Nickerz, I was disappointed in Cecil’s answer. The point of the problem is not simply to get a “correct” answer, but to to get some insight into probabiolity theory.
How would you answer this problem (which I believe is roughly equivalent):
You are offered an opportunity to play a gambling game where the amount you could win is twice the amount you could lose. You do not know what the odds are. Should you play?
I think Cecil’s answer was complete and correct. A mark of true genius is the ability to sniff out a red herring, and act like you didn’t smell anything. The denomination of the gold coins has nothing to do with their value. Their value is set by the mass of the coins. Even if these coins were minted during the gold standard, it seems highly likely that a full box of gold disks would be more massive than a half full box of larger disks.
The denomination of the coins has everything to do with their value. Their value is not necessarily set by the mass of gold. Their “real” value is set by whichever is greater - denomination, or gold value by mass. That is, if a $20 coin contains 1 oz of gold, and gold is selling for $25/oz (silly numbers, obviously), then the coin is to me a lump of gold worth $25, and I’ll treat it as such. But if gold is selling for $15/oz, then the coin is just a coin, the material it’s made of is irrelevent, and it’s worth $20 to me.
Cecil’s answer assumes that at least one of the following is true: (1) Only the volume of gold - not the denomination of the coins - matters. In that case, a full box of gold is always worth less than half a box of gold. Or: (2) $20 gold pieces occupy more volume per coin than do $10 gold pieces. In that case, even if it’s the denomination (rather than the amount of gold) that matters, there are at LEAST twice as many 10’s in a full box as there are 20’s in half a box. I believe this is what he means by “assumptions about what constitutes reasonable behavior on the part of the mint”.
Assumption (1) is certainly plausible, and (2) is pretty likely - but only because we happen to have “outside information” that that’s the way mints typically make GOLD coins. However, they certainly don’t make ALL coins that way. For example, I’d much rather have half a box of dimes than a full box of nickles (and that has nothing to do with the materials they’re made of, which are of trivial worth for both).
Thus, Cecil’s assumptions, though certainly not unreasonable, firmly remove the question from the realm of logic problems, and place it into the rather different realm of theorizing what decisions have been made by a particular group of people (the coin designers at the unidentified mint) based on our experience with similar decisions made by similar groups (other mints we’re familer with). If we’re to rely merely on logic, and not our sense of what a mint is likely to do, then the answer to the question is “not enough information given”
The general point Cecil was addressing seems to how one should answer an inadequatly-defined question. He says, “…Which box would you choose, given the information that you have? One has to make certain assumptions…”
When confronted with a vague question, Cecil’s approach is to make assumptions that permit an answer. Not only is Cecil an expert on every topic, but he is an expert’s expert on the topic of how to answer questions.
However, there are situations where the greater wisdom may lie in recognizing that the facts provided really offer no answer. Arbitrary choices would offer trivial examples. E.g., a hotel has just two vacant rooms available, one on the 15th floor and one on the 14th. Which should you choose? Or, suppose one of the rooms was on 13? It’s easy to say you shouldn’t avoid the room on 13 because of “bad luck”, but knowing only the floors is not enough information to prefer one room to the other.
There are real-world questions that must be fleshed out with your beliefs or assumptions. E.g., is a National Health plan a good idea? The answer depends on how you assume the plan will be constructed, how well you assume it will work, what you believe the alternative would be, etc.
BTW note that Cecil is better served by adding assumptions that lead to an interesting discussion. It would be boring if his column frequently consisted of some question followed by a response of, “That question includes too little information to be answered.” (By the way, would this be an example of a “stupid question?”)
I’m a bit nervous about December’s use of Baynesian probability theory on the two envelope problem.
I leave aside ‘game show’ circumstances (where you have some extra information).
I state that by opening an envelope, you don’t learn anything, so the probabilities don’t change. It’s 50-50 that you’ll gain or lose the SAME AMOUNT by switching.
If you find $8 in the first envelope you calculate that you can get either $4 or $16, and switch. But now imagine you decide to switch first, then open the envelope. You find $8, do the same calculation and swop back to the first envelope.
What this means is that your calculation is wrong, not that it’s worth switching!
Try this. At some point you open an envelope and find $8. Now the envelope stuffer either (50%) put in a total of $12 (8+4) or (50%) put in $24 (16+8). In the first case, you would win ON AVERAGE $6 over several plays, so swopping or not swopping gains or loses $2 (8-6=+2, 4-6=-2).
