Two boxes are placed in front of you. One box contains nine \$1 bills and one \$5 bill, while the other box contains two \$1 bills and one \$100 bill. You choose at random one of the two boxes and two randomly picked bills are taken out of the box. It appears that these two bills are \$1 bills. Next you get the opportunity to pick a third bill out of one of the two boxes. Should you stick to the box chosen before or should you switch to the other box when you want to maximize the probability of picking the \$100 bill?

This seems like a textbook cond prob problem. The answer seems clear without calculation. Since it's much more likely to draw 2 $1 bills from a box with 9/10 of $1 bills than from a box with 2/3 of $1 bills, probability of getting a box with $100 bill after the event is much higher after switching.

It is a brainteaser, not just a textbook exercise.

Let's call:A: Event of choosing first boxB: Event of choosing second boxC: Event of drawing two $1 billsp(A/C)>p(B/C) if p(A)=p(B)p(A/C)=0.8*p(A)/(0.8*p(A)+p(B)/3)The gain if you draw the third bill in box 1 is 5/8+7/8=1.5=G(1)The gain if you draw the third bill in box 2 is 100=G(2)So sticking gives you 1.5*p(A/C)+100*p(B/C) and switching gives you 100*p(A/C)+1.5*p(B/C) But as Zerdna said, it's obvious that G(1)<<G(2) and p(A/C)>p(B/C)

Vanubis1, the criterion is to maximize the probability of getting the $100 bill and not to maximize the expected dollar value of the third bill. By the way, I think the expected dollar value when switching should be (1/3)*100*P(A\C)+1.4*P(B\C) rather than 100*p(A/C)+1.5*p(B/C) . As I said before , it is a brainteaser.

So sticking gives you p(B/C) and switching gives you p(A/C).My results for the gain are right because you have removed 2 $1 bills

It's a trick question. It's not the probability the box contains the $100, it's the probability you pick the $100 out of the boxYou should keep the same box.Box 1: 9 x $1 and 1 x $5Box 2: 2 x $1 and 1 x $100 P(picking both $1s from Box 2) = 1/3=5/15P(picking 2 $1's from Box 1) = 8/10=12/15so there's a 5/17 probability that you picked both $1's from Box 212/17 probability that you picked both $1's from Box 1If you keep the same box, there's a 5/17 probability that it contains the $100 and it's the only bill in there, so there's a 5/17 probability you pick the $100If you switch the box, there's a 12/17 probability that it contains the $100 but there are 3 bills in there, so there's a 1/3 * 12/17 =4/17 probability you pick the $100 i.e. keep the same box, 5/17 you pick the $100if you switch, 4/17 you pick the $100 and 8/17 you pick a single

I agree. I have read it too quickly.