Expected number of sixes tossed before success

Question

Given a fair dice, what is the expected number of sixes that get tossed before throwing 2 sixes in a row? Would I start by finding all possible sets that occur without two sixes in a row and multiply them by the corresponding probability? ie $$\sum_{k=0}^\infty (A+B)^k $$ where A is all sequences that don't contain a six and B is all sequences that contain just one six

Thanks for your help

Answer 1

The set of sequences prior to the final two sixes that end the game is actually given by $$\sum_{k=0}^\infty (A+BA)^k $$

The generating function that counts the number of sixes and non-sixes prior to the final two sixes is
$$C(a,b)=\sum_{k=0}^\infty (a+ba)^k={1\over 1-(a+ba)}.$$ You turn this into a joint probability generating function by replacing $a$ with $5a/6$ and $b$ with $b/6$, then multiplying by $1/36$ to account for the final two sixes. This gives $$P(a,b)={1\over 36-30a-5ab}.$$

The marginal probability generating function for the number of sixes is found by plugging in $a=1$, to obtain $$P(b)={1\over 6-5b}. $$

Finally you get the expected number of sixes by differentiating and setting $b=1$. The expected number of sixes prior to the final two sixes is 5.

Answer 2

You may reduce this problem to a much simpler one by asking how many times will you have to throw the dice before getting a 6.
This reduction is equivalent to counting only the dice thrown after getting the first 6.
You throw the dice. If it falls on anything but 6 you ignore it. If you get 6 then you ignore it and look at the next number thrown. If it was 1-5 then the previous 6 is a 6 before the 2 consecutive throws of 6. If it is a 6 then you stop counting. Hence, the question you were asked is equivalent to asking "Expected number of non 6 before getting a first 6"

Answer 3

Denote by $E_\ne$ the expected number of additional sixes until the game ends when there was no six immediately before, and by $E_=$ the expected number of additional sixes when there was a six immediately before. Then $$E_\ne={5\over6}E_\ne+{1\over6}(1+E_=),\quad E_= ={5\over6}E_\ne+{1\over6}\ 1\ ,$$ from which we obtain $E_\ne=7$, $\>E_= =6$. It follows that we have to expect $5$ sixes to be thrown in vain.