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Say that I have a dice game. You can roll the die first and then have two choices.

First, take the dollar amount of the number that shows up (if you rolled a 5, you get $5 ).

Second, you can re-roll. However, if you decide to take the second choice, you are obligated to take whatever you get in the second roll (if you rolled a 1 in your second try, you must take $1, regardless of what you got in the first roll).

You need to pay for this game before you enter the game. Assuming that you are risk neutral, what's the maximum amount that you would like to pay for this game?

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  • $\begingroup$ This reads a bit like a rather routine textbook question; is this for some subject? $\endgroup$
    – Glen_b
    Commented Oct 26, 2014 at 22:12
  • $\begingroup$ Wut? The question does not give me any way to pay a bet or lose money, why do I need to break even? I'll just keep rolling and get whatever in the first roll in order to maximize my gain in limited time until the dealer finds out this is a bad deal. $\endgroup$ Commented Oct 26, 2014 at 22:14
  • $\begingroup$ @Penguin_Knight It's actually an interview question. Let's assume that you need to pay for this game before you play. What's the maximum amount that you'd like to pay, assuming that you're risk neutral. $\endgroup$
    – Paul Smith
    Commented Oct 26, 2014 at 22:16
  • $\begingroup$ Yes indeed. Great explanation! $\endgroup$
    – Paul Smith
    Commented Oct 26, 2014 at 22:32

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Under the additional assumption of risk neutrality (and the now much less remarkable assumption that the player will seek to maximize expectation rather than do something else), the player will re-roll on a 1,2 or 3.

So you have a 50% chance of retaining the first roll (with a conditional expectation of 5, given the first roll was retained) and a 50% chance of getting the second roll (with a conditional expectation of 3.5, given the die was re-rolled).

Or, more technically, you can think of it as "just apply the law of total expectation and you're done". Specifically, see the form:

$\text{E}(X) = \sum_{i=1}^{n}{\text{E}(X | A_i) \text{P}(A_i)}\,,$

at that link (where $A_i$ here are the two possibilities 'roll retained' and 'die rerolled' for $i=1,2$ respectively, and which cover all possible cases of $A$) and $X$ is the amount won in the game. All the probabilities and expectations for that formula are given above.

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