One of the most elusive problem of 20th century is the Two-Envelope problem(even 21st, yeah, it remains unsolved). Well if you already know about the problem and are about to drop the idea of going any further, hang on. This article is not about the problem as per se but about it’s quite unknown derivative. Although being from a knotty background, it is less complex and has an astounding significance.Be with me till the end and you would not regret. Let me serve it first :

The game contains a pair of identical envelopes, a game-master and a player. The envelopes contain certain Integer amounts such that one envelope holds double of what the other hold; the game-master knowing everything.For example, if one envelope contains A then the other contains 2A, where A is a positive Integer. The player has to choose one envelope, look inside it and then has two choices, either to swap or keep the one he has in his hand. Isn’t this simple ? Now let’s say that the envelopes had sums of 2 and 4 rupees. If the player chooses the envelope with 4 rupees then the other envelope may contain 2 or 8 rupees. There is an equal chance of winning or loosing. But this is monotonous and boring, let’s look at the problem from a more interesting perspective, the game-master’s perspective (Here comes the bit which will blow you ).

Imagine you’re the game-master and you decide to choose the value of A to set up the game. Lets take A=1, the two envelopes will contain 1 and 2 rupees. Now imagine the player selecting the envelope having 1 rupee in it. As soon as he opens it he will know that the other envelope has 2 rupees, why? Because the other envelope can’t have 0.5 rupees (it’s not an integer). A=1 ruled out; and for similar reason any A=odd number is also ruled out because the division by 2 will not result in an Integer. Okay, maybe one of the few games where the game-master has to be so meticulous. Let us take A=2. Very promising indeed but there is a problem. The player also knows that A is neither 1 or any odd number (even he can think like the game-master) . So as soon as he finds the envelope having 2 rupees he knows that the other envelope can’t have 1 rupee. This again results in a biased game. In-fact if u keep choosing different numbers and evaluate them as above ,you find that no number fits. This is incredible .The game check-mates the game-master. The game master has to play a game before setting up a game, a game he can’t come out of.

The above game can be played practically without thinking about the game-master’s perspective but on paper it will remain biased. The above analysis is not obvious and do tell us to give more time to the problems which concern us. Thinking from various ‘angles’ is always very insightful. Moreover it was fun looking at the game from the master’s point of view, Wasn’t it?

Infinite regression for the win.

That is the “so-called” solution for actual Two-envelope problem. This is a different problem because here you can use only integer sums.Moreover we are not looking at the solution.

Anyways great you mentioned it 🙂

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Thank You 🙂

If we consider A=4n where n is an integer; we see that the problem does not arise. In any case switching proves to be best option. e.g. consider A=4; in this case if the person chooses not to switch then his expected return is 4; if he chooses to switch then his expected return is 1/2*2+1/2*8=5. If you choose to switch; your expected return is always more. Hence in the long run of the game it is better to switch.

OK maybe u are not satisfied with the explanation that no number can be used for the value of A.

Let me write another post to answer it in detail.