probability involving rerolling dice

Question

The question go like the following.

You roll one 6 sided dice one time. You get the same money as the number of dots face up. But if you roll a six, you get nothing and re-roll. If your re-roll gets a six again, you reroll again and again. What is the expercted value of money get.

I guess i should go like the following

$$E(x)=\frac16+\frac26+\frac36+\frac46+\frac56+\frac16 E(x)$$

And if you go like the following, what result do you get in the end? — Arthur, Dec 14 '17 at 10:19

Siong Thye Goh · Accepted Answer · 2017-12-14T10:37:27.663

3

Good job, your working is fine.

It is as if you are tossing a dice with $5$ sides since it is equally likely to get any numbet between $1$ and $5$ inclusive by symmetry.

edited Dec 14 '17 at 10:37

answered Dec 14 '17 at 10:19

Siong Thye Goh

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but if i want to calculate the variance, when i go to the term which 6 face up , should it be $\frac16(6-3)^2$ or $\frac16(3-3)^2$ – Ricky Ng Dec 14 '17 at 10:23
You can check that the probability that you will get $i$ dots where $i \in { 1, \ldots, 5}$ would be $\frac15$. From there, you can compute the variance. – Siong Thye Goh Dec 14 '17 at 10:26

probability involving rerolling dice

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