Originally posted by David113
Two envelopes, connected by a string, are in a box. One envelope contains $1 and the other one $10. Now a coin is tossed. If the result is "head" another pair of envelopes is put into the box, this time with $10 and $100; otherwise the experiment is stopped.
If the experiment was not stopped, again a coin is tossed. If the result is "head" another pair of is "always swap envelopes", then why not take the other envelope in the first place?
Here's my go:
If you create a probability tree for the envelope pairs, you can see that the probability of the maximum cash value in the box being 10^n is 0.5^n (i.e. P(1/10) = 0.5^1, P(10/100) = 0.5^2, etc...). Now when you choose an envelope, you discover that the value inside is 10^n. The probability that you are dealing with the 10^(n-1)/10^n envelope is simply 0.5^n, and the probability that you are dealing with the 10^n/10(n+1) envelope is 0.5^(n+1). Since all other options are excluded once you pick a given envelope, the normalized probabilities are:
P(10^(n-1)/10^n envelope) = 0.5^n / (0.5^n + 0.5^(n+1))
P(10^n/10^(n+1) envelope) = 0.5^(n+1) / (0.5^n + 0.5^(n+1))
Say you switch; to determine the expected value of switching we multiply the outcomes by their respective probabilities and add them up as follows:
E(switch to 10^(n+1)) = 10^(n+1) * 0.5^(n+1) / (0.5^n + 0.5^(n+1))
E(switch to 10(n-1)) = 10^(n-1) * 0.5^n / (0.5^n + 0.5^(n+1))
Adding these two, we find that the expected value of switching is:
E(switch) = (0.5*10^(n+1) + 10^(n-1)) * 0.5^n / (0.5^n + 0.5^(n+1))
After crunching a few numbers on Excel, it turns out that E(switch) given an initial value of 10^n is equal to 34*10^(n-1). Now, the expected value of staying is simply 10^n. Comparing the two, we find that:
E(switch) = 34 > E(stay) = 10
E(switch) = 340, E(stay) = 100
and, more generally:
E(switch) = 34*10^(n-1) > E(stay) = 10^n
Therefore, it is always advantageous to switch.
I'm not quite sure how to answer the second part: if it's always advantageous to switch envelopes then why not choose the other envelope in the first place? My first instinct is to say that if you continually second-guess your choice and therefore never make one, the expected value of your action is simply 0. Therefore it is always better to choose an envelope and then switch than to choose neither. I'm not sure how satisfying this answer is...but at least it's more satisfying than the question.
Another suggestion I've heard about this kind of problem is that people do not evaluate their desire for increasing sums of money using simple subtraction or division - a log scale is much more representative. For example, the desire for $10,000 would be closer to log(10,000) = 4 as compared to the desire for $100,000 which would be log(100,000) = 5. To be sure, the larger sum is always desired more, but if we assume a human threshold where the perceived risks and rewards balance out in the chooser's head (say, if the ratio of the logs falls below 1.10:1, a difference of less than 10% ), then we can decide whether or not the chooser would want to switch given a certain sum. In this problem, if the chooser found $1,000,000, the ratio of the log of E(switch) = log($3,400,000) = 6.531 and the log of E(stay) = log($1,000,000) = 6 is 6.531/6 = 108.9%, and the chooser would stay with $1,000,000.