Tricky dice problem

Question

We simultaneously roll two dice in one round until at least one of them shows 3 once. What is the expected number of rolls needed?

I've calculated probability of getting 3 on the 1st roll I am really bad in math, sorry

$= (1/6)^2 + 2(5/6 * 1/6) = 11/36$

Then on the second

$= (1 - p(\text{on the 1st roll})) \cdot (11/36)$

Third

$= (1 - p(\text{on the 2nd})) \cdot (11/36)$

And so on, then I've just calculated the expected number of this distribution and reached the conclusion that I'll need at least $4$ rolls, since we operate only with integers.

I am really confused, if this approach is correct? Or should we use the geometric distribution and to get the expected number of rolls we just have to divide $1$ over probability of success, i.e. $11/36$?

And consequently how can we find the minimum number of dice needed to get $3$ at least once on at least one dice if we can simultaneously roll them no more than $2$ times?

Thank you in advance.

Answer 1

Or should we use the geometric distribution and to get the expected number of rolls we just have to divide 1 over probability of success, i.e. 11/36?

Yes, exactly that! Hence, the number of expected rolls is $\frac{1}{\frac{11}{36}}=\frac{36}{11}$ ... which is a little over $3$.

And consequently how can we find the minimum number of dices needed to get 3 at least once on at least one dice if we can simultaneously roll them no more than 2 times?

Not sure I understand your question here ... no matter how many dice you use, you will not be guaranteed to get at least one $3$ ...

I suspect the question is: what is the minimum number of dice so that the expected number of rolls with all those dice until you get at least one $3$ does not exceed $2$?

For this, assume that you have $n$ dice, and generate a formula for the probability of getting a $3$ on a single roll with all those dice, and then use the same idea of taking $1$ over that probability, and see for what value of $n$ it will get below $2$

Answer 2

$\frac {11}{36} + (2)(\frac {25}{36})(\frac {11}{36}) + (3)(\frac{25}{36})^2(\frac {11}{36})+ \cdots +(n+1)(\frac {25}{6})^n(\frac {11}{36})\cdots$

That is a $3$ on the first roll, or no $3$ followed by a $3,$ up to $n$ no 3's followed by a $3.$

You have something that resembles a geometric series. It is not quite, but I forget what it is called.

$E[Y] = \frac {11}{36}\sum_\limits{n=0}^{\infty} (n+1)(\frac{25}{36})^n$

Now look what happens when we subtract $\frac {25}{36} E[Y]$ from $E[Y]$

$E[Y] -\frac {25}{36} E[Y]= \frac {11}{36} + (2)(\frac {25}{36})(\frac {11}{36})-(\frac {25}{36})(\frac {11}{36}) + (3)(\frac{25}{36})^2(\frac {11}{36}) - (2)(\frac{25}{36})^2(\frac {11}{36})+ \cdots +(n+1)(\frac {25}{6})^n(\frac {11}{36})-(n)(\frac {25}{6})^n(\frac {11}{36})\cdots$

$E[Y](1 -\frac {25}{36}) = \frac {11}{36}\sum_\limits{n=0}^{\infty}(\frac{25}{36})^n$

and that is a more ordinary geometric series. And we perform a similar operation and we get.

$E[Y](1 -\frac {25}{36})^2 = \frac {11}{36}\\ E[Y] = \frac {36}{11}$

Slightly anti-climactic, isn't it?