarded outcomes having no effect on the probability, although I'm not thinking about that now).
So the simplest way to look at probability problems, at least for me, is to look at the effect of the eliminated outcomes. Unfortunately, this doesn't work as easily for the poker problem because we're not dealing with binary choices - Case 1 both adds a bunch of extra cases where there's more than one ace, and a bunch of extra cases where there's only one ace. But I can still think through this with a little bit more effort.
Add three cases: 3, 4, and 5. These are the same as Case 2, but with different suits.
Case 1 multiplies the number of single-ace outcomes from Case 2 by 4 (add up the single-ace outcomes from Cases 2-5), but it can't possibly multiply the number of multiple-ace outcomes by 4 (using the same technique, adding up the multiple-ace outcomes of Cases 2-5, that would necessitate a ton of overlap - you would have to, say, count "Ace of Diamonds & Ace of Hearts" twice, or "every Ace" four times, which of course doesn't make any sense). This might count as a calculation, because my explanation doesn't make sense if, for example, there are more suits than there are cards in a hand. Still, for the given rules, if Case 1 has four times as many single-ace outcomes as Case 2, and less than four times as many multiple-ace outcomes, the probability of a multiple-ace outcome would have to be greater in Case 2.
Does that make any sense?
Posted by: T | January 3, 2008 12:20 AM
Ok, I hope I get this right; I should really go to bed.
First off, every 5-card poker hand is unique and thus has 120 permutations. You can factor this right out of the picture. We can deal purely in combinations.
#1: "Given that the hand contains an ace, what is the probability that the hand contains another ace?"
I.e., given that you have greater than zero aces, what's the probability of you having greater than one ace?
Greater than zero aces is all poker hands, minus those that have no aces: (52,5) - (4,0)*(48,5) = (52,5) - (48,5) = 886656
Greater than one ace is the previous, minus those that have one ace: (52,5) - (48,5) - (4,1)*(48,4) = 108336
Probability = 108336/886656 ~ 0.12
#2: "Given that the hand contains the ace of diamonds, what is the probability that the hand contains another ace?"
I.e., given that you have the AD, what's the probability of you having greater than zero more aces?
A hand with the AD: (1,1)*(51,4) = (51,4) = 249900
A hand with the AD and more aces is the previous minus those that have zero more aces: (51,4) - (1,1)*(48,4) = 55320.
Probability = 55320/249900 ~ 0.22
Intuition tells us nothing about the truth of a/b > c/d when a>c and b>d. So we need to look at the numbers. It's not surprising that the number of hands with an ace is about four times as much as the number of hands with the ace of diamonds, at least not to me. What's surprising, though, is that the number of hands with at least two aces is only about twice as much as the number of hands with the ace of diamonds and at least one other ace.
That is until you realize that these possibilities are dominated by the hands with only a pair. As someone already pointed out, the number of ace pairs with the ace of diamonds is half that of the total number of ace pairs:
Hands with ace pairs = (4,2)*(48,3) = 6*(48,3)
Hands with the ace of diamonds and one other ace = (1,1)*(3,1)*(48,3) = 3*(48,3)
Thus, the denominator is diminished by a factor of four, but the numerator is only diminished by a factor of 2.
Posted by: John Moeller | January 3, 2008 3:24 AM
jeffk, I think you've nailed a very good analogy in your 9:39pm post, but you have the math wrong. If the dice are different colors, there are 36 possible rolls, and 11 of them have at least one "1" showing. So for case 1, the answer is 1/11, not 1/6. The answer for case 2 is indeed 1/6. This gives the same answer as the original ace question (the more specific case leads to a higher probability).
The analogy isn't perfect (it is a coincidence that the nearly-factor-of-2 difference is the same for both the die and the cards), but I think quite useful all the same.
