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).

  1. What is the probability that the players decide black but the deck is red?
  2. What is the probability that the players decide black?
  3. 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?

  • $\begingroup$ Where does this problem come from? $\endgroup$
    – coffeemath
    Commented Jan 19, 2023 at 11:30
  • 1
    $\begingroup$ Just to clarify the rules: Each player sees a single card, independently of the other players? Is that card then replaced into its deck before the next player sees their card? And the player just votes the color they saw, right? Seems like a lot of uneccessary structure...you could just skip the players and draw three cards (with or without replacement). $\endgroup$
    – lulu
    Commented Jan 19, 2023 at 11:31
  • $\begingroup$ Your $\frac1{5^3}$ is the conditional probability, given the predominately red pack is used, that the three players independently see a black card and call it. The issues include (a) it is not clear that the cards are provided with replacement (if they were dealt to the players without replacement then the conditional probability would be $0$) and (b) the question seems to ask for the joint probability not the conditional probability $\endgroup$
    – Henry
    Commented Jan 19, 2023 at 11:39
  • $\begingroup$ @coffeemath from my Probability Computer Science reader :) $\endgroup$
    – Max Jeltes
    Commented Jan 19, 2023 at 11:51
  • $\begingroup$ @lulu I think we can assume the players see a card independently from other players with replacement. Each player votes non-strategic so according to the card he/she saw. $\endgroup$
    – Max Jeltes
    Commented Jan 19, 2023 at 11:52

1 Answer 1


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 $$P(\text {deck two}\,|\,BBB)=\frac {P(BBB\cap \text {deck two})}{P(BBB)}=\frac {\frac {64}{125}\times \frac 12}{\frac {13}{50}}=\boxed {\frac {64}{65}}$$


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