A prisoner was about to be executed but was promised his freedom if he drew a silver ball from one of two identical urns. He was allowed to distribute 50 silver and 50 gold balls between the two urns in anyway he liked. The urns were then going to be shuffled around out of his sight and he was to pick one urn and draw one ball at random from that urn.
How did he maximise his chances of success? He could choose to put 25 of each colour into each urn; that way he would have a 50/50 chance whichever urn he picked. He could put more silver balls in one of the urns, but that would mean less in the other and he has no idea which urn he will select since they are identical.
Is there any way he can improve his chances?
2006-06-12 05:28:46 · 5 answers · asked by brainyandy 6 in Science & Mathematics ➔ Mathematics
He should set the ratio of silver to gold as follows:
Urn 1: 1 silver ball, 0 gold balls.
Urn 2: 49 silver balls, 50 gold balls.
First he is going to pick an urn, then he is going to pick a ball. Half the time he will pick the first urn and have a 100% chance of picking a silver ball. The other half of the time he will have a 49/99 chance (49.4949%, or almost 50%) of picking a silver ball.
Figuring the weighted averages:
0.50 x 1.00 + 0.50 x 0.4949
= 0.50 + 0.247474
= 0.747474
So with this arrangement he has a nearly 75% chance of surviving which is better than 50/50!
2006-06-12 06:02:16 · answer #1 · answered by Puzzling 7 · 2⤊ 0⤋
good answer bandf.
It is a good job this isn`t how justice is dished out all us mathematicians would have a considerable advantage, i agree with your calculation.
Here is a good follow up question : if the prisoner could also decide how many urns were used , could he make a choice which would increase his odds of winning any greater than just under 75% ?
2006-06-12 13:22:07 · answer #2 · answered by MARTIN B 4 · 0⤊ 0⤋
No.
It's fifty/fifty. The two urns are an irrelevance!
Edit: Nice one bandf!
2006-06-12 12:33:07 · answer #3 · answered by Anonymous · 0⤊ 0⤋
I'm pretty sure bandf has the correct answer. I've been trying to come up with a mathematical proof for a while now, but my multivariate calculus is more than a little rusty. :)
2006-06-12 16:53:22 · answer #4 · answered by Jay H 5 · 0⤊ 0⤋
sorry, fared very badly in probability.
2006-06-12 12:33:12 · answer #5 · answered by anna pavlova 2 · 0⤊ 0⤋