Posted by: Ken Wharton | January 3, 2008 9:46 AM
To make the dice analogy that Ian and jeffk raised as much like the original problem as possible, the two cases must be as follows:
1. We know that at least one of the dice is a 1, but we don't know whether it's the red die or the blue die (or both). There are 11 possibilities: six in which the red die is 1, six in which the blue die is 1, but we've double-counted the case where both are 1. Only one of those cases is snake eyes, so the probability is 1/11.
2. We specifically know that the red die is a 1. Now there are only six cases, and as Ian correctly pointed out, the probability is 1/6.
Having done this calculation, you can see that the inequality should go the same way in the original problem (which was contrary to my initial intuition that the probabilities should be the same). More hands will contain at least one ace of any suit than will contain the ace of diamonds, and this ratio is greater than the ratio between the number of hands with two or more aces of any suit and the number of hands with two or more aces, one of which is the ace of diamonds.
Posted by: Eric Lund | January 3, 2008 10:02 AM
I'm still not sure what's going on here. Has the generation of the sets question been settled? If not, maybe it would be better to say that you have 52 marbles of different colors. That way you eliminate the experiential issues dealing with dealing, which are really not relevant to the original question. I read that original question as being about an existing set of "cards" that had been generated in the past by an unknown but random method. Thus there is an existing set of N cards selected from 52 cards that is under examination. Is it necessary to know the value of N? That differs with different card games. Or are we poker snobs here?
I have an inkling of a rationale for why the probabilities might be different, but I would like to see the explanation, and then the discussion about why the explanation is wrong. Or right.
Posted by: Mark P | January 3, 2008 10:27 AM
I don't know if this got through last night; I posted it, but it did not appear.
First off, every 5-card poker hand is unique and thus has 120 permutations. You can factor this right out of the picture. We can deal purely in combinations.
#1: "Given that the hand contains an ace, what is the probability that the hand contains another ace?"
I.e., given that you have greater than zero aces, what's the probability of you having greater than one ace?
Greater than zero aces is all poker hands, minus those that have no aces: (52,5) - (4,0)*(48,5) = (52,5) - (48,5) = 886656
Greater than one ace is the previous, minus those that have one ace: (52,5) - (48,5) - (4,1)*(48,4) = 108336
Probability = 108336/886656 ~ 0.12
#2: "Given that the hand contains the ace of diamonds, what is the probability that the hand contains another ace?"
I.e., given that you have the AD, what's the probability of you having greater than zero more aces?
A hand with the AD: (1,1)*(51,4) = (51,4) = 249900
A hand with the AD and more aces is that minus those that have zero more aces: (51,4) - (1,1)*(48,4) = 55320.
Probability = 55320/249900 ~ 0.22
Intuition tells us nothing about the truth of a/b > c/d when a>c and b>d. So we need to look at the numbers. It's not surprising that the number of hands with an ace is about four times as much as the number of hands with the ace of diamonds, at least not to me. What's surprising, though, is that the number of hands with at least two aces is only about twice as much as the number of hands with the ace of diamonds and at least one other ace.
That is until you realize that these possibilities are dominated by the hands with exactly two aces. As someone already pointed out, the number of ace pairs with the ace of diamonds is half that of the total number of ace pairs:
Hands with ace pairs = (4,2)*(48,3) = 6*(48,3) = 103776
Hands with the ace of diamonds and one other ace = (1,1)*(3,1)*(48,3) = 3*(48,3) = 51888
Thus, the denominator is diminished by a factor of four, but the numerator is only diminished by a factor of 2.
Posted by: John Moeller | January 3, 2008 10:47 AM
My attempt. Compare: p(E2|E1)=p(E2)/p(E1), the probability of event E2, at least 2 aces in the hand, given the event E1, at least one ace in the hand; and p(E1H|E0H)=p(E1H)/p(E0H), the probability of event E1H, the ace of hearts (nicer than diamonds) plus at least one other ace in the hand, given the event E0H, the ace of hearts is in the hand, plus at least zero other aces. [The wording is such that E2 could be the probability of exactly 2 aces in the hand, and likewise for E1H, exactly 2 aces, one of them the ace of hearts, but I will discount that.]
