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Topic: A probability problem in chess competition (Read 1297 times) |
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Wonderer
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A probability problem in chess competition
« on: Dec 17th, 2005, 12:29am » |
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A probability problem in chess competition
Two teams A and B are playing against each other in the final show down of a chess competition. Each team has 5 players. Each player is given a number, from 1 ~ 9 according to his chess playing skill. 1 is the poorest while 9 is the strongest.
Team A’s players have skill numbers 3, 4, 5, 6, 7
Team B’s players have skill numbers 3, 6, 3, 4, 9
The probability of beating your opponent is directly resulted from the skill numbers. For example, if 4 vs 2, 4 has a 2/3 chance of winning. 6 will have a 3/5 chance of beating 4. etc. This chess game has NO DRAW.
The way the game goes is each team pre-arranges who their first player is, who their second player is … after they decide, the first players from each team face off. After winning, the first player will face the second player from the other team. This goes on until all 5 players in one team get beaten. Remember, there is NO DRAW. Every game will knock one player out.
The question:
What is the probability for team A to lose in the final? Can anyone solve this problem in 4 days? Thanks.
(Excuse my English, I hope I explained the question clearly. Sorry)
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JocK
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Re: A probability problem in chess competition
« Reply #1 on: Dec 17th, 2005, 2:18am » |
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Let's start with a more manageable problem first:
team A has two players with strengths 5 and 1.
team B has two players with strengths 4 and 2.
A player with strength p playing against a player with strength q has change of winning equal to p/(p + q).
If the team are lined up such that the strongest players meets first, team B will win with probability 18/35.
Team A can anticipate this and line up their weakest player first, such that 1 plays 4 in the first round. The result is that team A will win with probability 23/45.
Team B can not do better than this: lining up their weakest player first (such that 1 plays 2 in the first round) will even improve the likelihood of team A to win (probability 57/105).
So, the optimal line-up for both team is (1, 4) in the first round. Team A has a a probability 23/45 to win.
Increasing the number of players will make solving the problem much more tedious. I doubt wheter it would be interesting. The only additional phenomenon I can see happen is that you get a 'rock-scissors-paper effect'. This will cause both teams to adopt a probabilistic strategy for line-up of their players.
Sorry, I'm way too lazy for this...
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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Joe Fendel
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Re: A probability problem in chess competition
« Reply #2 on: Dec 17th, 2005, 7:39am » |
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Nice start, Jock. I agree that this seems tedious, but if I had to guess on how to extrapolate, it looks like perhaps it is favorable for teams with wide skill ranges to line up worst-to-best, and narrow skill ranges best-to-worst.
So based only on Jock's work, I'd line them up:
A: 7,6,5,4,3
B: 3,3,4,6,9
But I'm not gonna check the probabilities. 
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Joe Fendel
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Re: A probability problem in chess competition
« Reply #3 on: Dec 17th, 2005, 8:01am » |
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I lied! I checked the probabilities doing a simple Excel sheet. A will win with probability approx. 50.61729%
But what's more, the order doesn't seem to matter! I tried shuffling the players around and I always get the same number!
I rechecked Jock's work. I think he's mistaken. I get a 164/315 (about 52.0635%) chance of victory for the 4-2 squad no matter how either team lines up.
Not sure yet why lineup doesn't matter...
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Wonderer
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Re: A probability problem in chess competition
« Reply #5 on: Dec 17th, 2005, 3:11pm » |
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on Dec 17th, 2005, 2:54pm, towr wrote:
Hehehe.... kind of. 
Thanks for everyone's help.
This is what I have: 1 - 12768313/25225200 . exactly the same as Joe Fendel's result. The default player order from each team is used to get this answer. Now, I guess, the harder part would be why order matters not. 
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