In the second case, you would win ON AVERAGE $12 over several plays, so swopping or not leads to a $4 profit or loss.
Not to beat a dead horse (whap!), but there is only one chance here and no information on the motives of the stuffers. Here is my point of view: I would take the swap because, besides always wondering about the other envelope, I would loose much less then I would win. Half as much in fact For example, if my envlope had $4 and I traded and got $2, I lost $2. But if I traded and got $8 I gained $4. $2 loss – $4 win. My expected win for trading is greater then my expected loss. In other words, you would be just as likely to lose as win but winning would be sweeter then loosing bitter.
Of course I could just kick the stuffer in the shins and take both envelopes. Serves them right for provoking me so.
dan
Danno, you’re fine about all the side effects - you only have one chance; you might get annoyed with the envelope stuffer etc.
But I think your maths is wrong. Just because you know how much is in the envelope doesn’t mean you should switch! You don’t ‘win’ more than you would ‘lose’. Please see my earlier posting for my analysis.
Thanks glee, but I’m not sure you understand what I’m getting at. You would not win more times then you would lose, winning is just more rewarding then losing is punishing. Lets go over my math, point out any mistakes that you see please:
Envelope that you pick has $8
If you win you get $16
16 - 8 = 8
so $8 is your expected win
If you lose you get $4
8 - 4 = 4
so $4 is your expected loss.
8 > 4
expected win > expected loss
Therefor take the second envelope.
Danno,
I agree winning is more fun than losing, but I don’t agree that you make money by switching (though it’s tough to prove it!).
Imagine you and a friend both have envelopes. One has twice the amount of the other one. You both secretly look in your envelope. You both decide to switch! One of you loses, one of you gains - and it’s the SAME amount of money.
What this means is that your calculations are wrong, not that it’s worth switching!
Try this maths. At some point you open an envelope and find $8. Now the envelope stuffer either (50%) put in a total of $12 (8+4) or (50%) put in $24 (16+8). In the first case, you would win ON AVERAGE $6 over several plays, so swopping or not swopping gains or loses $2 (8-6=+2, 4-6=-2).
In the second case, you would win ON AVERAGE $12 over several plays, so swopping or not leads to a $4 profit or loss.
Hope this satisfies you - it’s a great problem.
Danno,
I agree that winning is more fun than losing, but not that you will win more than when you lose.
Try this:
You and your mate both have an envelope. One contains twice as much money as the other. You both secretly peek. Following your logic, both of you swap! Now one loses and one gains and it’s the SAME amount.
You need to consider the average over several trials to get the correct odds. Otherwise it’s like tossing a coin once and saying it always comes up heads.
See my earlier posting at the end of page 3 for a more maths.
Sorry about the near double posting - my browser denied the first message had got thru!
Glee, you’re right that the amount of money that could be gained by switching is twice as much as the amont it could lose.
Danno, you’re right that if one were to play this game repeatedly, always switching would have the same expected value as never switching.
This apparent paradox can be resolved by focusing on the probabilities. (Bear in mind that Expected Value is the based on amounts as well as the probability of achieving each amount.)
When you see $8, you know that the original envelopes contained a total of $12 or $24. The problem doesn’t tell you the probabilities of these two events. It may be reasonable in this case to assume that it’s 50-50. However, that does NOT mean that you can ALWAYS assume that the probabilities are 50-50, regardless of the amount of money you see. Here are three reasons why not:
Suppose amount of money in the opened envelope was an odd number. Then, you know that you have opened the smaller of the two envelopes.
Suppose the amount of money in the opened envelope is very large. Then it may be more likely that you opened the larger of the two envelopes.
If one assumes that the probability is always 50-50, then a mathematical contradiction occurs. The contradiction is precisely that you would both simultaneously be right. That is, you could prove that switching doesn’t help and also that it does help.
Glee, you’re right that the amount of money that could be gained by switching is twice as much as the amont it could lose.
Danno, you’re right that if one were to play this game repeatedly, always switching would have the same expected value as never switching.
This apparent paradox can be resolved by focusing on the probabilities. (Bear in mind that Expected Value is the based on amounts as well as the probability of achieving each amount.)