We want to see "intuitively" whether p(E2|E1) is greater or less than p(E1H|E0H), which is equivalent to wanting to know whether p(E0H)/p(E1) is greater or less than p(E1H)/p(E2). Intuitively, if we have one or more aces, the chance of one of them being the heart is less than if we have 2 or more aces (I hope that's right! I take it I'm not allowed to check), so p(E2|E1) is less than p(E1H|P(E0H).
For the dice problem, E2 is snake-eyes (1/36), E1H is also snake-eyes (1/36), E1 is "at least one die is 1" (11/36), E0H is "just the red die is 1" (1/6), so p(E0H)/p(E1)=6/11 is less than p(E1H)/p(E2)=1, and p(E2|E1)=1/11 is less than p(E1H|E0H)=1/6, which agrees with the above.
I wouldn't say this is intuitive exactly, the mathematical setup is wearying but essential, as always, but the last stage is perhaps more-or-less intuitive, and there's only one algebraic step to it, and I've obeyed the rule of not actually calculating the probabilities, right? Except not for the dice, sorry.
Indeed, I personally find probability never intuitive.
Posted by: Peter Morgan | January 3, 2008 11:17 AM
This is the same as the principle of restricted choice in card play. That principle states that if someone does not play a card when they could have, it affords an inference that they don't have that card. As an example, suppose there are two cards, say a queen and a jack, held between the two opposing players. You'd expect that 50% of the time each of the opposing players has one of the two cards, and 25% of the time the left-hand players has both and 25% of the time the right-hand player has both. (In reality, the probabilities are very slightly different because if a player has one of the two cards, he has fewer free places in his hand to hold the other card, but let's not worry about that) Then, at some point in the card play, your right-hand opponent has the chance to win a trick with one of these cards. Let's say he plays the queen. Now, where is the jack? Is it 50/50 that it is in either hand? No, it's not...remember, the chance that the right-hand players has either the queen or the jack or both is 75%. So, 75% of the time we are in the situation that the right-hand player will just have one a trick with one of the two cards. But, only 25% of the time does he hold both cards. So, only 1/3 of the time does he hold the remaining card, in this case the jack, and 2/3 of the time he does not hold that card. This is called the principle of restricted choice because if he held only the queen, he has to play the queen (his choice is restricted), but if he holds both the queen and then jack, he has a choice which one to play and only 50% of the time will he play the queen, which affords an inference that it is less likely that he holds the jack also. This is the same as David's problem.
Posted by: matt | January 3, 2008 11:46 AM
Kudos to Ken (jeez, you're up early!) and Eric for finding the correct dicing analogy. Personally, I tend to work with dice a lot since I partially developed an introductory physics lab on probability that uses them and, for whatever reason, I intuitively think that way (perhaps it is because I have a collection of very odd dice including a pair that are perfectly round...).
Posted by: Ian Durham | January 3, 2008 11:49 AM
oops, many typoes removed:
This is the same as the principle of restricted choice in card play. That principle states that if someone does not play a card when they could have, it affords an inference that they don't have that card. As an example, suppose there are two cards, say a queen and a jack, held between the two opposing players. You'd expect that 50% of the time each of the opposing players has one of the two cards, and 25% of the time the left-hand player has both and 25% of the time the right-hand player has both. (In reality, the probabilities are very slightly different because if a player has one of the two cards, he has fewer free places in his hand to hold the other card, but let's not worry about that) Then, at some point in the card play, your right-hand opponent has the chance to win a trick with one of these cards...and suppose either one will work equally well to do this. Let's say he plays the queen. Now, where is the jack? Is it 50/50 that it is in either hand? No, it's not...remember, the chance that the right-hand player has either the queen or the jack or both is 75%. So, 75% of the time we are in the situation that the right-hand player will just have won a trick with one of the two cards. But, only 25% of the time does he hold both cards. So, only 1/3 of the time does he hold the remaining card, in this case the jack, and 2/3 of the time he does not hold that card. This is called the principle of restricted choice because if he held only the queen, he has to play the queen (his choice is restricted), but if he holds both the queen and then jack, he has a choice which one to play and only 50% of the time will he play the queen, which affords an inference that it is less likely that he holds the jack also. This is the same as David's problem.
