# Ancestral Monty

You might have noticed that blogging has been a bit erratic lately, and I have fallen off my usual pace of updating every weekday. There's a reason for that! Regular readers of this blog are aware that I have a small obsession with the Monty Hall problem. I managed to convince Oxford University Press that a book on the subject would be a good idea, and now I'm supposed to submit a manuscript to them by New Year's. And since there is a limit to how many hours a day I can spend typing away at a computer, well, you get the idea.

I've also learned something else lately. Writing a book is hard! I always pictured it as just writing a long essay, and since I write long essays at this blog all the time, usually just making it up as I go along, I figured writing a book would be just like that, only more so. But it turns out you actually have to do reasearch in writing a book. That means not just obtaining copies of lots of obscure journal articles but also reading them! I even had to learn how to use a microfilm machine again (The New York Times had an informative front page story on the MHP back in 1991). They've gotten more sophisticated since the last time I used one back in high school.

And every time I think I've gotten all my references together, I discover a treasure trove of other references I need to go through. For example, the Monty Hall problem was living a parallel life in the technical literature in the form of something called the three prisoners problem. Suddenly all the time I spent typing “Monty Hall problem,” “game show problem,” “cars and goats,” and every other string I could think of into various search engines wasn't good enough.

But maybe I should start at the beginning.

Chapter One of the book describes the history of the MHP. I originally envisioned this as a short little romp, but it currently stands at around ten thousand words and it is not finished yet. Here's the super condensed version.

It is typically Blaise Pascal who is given credit for founding the modern theory of probability. In correspondence with Pierre de Fermat in the seventeenth century, the two hashed out the “Problem of Points”. Imagine that two players, A and B, are involved in the following game. A coin is tossed. If it comes up heads then A gets a point. If it comes up tails then B gets a point. The winner is the first to get to ten points, and the prize is a pool of money to which both players contributed equally. Let us suppose the action is interrupted at a moment when the score stands 8 to 7 in A's favor. How should the prize money be distributed between the players?

Of course, this is not strictly speaking a problem in mathematics. Some standard of fair play will have to be used in judging the equity of a proposed solution, and math has little light to shed on what that standard ought to be. Nonetheless, if we simply assume that people know a fair division when they see one, we can still hope to arrive at a definitive answer.

Fermat got the ball rolling by pointing out that if A needs x points to win and B needs y points to win, then the game will be over in no more than x+y-1 further tosses. Since these tosses can come in 2x+y-1 equally likely ways, a fair division can be reached by counting up the number in which A wins and the number in which B wins. The money should then be divided in accordance with this ratio.

For instance, in the concrete example given above we have that A needs two more points and B needs three more points to win. The game will consequently end in no more than four more tosses, which can come up in 16 different ways. Since 11 of those ways lead to A victories, while 5 of those ways lead to B victories, the money should be divided up in the ratio 11:5.

Pascal agreed that this was a fine way of doing things, but observed that it quickly becomes unwieldy to write out all the possible scenarios. He developed some counting techniques based on his famous arithmetical triangle that provide a general method for working out problems of this sort.

By reasoning in this way, Pascal and Fermat realized some very important points that had eluded previous proposed solutions to this problem. They realized that what is important is not so much the current score, but rather how many further points were needed for each player to win. They realized the significance of the fact that all future outcomes were equally likely. And they, especially Pascal, recognized the importance of elementary combinatorics in solving these sorts of problems. Elementary considerations by modern standards, but several other mathematicians failed to realize their significance.

Pascal actually made another contribution to probability theory. I refer, of course, to Pascal's wager. Pascal's argument here was considerably more subtle than is sometimes presented (for example, he was not making an argument for the existence of God, but rather an argument for why one ought to behave as if there were a God), but for all of that was still unpersuasive. For our purposes the significance of the wager was that it represents the first time that probabilistic considerations were invoked in the context of making a decision in the face of uncertainty.

Enumerating sample spaces, recognizing equally likely scenarios, using probability as a guide to making decisions under uncertainty. The Monty Hall problem in a nutshell.

Of course, in crediting Pascal with founding modern probability, I don't mean to slight the considerable achievments of other folks like Huygens and Liebniz. I am merely following the example of Ian Hacking, who wrote, in his book The Emergence of Probability:

Probability has two aspects. It is connected with the degree of belief warranted by evidence, and it is connected with the tendency, displayed by some chance devices, to produce stable relative frequencies. Neither of these aspects was self-consciously and deliberately apprehended by any substantial body of thinkers before the time of Pascal.

