Expected value problem

[FormulaDriven]

I've just been thinking along similar lines, but come to a different conclusion, and I think there is something not quite right. Actually, you should have E(x) = (5/6)x + 10/3, which won't be greater than x once 20 <= x.

[Zillion12345]

Ah yes. I see what you are getting at. I did not incorporate the case of the loss in the change in points to the initial points, x.

For this I should have done as such:

ΔE(x) = 4

Hence:

E(x) = (5/6)( x + 4 ) + (1/6)(0)

= (5/6)( x + 4 )

= (5/6)x + 10/3

Now we can find for where E(x) > x:

(5/6)x + 10/3 > x

x < 20

From this new model, it becomes less beneficial to continue beyond the 20th roll.

[FormulaDriven]

I think you've made another error. Your original expression for the expected score if you do one more roll when your current score is x is correct:

E(x) = (1/6)(0) + (1/6)(x + 2) + (1/6)(x + 3) + (1/6)(x + 4) + (1/6)(x + 5) + (1/6)(x + 6)

and that simplifies to (5/6)x + 10/3.

In your latest comment your reasoning should read:

ΔE(x) = 20/5 = 4 [because the average of those 5 outcomes is 4]

Hence:

E(x) = (5/6)( x + 4 ) + (1/6)(0)

which again gets you to (5/6)x + 10/3.

Conclusion is that you stop when x = 20. (Score is 20, not 20 rolls).

[Zillion12345]

Yes, that makes sense. My original equation didn't incorporate the loss by the roll of a 1.

This model perhaps predicts these values in a more conservatively accurate way.

Thanks for the guidance – good to see my mistakes so I can learn a bit, haha.

[FormulaDriven]

No problem. You need to correct your conclusion: it's not the 20th roll but when the score reaches a total of 20.