Dice probability

Question

I had a rather interesting yatzee experience. This is Maxiyatzee, 6 dice game. He had 13 throws. All together 78 dices. He only managed to throw one 2 on all those throws. And I am trying to find out the probability for that to happen. To manage none at all is easy. That would be (5/6)^78. A probability of 7 out of 10 millions. But I'm trying to beat my head around how to calculate the probability to manage one. Somebody who can help?

Answer 1

The probability of rolling exactly one $2$ in $78$ dice is

$$ \binom{78}{1} \left(\frac{5}{6}\right)^{77} \left(\frac{1}{6}\right)^1 = \frac{(78)(5^{77})}{6^{78}} \doteq 0.000010399 $$

or about $1$ in $96000$. The probability of rolling exactly one of some value between $1$ and $6$ is approximately (but not exactly) six times that, or about $1$ in $16000$.