When you see $8, you know that the original envelopes contained a total of $12 or $24. The problem doesn’t tell you the probabilities of these two events. It may be reasonable in this case to assume that it’s 50-50. However, that does NOT mean that you can ALWAYS assume that the probabilities are 50-50, regardless of the amount of money you see. Here are three reasons why not:
Suppose amount of money in the opened envelope was an odd number. Then, you know that you have opened the smaller of the two envelopes.
Suppose the amount of money in the opened envelope is very large. Then it may be more likely that you opened the larger of the two envelopes.
If one assumes that the probability is always 50-50, then a mathematical contradiction occurs. The contradiction is precisely that you would both simultaneously be right. That is, you could prove that switching doesn’t help and also that it does help.
December, I think you’ve got Danno and my positions reversed.
In any case I’d like to concentrate on the pure problem (I agree it makes a difference if you’re on a game show etc.)
I say it doesn’t make any difference to either your chances of winning or the amount you win / lose whether you open the envelope or not!
Your chances of winning on any one play are 50%. The amount you win or lose in the long run is $0.
The maths is precise (it always is) - the apparent contradictions come from either a loose statement of the problem or a misunderstanding of the maths.
(Please don’t be offended - I myself find this stuff both difficult and fascinating).
Try thinking about it like this. You hold one envelope. Another guy holds the other one. (He doesn’t do anything - you make all the choices).
You peek in your envelope and find $8. You decide to swap. Suppose the other guy has $16. You ‘win’ $8. But the other guy ‘loses’ $8. The game balances.
Now imagine you’re the one with $16. You swap (according to your theory). Now you lose $8 and the nonentity wins $8. The game is still balanced.
Perhaps you have chosen bad examples, because looking in an envelope and saying you’ll always swap leads nowhere.
You guys are unbelievable. CKDextHavn correctly explained this problem in May. It’s August now. All this baloney about how much you’d “win” by switching is irrelevant. Go outside and play.
I am honoured that the man himself has answered my posting (even if was only to tell me I’m months late!)
I realise the right answer appeared in May, but December doesn’t agree, so I thought I’d try my hand at it.
Also I’m relatively new at this Internet stuff. On another board, because I was also a recent poster, most people assumed I was just an alias of another user…
P.S. I really like your column - there’s far too much waffle and pseudo-science elsewhere on the Internet.
I don’t know what all the answers given here were to the problem tubby originally posted here, but given nothing in his statement of the problem to make one believe the original choice of envelope contents was not 50/50 as to whatever their differing contents might be – I say that, whether or not you look at the contents of the first envelope (this is not quantum mechanics), you have no reason to believe there should be any preference, probabilitiwise, as to which envelope you should accept, period.
Ray
Cecil, my rapture that you’re reading my posts is tempered by your endorsement of CKDextHavn’s probabilistic naiveté.
CK started well on 5/18 by assuming that the two envelopes contain X and 2X. The contestant has two unknowns. He doesn’t know whether he opened the larger or smaller envelope, and he doesn’t know the value of X. A model is needed that reflects both unknown quantities. However, CK’s 50-50 probability assumption focused only on whether the contestant has the larger or the smaller envelope. This limited model was insufficient to resolve the contradiction, so CK made up an excuse:
“The seeming paradox arises because you are applying the expected value to a single trial…you have to think of it as the expected results after many, many trials.”
(See my 5/26 post for an explanation of the paradox and its proper resolution.)
On May 19, CK again started well by saying:
“The problem is that your perception of the probabilities and the actual probabilities are different.”
By “Your perception of the probabilities” CK was referring to the contestant’s subjective probability, I assume. This is good.
However, there are no “actual probabilities.”
Probability is not an objective physical quantity like mass or temperature. Probability and expected value are defined relative to a particular model or set of assumptions. (An exception might be a special case where symmetry makes the probabilities clear, like dice. Even here, one makes the assumption that the dice are not loaded.)
In the two envelope problem, subjective probability is the only type of probability available.
CK later says:
“Now, suppose the contestant is allowed to open the envelope and switch if he wants. The fact is that the contestant gets no new information by having opened the first envelope.”
Not necessarily. The contestant may gain information by opening the envelope. E.g., if the envelope contains a very large amount of money, the contestant is likely to have chosen the larger one. Or, if the amount of money is an odd number, then the contestant certainly has opened the smaller envelope.
Later, CK says:
"This is an erroneous calculation, because the chance of $2 is in fact zero, but the contestant doesn’t know that. "
In a trivial sense, there was never any randomness, since the donor had already decided how much to put in the envelopes,
and the contestant had already chosen one. However, both of these events are unknown to the contestant, so he can use probabilities to discuss and analyze them.
