Posted on 07/19/2010 6:11:38 AM PDT by mattstat
Another probability “paradox”, the two-envelope problem1, goes like this:
Before you are two envelopes, A and B. One of them contains $X and the other $2X (which is equivalent to $Y/2 and $Y). You pick one envelope and are (1) asked if you would like to keep it or switch, or (2) open it, view its contents, and then asked if you would like to keep it or switch. Which strategy, keeping or switching, is likeliest to win you the big bucks?
No peeking solution
The traditional paradoxical solution to (1) is to argue this. Suppose you pick A, which can be said to have $X (recall you do not peek). There is then a 50% chance that B contains 2X (dropping the dollar signs, but understanding they are still there, lurking) and a 50% chance that B contains X/2. The “expected value“2 of B is said to be
0.5 * 2X + 0.5 * X/2 = 5/4 X.
So clearly you should switch, since the expected value of the envelope you did not pick is larger than X, which is the value of the envelope you hold.
But that’s nuts, because you could have picked up B, in which case the expected value of A is 5/4 X, too. Oh, what to do!...
(Excerpt) Read more at wmbriggs.com ...
The answer is the same.
He's right in that you SHOULD switch. The expected value of 5/4X is not illogical.
Yes it is. The probability is falsely stated. The envelopes contain X and X/2, not 2X and X/2)
Absent further information the "expected value" of either envelope is
0.5*X + 0.5*X/2 = 0.75X
That value is not affected whether the envelope is in your hand or not.
Best option is to peek and switch of the contents are odd - that is using additional information to make the choice.
Pull out your gun and take both envelopes.
If P(A) = P(B) = 0.5 and A contains M(A) and B contains M(B), and you keep picking envelopes, your average gain would be:
M = M(A)*P(A) + M(B)*P(B)
Substituting with riddle's values, we get that you draw on average: 0.5*200 + 0.5*50 = $125.
About switching of an unopened envelope. Envelopes A and B are indistinguishable in every sense. There is no reason to pick A and switch to B because you could just as easily have picked B and switched to A. It can be easily shown by switching 'A' and 'B' in the formula above:
M = M(B)*P(B) + M(A)*P(A)
The answer would be the same.
There are other ways to prove this. For example, you pick A and consider switching to B:
Probability of gaining money = 0.5, average gain = 0.5*(200-50) = $75
Probability of losing money = 0.5, average gain = 0.5 * (50-200) = -$75
To get the total effect of the switch you add those two choices and get exactly $0. The switch has no effect.
You must be from Chicago or New Orleans huh ;^)
The difficulty with these calculations is that they are for the “average” solution. Which, of course, only exists in the “long run.” We want the optimal solution for playing just once. See the full post.
No one is going to offer me free money because I’m a middle aged white guy.
Not exactly. The average is only the limit of the monetary gain function. The second method that I mentioned specifically deals with playing just once - it proves that for any M(A) and M(B) your likely gain is equal to your likely loss.
Specifically, you made your first pick and are holding an envelope in your hand. Your complete set of choices is (a) keep the envelope or (b) change it. To determine your most beneficial action you use the probability theory.
As I demonstrated earlier, each of those choices offers you a certain gain (positive or negative.) Such a gain would indeed converge to $150 (if $100 and $200 are in play) only if you do infinitely many tries. But we aren't calculating how much you'd win in a billion years! We only want to know what choice is most beneficial to you in each and every round of the game. Therefore it's not necessary to try one billion times, because advantages of one choice (if there are any) are present in every round of the game. For example, if you are playing Russian Roulette and given a choice of playing with a dummy round or a live round, the advantages of a dummy round are present each time you pull the trigger.
To illustrate this particular case: If you keep the envelope you have 50% chance that you get $100 and 50% that you get $200. If you change the envelope ... you have the same 50% chance that you get $100 and 50% that you get $200 (because the envelopes are identical and you haven't learned a thing from the fact that you held one in your hand.)
Since both options result in identical outcomes, the decision is irrelevant. Change or keep, you won't be any better off with any of those decisions. They are merely chairs on the deck of Titanic, they are of use only in a TV game, when the host wants to create an illusion of choice :-)
You can also think of it this way. Why do you say "you made your choice" when someone picks an envelope? Extend this phase; let the player to pick an envelope, then put it back (without looking), then pick another one ... for as long as you want. Once you are tired of picking you open what you have and get whatever money is inside. This way of seeing the problem eliminates the choice of "picking a different one"; it is truly a meaningless act in this scenario. Though if you wanted you could come up with a game that depends on such a choice - if, for example, each selection beyond the first one costs you some money.
Such riddles are easily solved by using standard mathematical methods, but without such a formal approach the solution may be not obvious. This is because humans depend a lot on their "gut feelings", small acts that are random and unjustified just because we aren't a donkey stuck between equally appealing food items.
This reminds me of a short story ("Chess at the bottom of the well" by Alexander Kazantsev) that I read recently. An Egyptologist discovers a well with a mathematical riddle. He has no clue how to measure the diameter of the well with just two dry reeds, but ancient candidates to priesthood had to do that to become priests - or to die trying.
His friend, a well educated guy, shows up, gets interested in the riddle and quickly solves it. He uses a quartic function to find the answer, and he is puzzled because these were not known in ancient Egypt. He pays no attention to this, though.
A parallel story line introduces a young candidate who is led into the well, the door sealed, and he is asked to find the solution or to die. He finds the solution by using a completely different approach.
This story is of interest here because modern math (of which what we used here belongs somewhere in sixteenth or seventeenth century) is a powerful set of tools that allow you to solve problems without trying to comprehend them. Most modern problems can't even be understood by a human mind, since they deal with infinities. Human mind is not a good tool to "guess" even a simplest surface integral, though a minimal toolset of calculus lets you do it just for fun :-)
Well, my dad was from Chicago, and I grew up in New Jersey. I guess it shows!
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