Google Answers Logo
View Question
Q: Probability of Winning a Game (Revised) ( No Answer,   2 Comments )
Subject: Probability of Winning a Game (Revised)
Category: Science > Math
Asked by: wobbleone-ga
List Price: $25.00
Posted: 29 Sep 2006 21:12 PDT
Expires: 01 Oct 2006 08:21 PDT
Question ID: 769624
Two friends say Bob and Sue play a game where Bob selects one of three
boxes containing three numbered ping-pong balls: Box A (2,6,7); Box
B(1,5,9); Box C(3,4,8).Then Sue chooses one of the remaining two boxes.
Each randomly select one ball from their boxes and the higher number
wins the game. If picking a ball has equal probability, who has a
higher chance of winning, and why?

p.s. Sorry to ansel001, I went to adjust the question but deleted it
instead. After revising, Box B contains balls numbered (1,5,9), was
(1,5,6). This gives an equal chance of winning from each box, that
being 4. I believe the key to the problem could be to consider the
Monty Hall effect, but I may be wrong. Sorry again, for the confusion.
There is no answer at this time.

Subject: Re: Probability of Winning a Game (Revised)
From: mathisfun-ga on 30 Sep 2006 02:13 PDT
Box A vs B:
     B(1) B(5) B(9)
A(2)  W    L    L
A(6)  W    W    L
A(7)  W    W    L

Box A vs C:
     C(3) C(4) C(8)
A(2)  L    L    L
A(6)  W    W    L
A(7)  W    W    L

Box B vs C:
     C(3) C(4) C(8)
B(1)  L    L    L
B(5)  W    W    L
B(9)  W    W    W

If Bob chooses A then Sue will pick C and win 5/9 times
If Bob chooses B then Sue will pick A and win 5/9 times
If Bob chooses C then Sue will pick B and win 5/9 times

Thus both playing smart the expected value of Sue winning is 1/3*3(5/9)=5/9

The "Monty Hall effect" would come into play if the boxes were (1,1,1)
(1,1,1) and (9,9,9) and Bob let Sue pick a closed box, then opened one
of the remaining boxes to reveal (1,1,1) and gave her the chance to
change, which she would then take.
Subject: Re: Probability of Winning a Game (Revised)
From: ansel001-ga on 30 Sep 2006 15:06 PDT

Yes, your revision is more like what I was expecting the first time. 
Mathisfun is correct.  The second person has a 5/9ths chance of
winning no matter what the first person picks.  This does not violate
the law of transitivity however.  Each box has an average expected
value of 5.  Note that when A beats be with probability 5/9, the
average margin of winning is 3 and the average margin of losing is
-3.75.  The overall margin of winning or losing is zero as would be
expected, since they both have the same average expected value.

Important Disclaimer: Answers and comments provided on Google Answers are general information, and are not intended to substitute for informed professional medical, psychiatric, psychological, tax, legal, investment, accounting, or other professional advice. Google does not endorse, and expressly disclaims liability for any product, manufacturer, distributor, service or service provider mentioned or any opinion expressed in answers or comments. Please read carefully the Google Answers Terms of Service.

If you feel that you have found inappropriate content, please let us know by emailing us at with the question ID listed above. Thank you.
Search Google Answers for
Google Answers  

Google Home - Answers FAQ - Terms of Service - Privacy Policy