The next significant development would surely be the work of Thomas Bayes, of “Bayes' Theorem” fame. He developed the most comprehensive solution up to that time to the problem of revising a probability assessment in the face of new data, and his theorem is of singular importance to solving certain versions of the Monty Hall problem. Nonetheless, in he interest of keeping this post to some sort of reasonable length, we will skip over it.

The first occurrence of a Monty Hall like problem of which I am aware came in 1889. In that year the French mathematician Joseph Bertrand published his book Calcul de Probabilites. On page one of this book he discusses the idea that a probability is the ratio of favorable occurrences to possible ooutcomes. On page two (literally!) he pondered a chest of three drawers, one of which contained two gold coins (literally “medallions” in the original text), one of which contained two silver coins, and one of which contained one gold and one silver. If we pick a drawer at random what is the probability that the drawer contains different coins? Obviously, the answer is one out of three.

But now suppose we pick a drawer at random and remove a coin. We do not look at it. What is the probability now that the drawer contains different coins? We might reason that regardless of whether the coin in our hand is gold or silver, there are still only two possibilities. The other coin is either the same or it is different. Thus, it would seem the probability is now 1/2.

Surely an absurd result. Our probability jumps from 1/3 to 1/2 just because we removed a coin? Ridiculous! Bertrand went on to explain what went wrong, that the two possibilities (same kind of coin, different kind of coin) were not equally likely. For example, if the coin in your hand is gold, then you might have reached into either the two golds drawer or the one gold one silver drawer, but it is more likely that you did the former than the latter. This was intended as a cautionary tale in what happens when you are too cavlier in listing your possible outcomes, without giving due consideration to the likelihood of each one.

This is nowadays described as the Bertrand Box Paradox, though what modern probability texts describe by that name is actually a bit different from what Bertrand wrote. Nowadays we usually ask, “Assuming that you reach into the drawer and pull out a gold coin, what is the probability that the other coin in the drawer is gold as well?” You want to say 1/2, because the silver/silver drawer has been eliminated and the remaining options are equally likely. But the correct answer is 2/3, because you are twice as likely to remove a gold coin from the gold/gold drawer as you are from the gold/silver drawer.

The relationship to the Monty Hall problem is clear. In both cases a counter intuitive result is obtained because in updating the probability of X in the face of new data Y, people tend to ignore the importance of assessing the probability of obtaining Y under the assumption of X.

We fast forward now to 1959. Martin Gardner, writing in Scientific American, asks us to consider the plight of three prisoners. One is to be freed, the others are to be executed. Each is equally likely to be the one freed. A asks the warden to tell him the name of one of the other prisoners who will be executed. The warden reasons that this information can't possibly be of any use to A, so he says, “B will be executed.” A chuckles, Since it is now between himself and C to see who will be freed, his chances of survival have gone from 1/3 to 1/2. Has A reasoned correctly?

No, he hasn't. This, now, is the Monty Hall problem in all but name. Gardner gave the correct solution, that A still has a mere 1/3 chance of being freed while C's chances have now improved to 2/3, in the following issue. By doing so, he kicked off a steady trickle of professional papers discussing the glories of the three prisoner problem. In a small 1965 book called Fifty Challenging Problems in Probability and Their Solutions, the Harvard mathematician Fred Mosteller included the prisoner problem as number 13. Foreshdowing the Monty Hall wars to come, Mosteller commented that this problem more than any other attracted lots of mail wherever it was published. Throughout the seventies and eighties various technical papers appeared from mathematicians, philosophers and cognitive scientists discussing various aspects of the problem.

The game show Let's Make a Deal premiered on American television in 1963, but it was not until 1975 that the phrase “Monty Hall problem” appears in print. A statistician named Steve Selvin published a letter to the editor about in The American Statistician, presenting the correct solution via an enumeration of the sample space. Predictably, many angry letters ensued claiming the solution was wrong. Selvin's lucid analysis made short work of the critics. Had more mathematicians been aware of it, some later embarrassment could have been avoided.

Then came l'affaire Parade. Marily vos Savant tackled the question in 1990 when it was posed by a curious correspondent. She gave the correct answer that the contestant doubles his chances of winning by switching. Over the next few months some ten thousand letters would come in, many from professional mathematicians boasting of their credentials in the subject, lecturing her in snotty, condescending tones, about how she had gotten things wrong. Overall, vos Savant estimated that over ninety percent of the letters disagreed with her solution.

vos Savant devoted three subsequent columns to the problem. Things were later adjudicated in her favor when classrooms across the country did a simulation of the problem, thereby generating huge quantities of data to look at. (In math jargon we would say they used a Monte Carlo simulation to tackle the problem). The empirical data was unambiguous, and Marilyn was completely vindicated.

