Envelope Problem

[London Colin]

I would like to point out that there is a variant of this problem for which the switching argument is correct and it is exactly analogous to Meistros altered roulette game.

I feel compelled to point out that I have already pointed this out! (Twice, in fact!) - Envelope Problem

I'm in the process of concocting one last post that, in part, highlights the differences between these two games.

 

[gronbog]

So you did. Well done!

 

[London Colin]

I've been reading another of the references - Bruss, F. Thomas (1996). "The fallacy of the two-envelope problem". Mathematical Scientist. 21 (2): 112–119. There's no link provided, but I tracked it down (in the form of a scan) to - https://www.researchgate.net/publication/266706577_The_fallacy_of_the_two_envelopes_problem

It's helpful stuff, and below is my attempt to rework elements of it into a useful overview, and apply it to our specific example (100,?).

-------------------------------------------------

First, let's look at a game that does indeed give the infamous 5/4 expectation from switching envelopes -

There are two envelopes, labelled A and B.
An unknown amount, call it U, is placed in envelope A.
Based on the toss of a fair coin, one of two possible amounts (2U and U/2) is placed in envelope B.

The player is given envelope A and the option to switch. The expected value of envelope B is:

E(B) = 0.5(2U + U/2) = 1.25U.

So, in our example, A is revealed to contain 100 (U=100), and E(B) = 125.
Switching from A to B gives an average gain of +25.

-------------------------------------------------

But now let's look at the actual two-envelope problem -

There are two amounts, S (for 'smaller') and 2S (twice as large).
They are placed inside identical envelopes, one of which is randomly given to the player.
Let's call the unknown amount in that envelope U.
There is a 50% chance that U=S and a 50% chance that U=2S. -
P(U=S) = P(U=2S) = 0.5

The expected value of the randomly chosen envelope the player currently holds is -
E(U) = 0.5(S+2S) = 1.5S

The expected value of switching to the other envelope (replacing U with S half the time, and with 2S the other half the time) is -

E(switch) = 0.5(S-U) + 0.5(2S-U) = 1.5S - U

We've seen that the expected value of U, E(U), is 1.5S. So E(E(switch)) = (1.5S - 1.5S) = 0. [The expected value of the expected value of switching is zero.]

We also, of course, knew from the beginning that the expected value of both envelopes was 1.5S, there being nothing to distinguish between them. But that's beside the point.


In our example, the player has opened the envelope to reveal that U=100. Half the time this will be the smaller amount (U=S=100), half the time it will be the larger amount (U=2S, U=100, S=50).

At this point, there is an opportunity to make the well-known blunder all over again. It's tempting to take the bottom-line formula, 1.5S - U, and evaluate it when U=100, like so -

When U=100,S=50, contribution to E(switch) = (1.5S - 2S) = 75 - 100 = -25
When U=S=100, contribution to E(switch) = (1.5S - S) = 150 - 100 = +50
Summing these gives an overall EV of +25. D'oh!

We have once again blended the two distinct scenarios. We should instead say -

E(switch) = 1.5S - U
When U=2S, contribution to E(switch) = 1.5S - 2S = -0.5S
When U=S, contribution to E(switch) = 1.5S - S = +0.5S
So E(switch) = 0, regardless of the value of S.

-------------------------------------------------

 

[gronbog]

Nicely done!

 

[sagefr0g]

Sagefrog, I would not worry too much about running across something, like this in real life. These kinds of situations are carefully crafted and are unlikely to come up by accident. You would probably smell something fishy. Having said that, the Monty Hall problem on Norm's forum is a real life example of another non intuitive result. However, it could be argued that the situation there is far from typical.

as confirmation bias would have it, it seems I've stumbled upon a similar problem, not a bet, but the computation of an average bet. it did indeed smell fishy, as a math check refused to confirm the validity of an average I tried to compute (in two different ways [one of which must be an incorrect procedure], the correct procedure I believe i'm cognizant of, the incorrect procedure, not so much), I believe the confliction has to do with the issue of dependency and independency, but really I fail to fully understand the issue..
anyway the paradox seems similar to the two envelope problem, as it's likely a result of averaging numbers that shouldn't be averaged, sorta thing.
later, i'll try and post an example (can't do it now, as the joints are calling and time is of the essence).
hopefully someone can point out the error of my ways.

 

[gronbog]

When you do post about it, a new thread would probably be best, so it can be more easily found for future reference.

