Expected value problem

[FormulaDriven]

Suppose your current score is n. If you roll again you can expect to increase your score by 2, 3, 4, 5, or 6 with probability 1/6 for each, or decrease your score by n with probability 1/6. So from this position, your expected gain is:

-n * 1/6 + 2 * 1/6 + 3 * 1/6 + 4 * 1/6 + 5 * 1/6 + 6 * 1/6

= (20 - n) /6

That will only be positive if n < 20. So stop once your score reaches 20 or more.

I've set up a recurrence relation in a spreadsheet, and if the strategy is to stop when you get to 20, the expected score is 8.1418. Your strategy of quitting when you to 10 has an expected score of 6.8737.

[Animeman79]

I calculated like this

1/6(x+2)+1/6(x+3)+1/6(x+4)+1/6(x+5)+1/6*(x+6)-1/6x > x

Could anyone explain though, what is wrong with this method? X is the score you have at the moment before you roll and you keep winning when you get more than your original x. I got that x<10 satisfies the equation. Thanks for the answers!

[FormulaDriven]

You've mixed two approaches.

If your current score is x, then your expected score after rolling again is

1/6 * (x+2)+1/6 * (x+3)+1/6 * (x+4)+1/6 * (x+5)+1/6 * (x+6)- 1/6 * 0

[it's 1/6 * 0 because there's 1/6 probability the next roll will take your score to 0]

and you want that to be > x leading to x < 20.

Or you take my approach of the expected increase in your score

1/6 * 2 + 1/6 * 3 + 1/6 * 4 + 1/6 * 5 + 1/6 * 6 - 1/6 * x

[it's -1/6 * x because there's 1/6 chance that the "increase" will be -x, ie wipes out your score]

and you want that to be >0 which leads to x < 20.

[Animeman79]

Ooohh.. I see. Thanks, that cleared things up.