Posted by: matt | January 3, 2008 11:51 AM
Hey Matt, I wonder if bridge players develop a better intuition for probabilities like this because of their exposure to the principle of restricted choice?
Posted by: Dave Bacon | January 3, 2008 12:03 PM
I think so, especially since they don't get to stop and calculate the probabilities.
Interesting computer science problem: a good program for playing bridge. There exist extremely fast programs to solve for best play given knowledge of all the cards. These lead to Monte Carlo attempts to solve the problem of unknown cards based on dealing out lots of hands which fit what you do know about the opposing cards and then picking the play which works in the majority of situations. These have the problem, though, of dealing with choices in the future: the program has a tough time knowing that it will have to take a guess in the future, and it will tend to assume that it takes any guess which will occur on a future card play correctly, since it just finds best play for each hand it deals. Simple situations like the one above can be dealt with, but to my knowledge there is no algorithm comparable to those in chess. Namely, the chess algorithms have the property that you run them on a faster machine and they get better and better, but I think there are certain situations which are just beyond the ability of existing bridge programs to solve. (that's just in the card play phase, not even in bidding, and not even worrying about deceptive play and so on)
Posted by: matt | January 3, 2008 12:34 PM
Great problem!!!
Posted by: Gil Kalai | January 3, 2008 12:55 PM
John Preskill's comment is correct, but I thought I'd add in an explanation of why the problem is confusing. It has to do with how the random hands are generated. Let's consider the given-an-ace-of-diamonds scenario in the following two cases:
Case 1:
I deal a random five card hand, then look at it. If it does not contain the ace of diamonds, I reshuffle and redeal.
Case 2:
I deal a random five card hand, then look at it. If it does not contain at least one ace, I reshuffle and redeal. Otherwise, I randomly pick one of the aces in the hand and announce its suit (which we suppose happens to be diamonds in a particular instance).
In case 1, Preskill's analysis applies. In case 2, knowing the suit provides no additional information about the probability of having a second ace, and thus in case 2, problem 2 has the same probability as problem 1. This should be verifiable with Bayesian statistics, but I'm too lazy to actually check.
Posted by: Travis | January 3, 2008 1:05 PM
The dice version (Eric Lund 10:02) also has a simpler setup (which I learned here, right beneath the statement and solution of the card version).
Person A and Person B each have two children. Person A tells you she has a son named John while person B tells you that his older son is also named John. Which of the two is likelier to have two sons?
Posted by: Joe Renes | January 3, 2008 1:16 PM
For anyone who wants to see the exact answer, I suggest you follow the link Joe posted in the last comment.
Posted by: Dave Bacon | January 3, 2008 1:43 PM
If person B's older son is named John, person B has a 100% chance of having two sons.
Posted by: Caledonian | January 3, 2008 2:22 PM
Caledonian, I knew I would screw that up somehow. Person B tells you his older child is also named John (presumably a boy).
Posted by: Joe Renes | January 3, 2008 3:00 PM
Again there is ambiguity is the son problem.
In the first case you could say that there are three possibilities: older son and younger daughter, older daughter and younger son, older son and younger son. One out of three then yields two sons. However, another interpretation is that if one child is a son, then there is a one in two chance that the other is a son.
Therefore in the first interpretation the probabilities of the two cases are different and in the second interpretation they are the same.
Conditional probabilities assume the first interpretation and this is what gives our intuition problems. For example suppose I ask if Person A has a son and she says yes. Then I say there is a 1/3 change of her having two sons. But then I ask if that son is the older or younger child. Regardless of her answer, I will raise my probability to 1/2.