BTW, I am a former Chairman of the Examination Committee of an Actuarial Society, so my posts on this topic ought to have some validity.
December
December, please don’t take this badly, but you’re wrong!
Probability is the % chance of somehting happening. It’s derived mathematically. In the case of something straightforward like coin-tossing or the dice game craps, you can work it out precisely. There are more complicated cases, such as predicting the result of something from a sample, where you express the chances to a level of confidence.
There’s no mathematical concept called ‘subjective probability’. You can watch gamblers saying ‘Red’s come up on the Roulette wheel 5 times - it must come up Black next spin’. Ignoring zeroes on the Wheel, the chances are still 50%, but the gamblers don’t understand that. This mistaken belief could certainly be called ‘subjective probability’, but it’s nothing to do with maths!
This envelope game turns out to be quite simple in probability terms, but is very easy to misunderstand initially.
I already agreed that if you’re on a game show, there might be reason to change the probability. But with two envelopes containing x and 2x (OK, I also state x is not odd!), then there is exactly a 50% chance of getting the larger amount and 50% of the smaller. Looking in an envelope doesn’t change these probabilities at all.
That’s it, honest!
P.S. You mention your post of 26/5. It states:
“1. The assumption that switching is always 50-50 leads to a contradiction. It would imply that one would get a better result by always choosing to switch envelopes.”
Switching is always 50-50; you can gain (2x-x) = x, or lose (x-2x) = x. The ‘contradiction’ is just in your thinking. WHY does this imply you get a better result by switching?
“2. The reason I said that this problem could not be tested is the same reason. In order to set up a test, you would need to know (or assume) what procedure the donor used to decide how much money to put into the envelopes originally.”
OK, he used a RANDOM procedure. This means that your chances of having either x or 2x are equally likely i.e. 50%. This is mathematically acceptable (honest!).
Finally your last post said:
"The problem is that your perception of the probabilities and the actual probabilities are different.
By ‘Your perception of the probabilities’ CK was referring to the contestant’s subjective probability, I assume. This is good."
I’m sure CK can defend himself, but while I’m here let me say I’m 100% sure that CK means YOU, December, have a perception that is different from the actual one. Not good, I’m afraid.
Finally although you held a distinguished post, it doesn’t means you’re right. Even Cecil can make a mistake (am I allowed to say that?!). The key to this is reading the arguments carefully.
Hey, December, I have a Ph.D. in math and I am an actuary (although I never wanted to be on one of those exam committees) so let’s not bandy credentials, eh?
The major point is one of perspective. And I am excluding the idea that you can tell by size (“If you open the envelope and it has a zillion dollars, you know the other is less”)… that’s kind of like assuming that the coin falls on its edge. The problem assumes that the contestant doesn’t know the rules for determining the amounts in the envelope, and that opening one envelope won’t give him a clue about the other.
From the perspective of the person who has set up the envelopes, there is X in one and 2X in another, and the expected value is 1.5X. This is pretty obvious, and this is a reflection of reality as well.
The contestant however doesn’t know this. He is forced to approximate probabilities, based on insufficient information. He opens an envelope; let’s say it contains X (he has the smaller amount, but he doesn’t know that.) He knows that the other envelope could contain 2X or could contain 1/2X. He estimates that his expected value is 50% (2X) + 50%(1/2 X) = 1.25X. This is incorrect, of course, but it’s the best guess he can make with the information he has.
It’s incorrect because the 50% is not correct. There is no play of the game that will give him 1/2 X.
The same logic applies if he has originally chosen the 2X (the larger envelope.) He estimates his expected value as 50% (X) + 50% (4X) = 2.5 X (which is actually the same estimate, it’s still 1.25 times the contents of his envelope.)
Now imagine a large number of players, all facing X and 2X (but they don’t know that). Whether a player switches or not, the group of players will (on average) come out with 1.5 X, the median amount. They will have calculated an expected value of 1.25 X but that is (as noted) based on insufficient information.
It seems to me that the sensible analysis before picking an envelope is for the contestant to note that he has a 50% chance of winning (getting the high amount.) Opening one envelope does not alter those odds. The calculations of 1.25X are bogus.
… but what are the odds of two actuaries being on this Message Board?
[Note: This message has been edited by CKDextHavn]