TITLE: Probability of 5 dices being 6s, given that I know 3 are 6s
QUESTION [0 upvotes]: Imagine a game with two players who each throw 5 dice and only they can see their own dice. If we roll and I see that I have 3 sixes, what are the probabilities that there are at least 5 sixes in total (all 10 dice P(X>=5))? I believe this is a cumulative binomial distribution, but I can't figure out how to calculate the conditional. 
P.S. I'm not asking for the probability that the other player has 2 sixes, but the probability, given that I have 3 sixes, that there are 5 in total.

REPLY [1 votes]: If player-1 has exactly 3 sixes, then the only way the two players can have 5 or more sixes is if player-2 has 2 or more.  Just use binomial distribution to calculate this.
$$\Bbb P(P_2 \geq 2) = 1-\Bbb P(P_2 \leq 1)= 1 - \binom{5}{4}\frac {5^4} {6^5} - \frac {5^5} {6^5}$$

However, if player-1 merely sees 3 sixes but may have more, then you need to use conditional probability.
$$\begin{align}
 \Bbb P(P_1+P_2 \geq 5\mid P_1\geq 3) & = \frac{\Bbb P(P_1+P_2 \geq 5, P_1\geq 3)}{\Bbb P(P_1\geq 3)}
\\[2ex] & =\frac{
 (\binom{5}{3}\binom{5}{2}+\binom{5}{4}\binom{5}{1}+1)\frac {5^5}{6^{10}}
+(\binom{5}{3}\binom{5}{3}+\binom{5}{4}\binom{5}{2}+\binom{5}{1})\frac {5^4}{6^{10}}
+(\binom{5}{3}\binom{5}{4}+\binom{5}{4}\binom{5}{3}+\binom{5}{2})\frac {5^3}{6^{10}}
+(\binom{5}{3}+\binom{5}{4}\binom{5}{4}+\binom{5}{3})\frac {5^2}{6^{10}}
+(\binom{5}{4}+\binom{5}{4})\frac {5}{6^{10}}
+\frac {1}{6^{10}}
}{\binom{5}{3}\frac {5^2}{6^5}+\binom{5}{2}\frac{5}{6^5}+ \frac{1}{6^5}}
\end{align}$$
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