Posted by: Matt Elliott | January 3, 2008 3:00 PM
Joe:
Thanks for the link to the solution. I was afraid that I got it horribly wrong. :-)
Matt posted his excellent bridge explanation while my posts were caught up in the queue, though. Oh well.
Posted by: John Moeller | January 3, 2008 3:03 PM
My intuition said that you can look at this as a set problem.
All the possible dealt options being the whole set,
options with two or more aces is a subset of this,
and options with two or more aces (one being the ace of diamonds) is a further subset of that.
Therefore the smallest subset has a higher probability as there are less options.
This suggests that with a fixed number of possibilities the more information you have about the system, the more accurate your guess about which one it is.
(But I am new to probability, so I may be way off here.)
Posted by: Nancy | January 3, 2008 3:31 PM
Nancy, in my experience you can look at almost anything as a set problem! But probabilities are particularly amenable to this since, historically, the two are linked in many ways (e.g. Venn diagrams, etc.).
Posted by: Ian Durham | January 3, 2008 4:09 PM
I find the second question a bit ambiguous. If I asked is there an ace? And you said yes, and I asked show me? It would be unremarkable whatever suit you showed me, i.e. the card had to have some suit? If I had instead asked "Does it contain an Ace of diamonds?", and the answer was yes, then isn't this a different kettle of fish. The second question is only true in the particular, and I know that cases with more than one ace would be oversampled relative to the first instance.
Posted by: bigTom | January 3, 2008 4:11 PM
I think the problem, as stated, is ambiguous. See, for example, Ambiguity in Probability Problems by I.G. Betteley.
Posted by: The Vlad | January 3, 2008 4:27 PM
Vlad: could you be a little more specific?
Posted by: Dave Bacon | January 3, 2008 4:41 PM
Nancy, the trouble with the sets you listed is that we are comparing p(E2|E1)=p(E2)/p(E1), the probability of event E2, at least 2 aces in the hand, given the event E1, at least one ace in the hand; and p(E1H|E0H)=p(E1H)/p(E0H), the probability of event E1H, the ace of hearts (nicer than diamonds) plus at least one other ace in the hand, given the event E0H, the ace of hearts is in the hand, plus at least zero other aces (all this copied from my post earlier).
Thus, there are four different events involved (indeed the definition in mathematics of probability spaces is in terms of sets, and events are defined as subsets of the sample space). All four of these sets have to be straight in your head.
Posted by: Peter Morgan | January 3, 2008 5:09 PM
Consider the following simplification:
Let the set of cards in the deck = {Ace hearts, Ace diamonds, Ace clubs, 2 spades}
Now define the following:
Ace hearts = {Ace hearts}
Ace not hearts = {Ace diamonds OR Ace clubs}
Ace = {Ace hearts OR Ace diamonds OR Ace clubs}
Assume that a hand contains 2 randomly-drawn cards (without replacement) from the deck, the 6 possible permutations being:
Ace hearts, Ace diamonds
Ace hearts, Ace clubs
Ace hearts, 2 spades
Ace diamonds, Ace clubs
Ace diamonds, 2 spades
Ace clubs, 2 spades
Adding the instances of various permutations according to a conditionality rule
Probability (Ace not hearts | Ace hearts) = 2/3 (where | denotes conditionality)
Probability (Ace not hearts | Ace not hearts) = 1/3
THE AMBIGUITY
Note that, according to the above formulation, the following is meaningless
Probability (Ace hearts | Ace not hearts)
Probability (Ace | Ace not hearts)
as we always count permutations containing Ace hearts as belonging to Probability (Ace not hearts | Ace hearts)
I think the ambiguity can be resolved by noting that the notion of conditionality involves a sense of temporal order, with one thing following (conditional upon the occurrence of) another. If one states the problem in terms of temporal order (one card is seen, then the second) the probabilities for the two cases will equal. Do you agree?
Posted by: The Vlad | January 3, 2008 5:12 PM