Tilted Forum Project Discussion Community - View Single Post - Another of those crazy paradoxes (similar to Monty Hall problem)

rsl12 · 12-11-2003, 06:15 PM

knifemissile, thanks for your reply. I feel i will have to argue with you, however.

and what i will argue is one single point: that the two games, (the original game and your game of being shown a box and then given a choice to change for double or 1/2 the amount) are NOT the same.

Let's start with the original game, and then modify it in mid-game to make it exactly like the 2nd game. Cup A contains $1, cup B contains $2. You can look at only one of them. There are 2 possibilities.

A. You look at cup A. Now, we can modify the game to be the same as yours by flipping a coin and deciding whether to leave the $2 in cup B or replace it with $0.50. Now you know the expected value in cup B is $1.25, and of course you should pick cup B.

B You look at cup B. Now we modify cup A by flipping a coin and deciding to leave $1 in it or change it to $4. The expected value is $2.50, and of course you should pick cup A.

There is a difference!!! in these two cases, it is OBVIOUS that switching gives you the higher return. But if you had left the game alone as it was without looking inside either cup, then it is obvious it should make NO DIFFERENCE which cup you choose--you have a 50/50 chance no matter what. And the fact that you get to look inside one cup does not, in any way, change your ability to get the larger amount of money. I'll show you--there are 2 possibilities:

A. You look under cup A, and see there is $1. Then you switch to cup B. You get $2.

B. You look under cup B, and see there is $2. You switch to cup A, and get $1.

And if you don't bother switching,

A. You look under cup A, and get $1.

B. You look under cup B, and get $2.

In both cases, your expected winnings are $1.50, so switching made absolutely no difference! Common sense should tell you this is the right way to look at the problem, and that the other way is wrong.

12-11-2003, 06:15 PM	#8 (permalink)
rsl12 On the lam Location: northern va	knifemissile, thanks for your reply. I feel i will have to argue with you, however. and what i will argue is one single point: that the two games, (the original game and your game of being shown a box and then given a choice to change for double or 1/2 the amount) are NOT the same. Let's start with the original game, and then modify it in mid-game to make it exactly like the 2nd game. Cup A contains $1, cup B contains $2. You can look at only one of them. There are 2 possibilities. A. You look at cup A. Now, we can modify the game to be the same as yours by flipping a coin and deciding whether to leave the $2 in cup B or replace it with $0.50. Now you know the expected value in cup B is $1.25, and of course you should pick cup B. B You look at cup B. Now we modify cup A by flipping a coin and deciding to leave $1 in it or change it to $4. The expected value is $2.50, and of course you should pick cup A. There is a difference!!! in these two cases, it is OBVIOUS that switching gives you the higher return. But if you had left the game alone as it was without looking inside either cup, then it is obvious it should make NO DIFFERENCE which cup you choose--you have a 50/50 chance no matter what. And the fact that you get to look inside one cup does not, in any way, change your ability to get the larger amount of money. I'll show you--there are 2 possibilities: A. You look under cup A, and see there is $1. Then you switch to cup B. You get $2. B. You look under cup B, and see there is $2. You switch to cup A, and get $1. And if you don't bother switching, A. You look under cup A, and get $1. B. You look under cup B, and get $2. In both cases, your expected winnings are $1.50, so switching made absolutely no difference! Common sense should tell you this is the right way to look at the problem, and that the other way is wrong. __________________ oh baby oh baby, i like gravy. Last edited by rsl12; 12-11-2003 at 06:31 PM..