What is the probability that the players decide black but the card deck is red? There are two decks, one with $8$ red cards and $2$ black cards, and one with $2$ red and $8$ black cards. Let us call the first deck the red, and the second the black deck. Any of the two decks is chosen with probability $1/2$. Once that has been done, there are $3$ players who get to see, each individually, one random card of the deck. In order to decide that the deck was black, the players must unanimously vote black. In all other cases, the result is red. Let us assume that players are non-strategic (i.e., follow their private signals).

*

*What is the probability that the players decide black but the deck is red?

*What is the probability that the players decide black?

*Suppose the players have decided black. What is the probability that the deck is really
black?

I came up with the following probabilities:

*

*$P\text{(red deck | red card)}=P\text{(black deck | black card)}= 8/10$

*$P\text{(black deck | red card)} = P\text{(red deck | black card)} = 2/10$

*$P\text{(red deck)} = P\text{(black deck)} = 1/2$
Then for question 1, I think the answer is
$$P\text{(red deck | black card)}^3 = 1/125.$$
But I have no idea how the approach question 2 and 3. Anybody has an idea?
 A: To summarize the discussion in the comments:
It's useful to strip down the problem a bit...in particular, the "players" serve no meaningful role.  Accordingly, we simply draw three cards from a deck chosen uniformly at random from the two choices.  We draw them independently of each other (from the single chosen deck) and we replace them after each draw.  If we see three blacks, so $BBB$, we declare "Black", otherwise we declare "Red".
So, the probability of drawing $BBB$ in this way from deck one is $\left(\frac 2{10}\right)^3=\frac 1{125}$.  And the probability of drawing $BBB$ in this way from deck two is $\left(\frac 8{10}\right)^3=\frac {64}{125}$.
We can answer the second question immediately.  By the Law of Total Probability, the probability of seeing $BBB$ is $$\frac 12\times \frac 1{125}+\frac 12\times \frac {64}{125}=\boxed {\frac {13}{50}}$$
To answer the first one, note that we are asking for a joint probability...namely the probability that you choose deck one AND you see $BBB$.  That answer is simply $$\frac 12\times \frac 1{125}=\boxed {\frac 1{250}} $$
The third question requires Bayes Theorem.  Specifically, we are asked to compute $$P(\text {deck two}\,|\,BBB)$$
(using what I hope is obvious notation).
By Bayes that is given by $$P(\text {deck two}\,|\,BBB)=\frac {P(BBB\cap \text {deck two})}{P(BBB)}=\frac {\frac {64}{125}\times \frac 12}{\frac {13}{50}}=\boxed {\frac {64}{65}}$$