The whole spectacle reached the height of journalistic importance, page one of the Sunday New York Times, on June 21, 1991.

Thoughout the nineties and even into the current century, various professionals started turning out a stream of technical articles on the subject. Mathematicians and statisticians discussed the probabilistic issues. Philosophers found relationships between the Monty Hall problem and other philosophical questions such as the famous Doomsday Argument. Physicists developed quantum mechanical versions of the problem. And cognitive scientists and social psychologists undertook some studeis to figure out why people find this problem so darned confusing.

Cognitive scientist Massimo PIatelli-Palmarini aptly described the situation by writing, “...no other statistical puzzle comes so close to fooling all the people all the time...The phenomenon is particularly interesting precisely because of its specificity, its reproducibility, and its immunity to higher education.”

So there you go. Just a taste of what you will find in the book. Assuming I manage to finish it before driving myself crazy!

Tags
Categories

### More like this

##### The Big Monty Hall Book Gets Reviewed in Science
And mostly favorably, too. You might need a subscription to read the review, alas. The reviewer is Donald Granberg, a sociologist (now retired) at the University of Missouri. He published several papers on the MHP during the nineties. I liked this part of the review: The author does a…
##### Another Way to Be Confused About the Monty Hall Problem
In the Monty Hall problem, you are confronted with three identical doors, one of which conceals a car while the other two conceal goats. You choose a door at random, number one say, but do not open it. Monty now opens a door he knows to conceal a goat. He then gives you the option of sticking or…
##### Monty Hall and Interpretations of Probability
If I say that X has probability p, what does that mean? What sort of thing is X, and what does the number p represent? Philosophers have spilled a lot of ink on this question, with no clear answer emerging. Instead there are a handful of major schools of thought on the issue. Each school…
##### Monty Hall Deniers?
My account of the big creationism conference will resume shortly, but I really must take time out to discuss this article by Brian Hayes of American Scientist. He is discussing the Monty Hall problem, you see. The story begins with this earlier article by Hayes. He was reviewing the recent book…

mgarelick,

Sorry I took so long to get back to you. What you were saying in your last comment is basically correct. Regardless of how many gold coins are in the "gold coin only" drawer, the draw will break down thus:
1/3 of the time you will pick a gold coin from the gold drawer.
1/3 of the time you will pick a silver coin from the silver drawer.
1/6 of the time you will pick a gold coin from the mixed drawer.
1/6 of the time you will pick a silver coin from the mixed drawer.

So you will be twice as likely to pick a gold coin from the gold drawer than from the mixed drawer. (The break-down before looking at the coin works out to the same as in my poorly expressed previous explanation:
2/3 of the time from a "homogeneous" drawer {1/3 + 1/3}
1/3 of the time from the mixed drawer {1/6 + 1/6})

This ratio will hold for the 2 drawer example with the break-down being:
1/2 of the time you will pick a gold coin from the gold drawer.
1/4 of the time you will pick a gold coin from the mixed drawer.
1/4 of the time you will pick a silver coin from the mixed drawer.

You are twice as likely to pick that gold coin from the gold drawer than from the mixed drawer.

Incidentally, your "three gold coins" explanation is the correct one for explaining the paradox as it is posed on the Wiki page that Jason linked to in his post.

Also, I've enjoyed this problem so much, I went and made a video about it and posted it on YouTube.

Have you tried to get in touch with Monty Hall? I think he is still alive. He really should write the introduction or at least a cover blurb for the book.

Jason,

I can't wait for your book to come out.

It's funny; I always saw the trap that the naysayers in l'affaire Parade fell into as a sort of "reverse Gambler's Fallacy." (though not precisely)

Even though I don't know much about the Monty Hall problem, the fact that you have a book in the pipeline is awesome! You have a guaranteed buyer right here.

"For example, if the coin in your hand is gold, then you might have reached into either the two golds drawer or the one gold one silver drawer, but it is more likely that you did the former than the latter."

