Jars Riddle
Posted by Rob Herman at November 23rd, 2006
Happy Thanksgiving! In honor of the holiday I have given this riddle a gratuitous and certainly unnecessary theme. Several of you may have seen this riddle before; I think it’s on the less difficult side.
You are a turkey slated to be eaten for Thanksgiving. However, it turns out you can talk and have pleaded for your release. Your captors are inclined to pardon you; on the other hand, they are still hungry, so they have proposed a game of chance:
You are provided with 100 balls, 50 white, 50 black; and two urns. You must put the balls into the urns. One of the urns will randomly be chosen, and then a ball will be randomly chosen from that urn. If a white ball is selected, you will be pardoned; but if a black ball is selected, you will be eaten. You can’t leave an urn empty.
How should you divide the balls to maximize your chance of survival?
I haven’t done any of the math to confirm that this is correct but intuition tells me that the best option is to put one white ball into one of the urns and all of the rest of the balls in the other urn. That means you have an automatic 50% chance of survival and you’ve minimized the possibility that they’ll pick a black ball out of the other urn while still fulfilling the requirement that neither urn can be empty.
I agree with Ephraims answer, adding that, while this probably violates logical riddle rules, if it was my life on the line, I’d make sure that I put all the black balls into the second urn first, covering them with the white.
Correct!