# Dice Expected Value - Game of chance [closed]

Suppose you repeatedly roll a fair six-sided dice until you roll a 1 (and then you stop). Every time you roll a 2, you lose x points, and every time you roll a 6, you win y points. You do not win or lose any points if you roll a 3,4, or 5.

What is the expected number of points (as a function of x,y ) you will have when you stop?

## closed as off-topic by Namaste, GNUSupporter 8964民主女神 地下教會, José Carlos Santos, Deepesh Meena, SaadSep 30 '18 at 0:09

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Let the answer be $$E$$. When we throw the die, we note four possible outcomes. Either you get a $$1$$, ending the game, or you get a $$2$$ and lose x points, or you get a $$6$$ and get $$y$$ points, or you get $$3,4,5$$ and just restart. It follows that $$E=\frac 16\times 0+\frac 16\times (E-x)+\frac 16\times (E+y)+\frac 12\times E\implies \boxed {E=y-x}$$

Informally: there are three relevant outcomes (ignoring the ones that change nothing) and all are equally likely. By symmetry, you expect it to take three (relevant) tosses to get a $$1$$. Thus you expect to get two relevant throws before the game ends. By symmetry, you must expect to get one of each of the two other relevant throws.

• since by 2 we lose x points the sign of x should be negative in your answer though. Thanks – Soyol Sep 29 '18 at 22:30
• but should it be -(E+x) or (E-x)? – Soyol Sep 29 '18 at 22:37
• @Soyol Yes, I missed that $x$ was a loss. I will edit. – lulu Sep 29 '18 at 22:37