Here's the question: Shrin is participating in a lottery where she's going to pull up balls from a bowl without looking at their colours. There are two white balls and eight black ones. She pulls up a ball and the colour is checked (not by her of course), and the ball is not put down in the bowl again. She gets to continue until she receives a white ball, then the game is over and she loses all her money. She gets to keep pulling out balls until she decides to quit. For each black ball she receives, one hundred dollars is given to her.

When should she quit playing the game, that is, when the risk of losing is larger than the chance of winning?

Here are my thougths: The chance of pulling two black balls is 8/10*7/9 = 62,2% But the chance of pulling three balls (8/10*7/9*6/8) is only 46.7%, therefore the risk of losing is larger than the chance of winning and she should quit after pulling two balls?

The only thing confusing me is that if you've already pulled two black balls, then the chance is still 6/8 (75%), right ? But then the answer would go down until there's only one black ball and two white ones left, and that can't be the true answer can it?


Maybe you did not realize a very important thing in this game. You said that when she pulls out, a ball is checked. Why did you add that it is not checked by her? It cannot be true because then you said that when she pulls a white ball the game is over (immediately). This means that after each pulling she must be told which color she pulled! So, after each pull she knows the probability of choosing a white ball in the next pull! And this means that after every pulling she can forget probabilities of choosing a black ball in all previous pullings. So your conclusions are partially wrong. Yes, it is true that the probability of pulling three black balls is 46.7 % but it is true only before the game starts! If she gets to the third round, she knows that she pulled two black balls in previous two pullings, so her chance of winning is now 6/8. But you forgot another important thing - in the third round, her chance of winning is 75 % but if she win, her reward will be only 100 dollars while if she loses, her lost will be 200 dollars (which she won in previous two rounds). In the fourth round, the chance of winning 100 dollars is 5/8, while the chance of losing 300 dollars is 3/8. And because 100*(5/8) is less than 300*(3/8), I reccomend her to quit after the third round.

  • $\begingroup$ Yes, with that comment I ment that she herself doesn't check the ball, but of course she's going to know that she pulled a black ball because nobody is going to stop her from playing. That is, as long as nobody says something, she pulled a black ball. Anyway though, I did not think about how much she would lose and win respectively, but why did you do 3/8*300 and 5/8*100? It sounds great but I just don't know exactly why you did it? By the way, in the fourth round, isn't the chance of winning actually 5/7 and losing 2/7 ?, Or am I missing something here? Thanks. $\endgroup$ – didnotcomeuptosomething Oct 5 '14 at 13:24
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    $\begingroup$ Sorry, of course, in the fourth round is the chance of winning 5/7 and the chance of losing 2/7. So (2/7)*300 is her expected lost (probability of losing times the amount of lost) and (5/7)*100 is her expected win. When the expected lost becomes greater than the expected win, it is time to quit the game. $\endgroup$ – Joseph Oct 5 '14 at 13:33

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