The probability of reaching into one of two drawers can only be 1/2 - absent something that causes one drawer to be preferred over the other. When three 'gold'* coins are distributed among two drawers finding that you have picked up a 'gold' coin and not a silver one is a more likely experience if you reached into the drawer that had two 'gold' coins in it. Thus, when considering what the known distribution of coins is and what having picked up a specific coin tells you about where you probably got it the number of drawers that the coins are distributed among is misleading. I think a large part of the difficulty with the Monty Hall problem is in how it is described.... (Marilyn Vos Savants' explanation always seemed to assume - without stating it - that Monty Hall was intentionally picking which doors to open, and not opening them at random. Otherwise her answer never made sense to me. If he was picking doors at random I do not see how knowing what was behind one door could give you any information about what was behind either of the other doors.)

* gold is in 'scare quotes' because a gold coin and a silver coin of the same size are noticeably different in mass. enough so that bumping one can enable you to tell which is which.

By oscarzoalaster (not verified) on 07 Nov 2007 #permalink

The first paragraph amazed me. "How do you pad a half page conditional probability exercise (which is all it took the first time I saw it) out to an entire book?" came the thought.

But, of course, once you start looking at history, and some of the background of probability and so on, and maybe look at why so many people get it wrong, then yes, of course.

oscarzoalaster: She was correct to assume that, because that's exactly how the game works. Monty *never* reveals the good prize when he shows what's behind one of the doors. That is, he deliberately reveals a bad choice.

Indeed, it's obvious this is the only way it CAN work. How does it work as an exciting TV game if Monty shows him the freaking PRIZE and then says "but think really carefully now - do you want to switch?". What's he going to do, offer ten seconds of think music so the contestant can decide between the two lame options?

"For example, if the coin in your hand is gold, then you might have reached into either the two golds drawer or the one gold one silver drawer, but it is more likely that you did the former than the latter."

The probability of reaching into one of two drawers can only be 1/2 - absent something that causes one drawer to be preferred over the other. When three 'gold'* coins are distributed among two drawers finding that you have picked up a 'gold' coin and not a silver one is a more likely experience if you reached into the drawer that had two 'gold' coins in it. Thus, when considering what the known distribution of coins is and what having picked up a specific coin tells you about where you probably got it the number of drawers that the coins are distributed among is misleading.

oscarzoalaster,

But the problem has one picking a coin from one of three drawers (2 gold; 2 silver; 1 gold/1 silver). The odds of picking a coin from the "1 gold/1 silver" drawer is 1/3 regardless of whether one picks a gold or silver coin. Knowing that it's a gold coin eliminates the "2 silver" drawer, but the odds of having picked from the "1 gold/1 silver" drawer is still 1/3. Therefore the odds of having picked the coin from the "2 gold" drawer must be 2/3.

I hope you will spend some time in the book on why so many people (myself included) have so much trouble "getting" the MHP. And I don't take this as primarily a mathematics/probability question. I'm not talking about people who don't "believe" it (that's the math, barring outright stupidity or pure stubbornness) but about those of us who accept and understand the math (and can even work it through) but who still don't understand why it's so. "Why" the problem doesn't "reset" to a "2-door problem" after Monty opens that door is a different question from "whether" it does.

By Jeff Chamberlain (not verified) on 07 Nov 2007 #permalink

I sorta think that the confusion is mostly caused by a perceived symmetry between your first choice and the new option, implying a toss-up as no new information seems to have been revealed about either door.

The wikipedia entry has a nice variant with 100 doors. It makes it clear that your first choice is as bad as it gets (1/100 chance), and it can't possibly be a toss-up between it and the remaining door.

As you are up to speed on the history, Jason, I'm willing to be corrected, but as I recall, Marilyn bollixed up the phrasing a bit in her Parade article. This left some ambiguity as to the correct interpretation.

Don't be short-changing your readers now, give them the full Monty.

By Tegumai Bopsul… (not verified) on 08 Nov 2007 #permalink

Sounds like a very interesting book.

By the way, there's a great treatment of the Monty Hall problem by the autistic narrator of the recent novel "The Curious Incident of the Dog in the Nighttime" (and it's a great quick read as well).

What's interesting is that everyone understands the correct answer if you use two doors:

Monty: Pick A or B.

You: I'll go with A.

Monty: A doesn't have the prize. Do you want to switch doors?

You: But then I'll automatically win!

Monty: Yup

Somehow, people understand the value of a new 1/N chance when N is 1, but not when N > 1...

By Clay Shirky (not verified) on 08 Nov 2007 #permalink

By the way, there's a great treatment of the Monty Hall problem by the autistic narrator of the recent novel "The Curious Incident of the Dog in the Nighttime" (and it's a great quick read as well).

