March 10, 2023
Bill has four opaque bags, each of which has three marbles inside. Three of the bags contain two white marbles and one red marble, while the last bag contains three white marbles. The bags are otherwise indistinguishable.
Ted watches as Bill randomly selects a bag and reaches in without looking to grab two marbles without replacement. It so happens that both marbles are white. Bill is about to reach in and grab the last marble in that bag.
What is the probability that this marble is red?
\(\frac{1}{2}\)
Explanation:
We can solve this puzzle using Bayes' Theorem. The key here is to realize that we are given that Billy picked out two white marbles. Given this information, the probability that Ted picks out the last red marble is actually higher than one might think because in three out of the four bags, the last marble would be red. However, the answer is also lower than 3/4 because if Billy picked the last bag containing three white marbles, then there will be more cases where he would pick out two white marbles since any combination would work. The actual answer lies in between those two wrong intuitions and can be correctly solves via Bayes' Theorem.
Let \(R\) be the event that Ted picked out the last red marble, and \(2W\) be the event that Billy picked out the two white marbles without replacement. Normally, events are denoted by a single letter but here I am using \(2W\) for better readability. The probability we are seeking is:
\begin{align} P(R|2W) \\ \end{align}
That is, the probability that Ted picks out the last red marble given that Billy drew two white marbles without replacement. According to the Bayes' Theorem, then, we have:
\begin{align} P(R|2W) &= \frac{P(2W|R) P(R)}{P(2W)} \\ \end{align}
The above equation helps us because the right hand side has three probabilities that we can determine either with inspection or a little bit of calculation. For the first piece, \(P(2W|R)\), we are asked the probability of picking out two white marbles without replacement given that the last marble is red. This is obviously 1, as we would be dealing with bags containing two white marbles and one red, and we have to pick out the two white marbles in order for the last marble to be red. For \(P(R)\), this is the answer to the first wrong intuition where we simply follow the stochastic path where Ted would pick out the last marble. First, Billy must select the bag with the red marble with probability 3/4. Then, the first pick must be the white marble with probability 2/3 followed by the second with probability 1/2 since there is no replacement. This comes to be (3/4) x (2/3) x (1/3) = 1/4. This would be the probability that Ted picks out the last red marble without prior knowledge that Billy picked out the two white marbles first without replacement. Finally, for \(P(2W)\), we calculate as follows:
\begin{align} P(2W) &= \left(\frac{3}{4}\right) \left(\frac{2}{3}\right) \left(\frac{1}{2}\right) + \frac{1}{4} \\ \\ &= \frac{1}{4} + \frac{1}{4} \\ \\ &= \frac{1}{2} \\ \end{align}
If Billy picks the bag with two white marbles, then picking out two white is 1/3. This case has a 3/4 chance of happening. If Billy picks the bag with three white marbles, then he is guaranteed two white marbles. This case has a 1/4 chance of happening. All in all, \(P(2W) = \frac{1}{2}\).
Putting everything together, we have:
\begin{align} P(R|2W) &= \frac{P(2W|R) P(R)}{P(2W)} \\ \\ &= \frac{1 \times \frac{1}{4}}{\frac{1}{2}} \\ \\ &= \boxed{\frac{1}{2}} \end{align}
So the correct answer indeed lies in the middle of our two estimates from the wrong intuitions.