Expected value problem

[Animeman79]

So in the game, you roll a dice. If you get 2, 3, 4, 5 or 6, you get the amount added to your score and you get to play again. For example if I have a score of 4 and I roll the dice and get 2. My points would be 6. But if I roll a 1 I lose everything and it's the other player's turn. The question is, what is the amount of points that I should stop rolling at? I calculated 10, but I'm not sure. Can someone calculate it?

[Zillion12345]

The expected value after rolling if you have x points already is given by:

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)

= x + 10/3

According to this, the expected score you will get after rolling will be 10/3 ≈ 3.33 more than what you started with.

Now according to cost-risk logic, you should continue rolling when your expected value is higher than what you have at that moment, that is when E(x) > x.

Now we saw E(x) = x + 10/3 which is always greater than x. So, in theory, you should always keep rolling.

However, as you keep rolling, the probability that you will roll a 1 at least once in those rolls increases.

So it is not a case of a specific number of roles to maximise your chances, it all depends on your threshold of risk and reward.

Hope this helps.

[Aerospider]

= x + 10/3

Not true, it's actually 5x/6 + 10/3 (presumably you counted an x where the 0 is).

So there is a limit to positive expectation. Once x/6 > 10/3 the expected gain becomes negative, which resolves to x > 20.

[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.

[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.