Hey, I just read that book on my flight back from California. Well worth the read. There are also some other nice math problems in there.

John McKay-

Interesting suggestion.

Tyler and Science Pundit-

Thanks for the encouragement!

oscar-

You're right that it is crucial to assume that Monty chooses randomly from among the goat concealing doors when he has a choice. Marilyn didn't emphasize that specifically, but it was certainly implied and I don't think that was generally the main cause of confusion.

glen-

Basically, the first chapter will be a discussion of the history of the problem (and probability generally!) Chapter Two discusses the classical Monty Hall problem. Chapter Three discusses certain variants where you alter the assumptions about what Monty knows and how he behaves. Chapter four discusses variations with more than three doors. Several other chapters will discuss still further variants (say, versions with infinitely many doors, or versions with n doors where it is not certain that Monty will always give you the chance to switch.) I don't have these chapters fully mapped out yet in my own mind. Then a final chapter discussing other topics related to the MHP, such as the quantum mechanical version, or some of the psychological research into why people have so much trouble with this problem.

Peter-

I'm such a Luddite! It never even occurred to me to check their online archive.

Jeff-

There actually is a substantial literature on how people think about probability problems, and I do intend to discuss some of this in the book.

Kora-

The hundred door version is a helpful way of thinking about the problem, but even there you have to be careful. If you're playing Deal or No Deal, for example, and the only two cases that remain have a million dollars and one dollar respectively, it is fifty-fifty which of the two cases contains the million.

NJ-

The original correspondent phrased the question in a way that was a bit unclear, but I don't think this was really the cause of the confusion. Marilyn didn't make certain points as clearly as she might have in her initial reply, but that wasn't the main point made by her angry letter writers. She also got clearer with each subsequent column.

Tegumai -

Heh!

Derek-

I've read that novel! I started reading it because I had heard there was a treatment of the Monty Hall problem in it, and I kept reading because I got hooked on it after a few pages. It will definitely be mentioned in the book.

Clay-

That's a clever way of looking at it, but I'm not sure if it really proves very much!

As I've mentioned before, there is a classic issue in the (card) game of Bridge, known as the "theory of restricted choice", which I think has similar mathematical underpinnings. I read somewhere that when it was first proposed, many found it counter-intuitive and did not accept it. You might want to consider it for your book, as a related topic.

Kora: I've never understood why people think that the "100 door" variant helps avoid the counter-intuitive nature of the MHP. In the 3 door original version, the intuition (or mine, anyway) is that after Monty opens a door the problem resets to become a 2 door problem In the 100 door variant, the intuition (") is that the problem resets 98 times, but still ends up seeming like a 2 door problem. In either variant, the "money choice" occurs when there are only 2 closed doors to choose between. If you don't "get" the 3 door variant for this reason, there is little likelihood that you will "get" the 100 door variant, for this same reason.

By Jeff Chamberlain (not verified) on 08 Nov 2007 #permalink

However, I've found that most people do indeed understand that it makes sense to switch when there are a large number of doors (I usually say a million) and Monty opens all of them but one, because in this case it is easy to appreciate that the probability of the door initially picked being correct is negligible.

Jeff:

The problem doesn't reset per se; the first event is the placement of the prize with the obvious probabilities, and the second event is the host's singling out of one of the remaining doors. The second event depends on the first as if you didn't pick the prize door the host must single it out, otherwise he can pick any other door randomly.

99 out of 100 times you picked a goat in the first event, so the host has to single the prize out. 1 out of 100 times you picked the prize, so the host has to pick a random goat, but we don't care which goat. So, what the host does may appear random, but it's irrelevant to us.

You may also want to think about this variant: there are 99 prizes and 1 goat. You make your choice and the host opens 98 prize doors. You should stick with your original choice.

The Science Pundit said:

The odds of picking a coin from the "1 gold/1 silver" drawer is 1/3 regardless of whether one picks a gold or silver coin. Knowing that it's a gold coin eliminates the "2 silver" drawer, but the odds of having picked from the "1 gold/1 silver" drawer is still 1/3. Therefore the odds of having picked the coin from the "2 gold" drawer must be 2/3.

Many people will have difficulty understanding how the probability of having picked one of two drawers can be 1/3 or 2/3, so I think it could be helpful to shift focus at this point. Instead of thinking about the two drawers, think about the three gold coins. There are two gold coins in the "2 gold" drawer, and one gold coin in the "1 gold/1 silver" drawer. Therefore, you had two chances to pick the coin in the "2 gold" drawer (there's your 2/3) and one chance to pick the coin in the "1 gold/1 silver" drawer (there's your 1/3).

