# Throw a fair die repeatedly until I have thrown three sixes

I play a game in which I have to throw a fair die repeatedly until I have thrown three sixes, after which I stop and note the total number of throws. What is the probability that I take six throws?

I did $(1/6)^3 \times (5/6)^3$ and this gave me $0.00268$ to 3.s.f.

This doesn't seem right as couldn't this be achieved in many ways by throwing the die to be 6 in different orders? E.g. $6,6,6$, non six,non six, non six OR 6,non six, non six, 6, 6, non six

Could anyone explain this?

• Your analysis of the issue is right. To get the right answer quickly, note that we get the third six on the $6$-th try if and only if (i) we have exactly $2$ sixes in the first $5$ trials and (ii) we get a six on the $6$-th trial. To find the probability of (i) is a probably familiar binomial distribution problem. Or else, equivalently but more slowly, enumerate all the different orders in which it could happen (you started doing that) and add up the probabilities. – André Nicolas Feb 11 '16 at 19:35

In the first $5$ throws exactly $2$ sixes must be thrown.

The probability of this event is $\binom52\left(\frac16\right)^2\left(\frac56\right)^3$.

After that a six must be thrown so we end up with a probability of: $$\binom52\left(\frac16\right)^3\left(\frac56\right)^3$$

• So the answer is 0.0536? – Simon Feb 11 '16 at 19:42
• I haven't a calculator at hand, but I think it is 0.026791838.;) – drhab Feb 11 '16 at 19:47
• You are very welcome. – drhab Feb 11 '16 at 19:59

You would need two $6$'s in the first five throws, and third $6$ in sixth throw. There are $\binom52$ ways to order the first five throws, so the final result is: $$\binom52\left(\frac56\right)^3\left(\frac16\right)^3$$

• So the answer is 0.0536? – Simon Feb 11 '16 at 19:42
• No, $\binom52=10$, so the expression evaluates to $0.02679...$ – Logophobic Feb 11 '16 at 19:47