# 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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Namaste, GNUSupporter 8964民主女神 地下教會, José Carlos Santos, Deepesh Meena, Saad
If this question can be reworded to fit the rules in the help center, please edit the question.

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.
• @Soyol Yes, I missed that $x$ was a loss. I will edit. – lulu Sep 29 '18 at 22:37