All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Probability
A birthday gamble (Posted on 2002-05-28) Difficulty: 3 of 5
For Cathy's birthday, her uncle decided to make her a deal. He took ten singles and ten one hundred dollar bills, and asked Cathy to divide them into two piles as she saw fit. He would then blindfold her, and thoroughly shuffle each pile of bills, so that the order was completely random. Finally, he would put each pile in a separate box.

Cathy is to pick one of the two boxes at random, and then pick out a random bill from that box (still blindfolded). She would get to keep whatever bill she pulls out.

Naturally, Cathy prefers to get a $100 bill. What strategy should she use in breaking up the bills into two piles to maximize her chance of getting a hundred?

See The Solution Submitted by levik    
Rating: 3.5000 (8 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
proof | Comment 3 of 8 |
To actually prove that that is her best strategy takes a littl more work.

First, consider that there are four basic "schemes" for dividing up the money:

1) All the $100 bills in one box; All the $100 bills in one box. There are four ways of doing this, but only two different results. Either all the money is in one box (Cathy's chances are 25% $100, 25% $1, 50% $0), or the Ones and the Hundreds are in separate boxes (Cathy's chances are 50% $100, 50% $1)

2) All the $100 bills in one box, some $1 bills in each box. Chances are 10/(2*[s + 10]), where s = the number of singles in the hundreds' box. This number is maximized at s=1, where it becomes 10/22=0.4545=45.45%

3) Some $100 bills in each box, all $1 bills in one box. Chances are 50% + {h/(2*[h + 10])}, where h = the number of hundreds in the singles' box. This number is maximized at h=9, where it becomes 50% + {9/38} = 73.68%

4) Some $100 bills in each box, some $1 bills in each box. Chances here are ½[h/(h+s)] + ½[10 - h)/(20 - h - s)], where s and h are the numbers of hundreds and singles in the first box. For any given h, the chances are greatest when s=(10-h), giving us ½[h/10] + ½[(10 - h)/10] = ½[10/10] = 50%

So the only scheme that increases the odds beyond 50% is number 3, and the best strategy in that scheme is to put everything but one $100 bill in the singles' box.
  Posted by TomM on 2002-05-29 03:31:25
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (1)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (17)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information