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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?

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closed as off-topic by Namaste, GNUSupporter 8964民主女神 地下教會, José Carlos Santos, Deepesh Meena, Saad Sep 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.

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

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