 

[sagefr0g]

When you do post about it, a new thread would probably be best, so it can be more easily found for future reference.

i posted the conundrum at this link: https://www.blackjackinfo.com/community/threads/average-bet-conondrum.55861/

 

[Gramazeka]

got to say, i'm thankful to Meistro for posting the two envelope problem. i'd never been aware of it before.
doubtless it's a vicious, malicious monstrous scenario, lol.
just me maybe, even if one knows the proper solution, it's easy to get mired in the wrong approach and its darn near impossible to convince one's self that it's wrong and then even when you realize the solution it's very difficult to understand how and why it's the correct way.
what really kind of scares me about a problem such as this, is the question of if and when one might run into a similar problem in the real world and just how difficult it would be to recognize it for what it is.

This is my thread ))-

https://www.blackjacktheforum.com/showthread.php?40269-Probability-theory

 

[Gramazeka]

As I wrote earlier, we have given two boxes, one X and the other 2X. THIS IS THE TERMS OF THE TASK !!! We opened one casket. In our problem, either X or 2X is always opened, and we do not know what exactly was opened - but it is NOT a fault to come up with some new values ??of sums not specified in the condition of the problem. Then, if this box X, then with a probability of 100% we will be given a replacement for the box 2X. If this box is 2X, then with a probability of 100% we will be given a replacement box for X.

But we CAN NOT say that we will ever be given a casket with 1 / 2X, as there is NO such box at all !!!

If someone wants to insist that after taking a box with X money we can open a casket with 1 / 2X money, then this someone MUST change the conditions of the task and say that we have two boxes X and 1/2. And then the casket with 2X money will cease to exist, which again shows the error in the solution selected above.

In other words, we do not have the right to name the amount of money in the chosen box as X and from this amount to determine 2 possible value propositions in the second box. The given conditions of our specific task DO NOT ALLOW us to do so - this is a mistake. The trick is that we really get 50/50 or more, in relation to the amount in the opened box, or a smaller amount, but these 50/50 should not be substituted in the formula for calculating the expectation of this action (the choice of the second box), because by choosing the first We are already bogged down in the given conditions of the problem! We do not have boxes with 1 / 2X, since the probabilities are dependent on us.

 

[Dummy]

As I wrote earlier, we have given two boxes, one X and the other 2X. THIS IS THE TERMS OF THE TASK !!! We opened one casket. In our problem, either X or 2X is always opened, and we do not know what exactly was opened - but it is NOT a fault to come up with some new values ??of sums not specified in the condition of the problem. Then, if this box X, then with a probability of 100% we will be given a replacement for the box 2X. If this box is 2X, then with a probability of 100% we will be given a replacement box for X.

But we CAN NOT say that we will ever be given a casket with 1 / 2X, as there is NO such box at all !!!

If someone wants to insist that after taking a box with X money we can open a casket with 1 / 2X money, then this someone MUST change the conditions of the task and say that we have two boxes X and 1/2. And then the casket with 2X money will cease to exist, which again shows the error in the solution selected above.

In other words, we do not have the right to name the amount of money in the chosen box as X and from this amount to determine 2 possible value propositions in the second box. The given conditions of our specific task DO NOT ALLOW us to do so - this is a mistake. The trick is that we really get 50/50 or more, in relation to the amount in the opened box, or a smaller amount, but these 50/50 should not be substituted in the formula for calculating the expectation of this action (the choice of the second box), because by choosing the first We are already bogged down in the given conditions of the problem! We do not have boxes with 1 / 2X, since the probabilities are dependent on us.

As pointed out, it all depends on how the problem specifics are presented making the logic is simple. You either understand the problem presented correctly and switching has no effect on your EV. Or you don't understand the specifics right and switching is plus EV. Given a choice of not switching which has no effect or not switching costs you EV for one set of specifics for the problem, and switching has no effect or it gains EV for the other specific problem there is one logical choice. You do not loss EV in either specific for the problem by switching. If you don't switch you either gain nothing or lose EV for each specific. Obviously the safe choice is to switch. You either lose no EV or gain EV depending on the problems specifics.

 

[Gramazeka]

The funny thing is that the very first action (when we take any box of two, one of which is X money, and the other - 2X) ALREADY has EV = 1.5X !!! Therefore, it should not be surprising that the EV decision to replace the box has the SAME 1.5X, that is, we take the contents of the first box or change it to the second one - the EV equally for that and for another case. That is the answer - it makes no difference whether or not to change.

 

[Dummy]

I read that in the thread. But that doesn't change common sense. I can't tell you how often the well educated are shown up by the ignorant person with lots of common sense. Why take the chance that you misunderstood the problem?

One choice, not switching, never gains EV but loses EV if you misunderstood the problem's specifics. The other choice, switching, never costs EV but if you misunderstood the problem's specifics you gain EV. Who would argue that not switching is the better of the two possibilities, interpreting the problem correctly or having misunderstood the problem. It doesn't matter which envelope you pick in the former. The latter the unopened envelope increases your EV. Sometimes people are just too smart to use their common sense.