By mgarelick (not verified) on 11 Nov 2007 #permalink

I can't overemphasize how important it is to distinguish between the "Monty Hall Problem" as an abstract probability problem and the problem facing contestants as the game is actually played. I would venture to say that it is impossible to understand the latter without knowing (1) that Monty knows where the car is and (2) that Monty will not open the door with the car.*

Jason, you pointed out, in response to Oscar

it is crucial to assume that Monty chooses randomly from among the goat concealing doors when he has a choice

I think it's more important to point out that when Monty has a choice, and thus chooses randomly, he's not telling you anything, but when he has a choice, and thus is not choosing randomly, he is telling you everything. Your problem is that you don't know which situation Monty is in -- choice or no choice -- but you do know the chances for each situation. He will have a choice when you have picked the car -- one out of three times, in the long run. Two out of three times, when you haven't picked the car, he has no choice, he's telling you that the car is behind the door he hasn't picked, and you should switch. Bottom line: you're twice as likely to win if you switch.

-------------
*There is a third assumption, but I'm not sure whether it's valid: that Monty always does the "open a door/offer the switch option" routine. If he is more (or less) likely to do it when you have chosen the car, then obviously the problem is changed.

By mgarelick (not verified) on 11 Nov 2007 #permalink

Many people will have difficulty understanding how the probability of having picked one of two drawers can be 1/3 or 2/3, so I think it could be helpful to shift focus at this point. Instead of thinking about the two drawers, think about the three gold coins. There are two gold coins in the "2 gold" drawer, and one gold coin in the "1 gold/1 silver" drawer. Therefore, you had two chances to pick the coin in the "2 gold" drawer (there's your 2/3) and one chance to pick the coin in the "1 gold/1 silver" drawer (there's your 1/3).

mgarelick,

Let's look at the following variation: drawer #1 has 100 gold coins; drawer #2 has 1 gold/1 silver; drawer #3 has 100 silver coins; each drawer has a dispenser which gives you a random coin without revealing how many other coins are in that drawer. Again, you take a coin from a random drawer. The odds that it came from the 1 gold/1 silver drawer are 1/3, while the odds that it came from one of the drawers with 100 coins is 2/3. Now you look at the coin and see that it's gold. Are the odds that it came from the 100 gold coins drawer 2/3 or 100/101?

Same scenario except the distribution is: 1 gold coin; 1 gold/1 silver; 1 silver coin. Are the odds that it came from the 1 gold coin drawer 2/3 or 1/2?

I say that in both cases the odds are 2/3. The number of coins in each drawer is actually irrelevant.

Science Pundit,

I think that your reasoning may be something like this: before you know the color of the coin, the chances that it came from drawer #2 are 1/3, and the chances that it came from drawer #1 or #3 are 2/3. Once you see that the coin is gold, you know it didn't come from #3. Since the chances of #1 or #3 were 2/3, and you now know that the chances of #3 are zero, then the entire 2/3 goes to #1. Am I correct that this is basically what you're thinking?

I agree that once you know that the coin is gold, the chances that it came from drawer #3, in either scenario, are zero. But the problem becomes -- how can you justify including that drawer in the sample space? What does the denominator in "2/3" represent? What are there 3 of?

So, it starts to look more like 1/2; now that you know it's gold, it came from one of two drawers. But wait -- are both drawers equally likely? It would seem not, since the chance of getting gold out of #1 is 1, while the chance of getting gold out of #2 is 1/2.

(I would agree at this point that 100/101 is not on the table. Everything else aside, it just seems like it can't be right. So, in this scenario, I would agree that you don't want to look at the absolute number of coins, but rather at their proportion of the coins in the drawers. The implication of this is that your two scenarios are really the same, which I think was part of your point.)

You know, I think I might be over my head a bit, or at least tapped out on brain power that I can divert from my actual job. I think I'll stop now, and maybe pick it up a bit later, maybe after you or someone else has weighed in.

Note about terminology: I believe the "odds" that you quote are actually "chances." Odds are a ratio of "event yes" to "event no" (or vice versa, depending on whether you say "the odds of X" or "the odds against X"). Chances are a ratio of "event" to "all possible events". So, in your penultimate sentence, the chances are 2/3, but the odds are 2:1.

By mgarelick (not verified) on 12 Nov 2007 #permalink