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

Home > Probability
This or that? (Posted on 2015-08-19) Difficulty: 3 of 5
Alan and Bob are playing a game of marbles. Alan has two marbles, Bob has one, and each rolls to try to come nearest to a fixed point. If the two have equal skill, what is the chance that Alan will win?

There seem to be two contradictory arguments. On the one hand, each of the three marbles has an equal chance of winning, and two of them belong to Alan, so it seems that there’s a 2/3 chance that Alan will win.

On the other hand, there are four possible outcomes:
(a) both of Alan’s rolls are better than Bob’s,
(b) Alan’s first roll is better than Bob’s, but his second is worse,
(c) Alan’s first roll is worse than Bob’s, but his second is better, and
(d) both of Alan’s rolls are worse than Bob’s.
In 3 of the 4 cases, Alan wins, so it appears that his overall chance of winning is 3/4.

Which argument is correct?

Source: J. Bertrand, Calcul des Probabilités, 1889, via Eugene Northrop, Riddles in Mathematics, 1975.

See The Solution Submitted by Ady TZIDON    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution My take on it | Comment 2 of 4 |
There are six equally likely possibilities, listing from closest to farthest:

A1 A2 B (a)
A1 B A2 (b)
A2 A1 B (a)
A2 B A1 (c)
B A1 A2 (d)
B A2 A1 (d)

The four lettered possibilities are not equally likely.

Prob. of:

(a) 2/6 = 1/3
(b) 1/6
(c) 1/6
(d) 2/6 = 1/3

The first argument was the correct one.

  Posted by Charlie on 2015-08-19 10:03:55
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 (6)
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