Throw dice till numbers are the same

Question

Let's consider the following experience : we throw two dices till same number on two dices appears. I want to calculate expected sum of all dices appeared.

My work so far

I wanted to do it (and also I found it very intuitive) to do it by wold's decomposition theorem. The sum then will be presented by formula $E[S_n]=E[T]E[X_1]$, where $T$ is the stopping time.

How to calculate $E[T]$ ?

$P(Y=1)=\frac16$

$P(Y=2)=\frac56 \cdot \frac16$

$P(Y=n)=(\frac{5}{6})^{n-1} \cdot \frac16$

So :

$$E[Y]=\sum_{i=1}^\infty i \cdot P(Y=i)=1 \cdot \frac16 +2 \cdot \frac56 \cdot \frac16+...$$

And it's type of series $\sum_{i=1}^{\infty} i(1-p)^{i-1}p$ where $p \in [0,1]$ and this sum equals to $\frac1p$. So my sum above equals to 6 and that's exactly $E[T]$

why it would be good to use wold's theorem? Because it's very easy to calculate either $E[X_1]$ or $E[T]$ (calculated above). However, I don't know how should I define set $T=\inf\{...\}$ that will show that equality $E[S_n]=E[T]E[X_1]$ holds. I'm not sure how to strictly justify it.

Answer 1

One way is to just manually take all sums. The first part is where all of them are equal sums, the second is where one of them is $1$ and other is between $2$ and $6$, so on..

The last part is for the rolls to continue with $\frac{5}{6}$ probability.

It can surely be optimized but given it is just two dice, this works.

$E(X) = \displaystyle \frac{1}{36} (2 + 4 + 6 + 8 + 10 + 12) + \frac{2}{36}(3 + 4 + 5 + 6 + 7) + \frac{2}{36}(5 + 6 + 7 + 8) + \frac{2}{36}(7 + 8 + 9) + \frac{2}{36} (9 + 10) + \frac{2}{36}(11) + \frac{5}{6}E(X).$

$E(X) = \displaystyle \frac{42}{36} + \frac{50}{36} + \frac{52}{36} + \frac{48}{36} + \frac{38}{36} + \frac{22}{36} + \frac{5}{6}E(X)$

$E(X) = \displaystyle \frac{252}{6} = 42$.

Answer 2

Let $E_1$ be the expected sum of a single throw given that the two dice are the same, and let $E_2$ be the expected sum of a single throw given that the two dice are different.

The probability that the two dice are the same is $\frac16$. So $E$, the expected total sum, satisfies $$E=\frac16 E_1 + \frac56 (E_2+E)$$

And by linearity of expectation, $E_1$ and $E_2$ are both equal to $7$. So you can easily solve for $E$.

Answer 3

From your work that $E[y]=6$ (on average it will take $6$ rolls of two dice):

You can say that you will have an average of $12$ single die rolls.

A single die roll has expected value $3.5$ (this is true even in the case that the die is part of a roll of doubles).

Can you take it from there?