Math Puzzle: Pulling Balls Out of an Urn
June 9th, 2009
| Categories: Math Puzzles, Probability
Now, I know what you’re thinking, but we don’t endorse those sort of games on this site. All of our puzzles are family oriented (if you come from a family of geeks, that is).
Here is today’s challenge. You are given an urn with 4 black balls and 3 white ones. We play a game: you can, each turn, draw as many balls as you’d like (not drawing is also an option); you can keep doing so until you decide to stop, or until there are no balls left in the urn. At the end I’ll pay you 1$ for every white ball in your possession, and take 1$ from you for every black ball.
Should you play this game at all? If not, why? If yes, how?
EDITED, 10/June/2009: A solution has been posted here.
I don’t even know if this is relevant but if you draw balls, are the payments reversed? (I.e. I get $1 for every white ball in your possession?)
I have no idea - I don’t think that’s part of the question!
Hello again!
At first I thought that this was a losing game, as your odds of getting a white ball are almost always lower than the odds of a black ball coming out, BUT then occurred to me that you can stop any time you want, so that should skew the odds a bit, no?
I think the best strategy in this game would be to quit when you are 1 white ball ahead (didn’t try with 2 white balls ahead, but I guess the chances aren’t worth the trouble) or keep playing until the bitter end (although you could stop when there is only one black ball remaining in the urn, but didn’t take that into account in my calculations).
With that strategy in mind, I analyzed the whole tree of possible actions (not sure if there is a better way to do this, but it’s just 7 balls) and calculated the probabilities of each branch leading to a 1 white ball advantage, and got this:
3/7 + (4/7)*(1/2)*(3/5)*(1/2)*(1/3) + (4/7)*(1/2)*(2/5) + (4/7)*(1/2)*(3/5)*(1/2)*(1/3) = 3/5 (I’ll hope nothing gets treated as an emoticon here :-))
So, you have a 60% chance of winning $1, a little less than 40% chance of losing $1 and a small chance of breaking even in each game, which is some quite good odds if you can play multiple games.
Your insight is spot on, but you’ve missed a part of the game tree - you can in fact make about 0.34$ per game. Take a look at the posted solution.