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Q: Probability of Winning a Game (Revised) ( No Answer,   2 Comments ) Question
 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. ```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 p(a>b)=5/9 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 p(a>c)=4/9 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 p(b>c)=5/9 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.```
 ```wobbleone, 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.``` 