Ball Game

Solution

Think of the decision like this: At any moment, compare these two choices:

- stop now and keep what you already have
- draw again and see whether the average value of continuing is actually better

If the expected value of continuing is less than the money currently in your hand, you should stop. Now let’s walk through it stage by stage.

You start with \(\$0\). If you draw one ball, there is a \(10/20 = 1/2\) chance it is red, in which case you now have \(\$5\), and a \(1/2\) chance it is blue, in which case the game ends and you get \(\$0\). So the expected value of taking the first draw is \[ \frac{1}{2}\cdot 5 + \frac{1}{2}\cdot 0 = 2.5 \] That is better than stopping immediately with \(\$0\), so the first draw is worth taking. Now suppose the first ball was red. You currently have \(\$5\). There are now \(9\) red balls and \(10\) blue balls left. If you draw again, then:

- with probability \(9/19\), you get a red and your total becomes \(\$10\)
- with probability \(10/19\), you get a blue and lose everything, ending with \(\$0\)

So the expected value of taking the second draw is \[ \frac{9}{19}\cdot 10 + \frac{10}{19}\cdot 0 = \frac{90}{19} \approx 4.74 \] But you already have \(\$5\) in hand. Since \[ 4.74 < 5 \] continuing is worse than stopping. So if you get one red, you should stop there.