4by12

Two Envelopes

Players A and B play a game. Player A somehow selects two different natural numbers and puts them in two envelopes. He then chooses one envelope at random and hands it to player B. Player B looks at the number, and has to guess whether the number in the other envelope is higher or lower.

The question is: How can player B guess correctly at probability > 0.5 on average?

There are some additional conditions: Player A knows the strategy player B uses to obtain the result, and he can use whatever counter-strategy he likes to select the two numbers. His only constraints are that the numbers must be different, and that the number that is revealed to player B is selected at random.

I know of two solutions to this riddle. I’ll present the first one here and leave the other to another post.

After player B receives the number, she selects a natural number using some distribution, making sure the new number is different from the number in the envelope, and making sure each natural number has a non-zero chance of being selected. Given the two numbers, this is what player B says:

1. If the new number is larger than the envelope number, the answer is ‘higher’
2. Otherwise, the answer is ‘lower’

Now, if both numbers in the envelopes are smaller than the number player B chooses, then player B is right with 0.5 probability (because the first envelope is selected at random). This is also true if both numbers are larger than the number player B chooses. But if the number she chooses is higher than one envelope number and smaller than the other, then player B is always correct.

Therefore, the chance of player B being correct is greater than 0.5 on average.

Note that the same solution would apply if we played with real or integer numbers instead of natural numbers.