# How do you explain this probability paradox?

Imagine there are two bags of money, and you are allowed to choose one. The probability that one of them contains $10^{n-1}$ dollars and the other contains $10^{n}$ dollars is $1/2^n$, $n\in\{1,2,3...\}$.

That is to say, there is $1/2$ probability that one of the two bags contains $\$1$and the other contains$\$10$; $1/4$ probability that one of the two bags contains $\$10$and the other contains$\$100$ , etc.

What's interesting is that, no matter which one you choose, you'll find that the other one is better. For example, if you open one bag, and find there are $\$10$in there, then the probability of the other bag contains$\$1$ is $2/3$ and the probability of the other bag contains $\$100$is$1/3$, and the expectation of that is$\$34$, which is better than $\$10$. If the other one is definitely better regardless of how much you'll find in whichever one you choose, why isn't choosing the other one in the first place a better choice? - This appears to be a variation on the "two envelopes" paradox. Wikipedia article – MJD Sep 23 '12 at 5:27 I didn't vote to close your question as a duplicate, but I still think the questions are similar enough that you should take a close look at the other one. – MJD Sep 23 '12 at 5:51 I'm sorry, I clicked the Wiki link and that is different. Your first link that I didn't notice is the same question. – Voldemort Sep 23 '12 at 5:54 If the se.math question I linked to is relevant, then the Wikipedia article is too. – MJD Sep 23 '12 at 5:55 The expected number of dollars in the bag you choose is infinite. So too is the expected number of dollars in the other bag. To say it is always worth changing it like saying$\infty \lt \infty\$ –  Henry Sep 23 '12 at 7:26

You only have two bags, not an infinite number of bags, so their probabilities should add to unity - so their probabilities are 2/3 and 1/3. The paradox conflates the population and sample statistics. In reality the other (third ;-) bag has a goat in it (it was a 50:50 choice).

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