[bushindo]

Saw this on another site.

Players A and B are taking part in a competition consisting of 11 different games. Each of the 11 games is always won outright by either A or B (i.e. a draw is not possible). 1 point is scored for winning a game, 0 points for a loss. So the overall competition has to be won outright by either A or B. Whoever has a higher score at the end of 11 games wins the competition.

Player A's win odds for all 11 games are known (see below), and it follows that Player B's win odds for any game n is [ 1 - P(A wins game n) ].

P(A wins game 1) : 0.9

P(A2) : 0.9

P(A3) : 0.9

P(A4) : 0.9

P(A5) : 0.7

P(A6) : 0.6

P(A7) : 0.5

P(A8) : 0.2

P(A9) : 0.2

P(A10): 0.2

P(A11): 0.1

What is the probability of A winning the overall competition (having a higher total score than B)?

Edited May 16, 2009 by bushindo

[Guest]

55.45%

[Guest]

Chance of A winning the first six straight appears to be about 81.66%. This seems to be good odds. The chance of A winning 6 out of 11 appears to be about 106.66%

Edited May 16, 2009 by PLMerry

[Guest]

Probability wise, A has a chance of winning 7 hands out of 11, where as B has only a 5 out of 11 chance. (I get 12 because of the 50/50 hand) So I would say A has about a 63% chance of winning it all.

I don't even know if I did that right after reading it again, but oh well I'll leave it there.

[Guest]

if they play 10 times at each game, player A will win game 1 9 times, game 2 9 times, and so on. They will have then played 110 games, of which player A will have won 61. Odds are 61/110 or just a bit under 55 1/2% that player B will come in second.

[bushindo]

if they play 10 times at each game, player A will win game 1 9 times, game 2 9 times, and so on. They will have then played 110 games, of which player A will have won 61. Odds are 61/110 or just a bit under 55 1/2% that player B will come in second.

61 is actually the average number of wins for A over 10 repetition of the tournament. Dividing that over the number of games does not give the expect change of winning a single competition. The chance of A winning a single tournament is actually higher than that.

[unreality]

P(A1) : 0.9

P(A2) : 0.9

P(A3) : 0.9

P(A4) : 0.9

P(A5) : 0.7

P(A6) : 0.6

P(A7) : 0.5

P(A8) : 0.2

P(A9) : 0.2

P(A10): 0.2

P(A11): 0.1

if a random game is picked out of the 11, A's average chance of winning is 6.1/11 = .5545, so B's average chance of winning is 4.9/11 or .4455 or so

say that 'a' is .5545 and 'b' is .4455

for A to win the competition, A needs to win 6,7,8,9,10 or 11 games

(a^6)(b^5) + (a^7)(b^4) + (a^8)(b^3) + (a^9)(b^2) + (a^10)(b^1) + (a^11) should give the right answer if I'm right (hell I'm probably wrong, maybe getting the average first was a bad idea). Also I might need to add the binomial coefficients from the 11th row of Pascal's triangle starting from the sixth and moving right, but I'll just try with coefficients of 1 and see if it's the correct answer. Probability isn't my strong suit

anyway I calculated the above as:

0.0056667911

or half of 1%.

Yeah that's got to be wrong. I bet the binomial coefficients have to be put on (but not the coeffiecients for .5 but the coefficients for .5545 I think). Oh well I tried

[bushindo]

if a random game is picked out of the 11, A's average chance of winning is 6.1/11 = .5545, so B's average chance of winning is 4.9/11 or .4455 or so

say that 'a' is .5545 and 'b' is .4455

for A to win the competition, A needs to win 6,7,8,9,10 or 11 games

(a^6)(b^5) + (a^7)(b^4) + (a^8)(b^3) + (a^9)(b^2) + (a^10)(b^1) + (a^11) should give the right answer if I'm right (hell I'm probably wrong, maybe getting the average first was a bad idea). Also I might need to add the binomial coefficients from the 11th row of Pascal's triangle starting from the sixth and moving right, but I'll just try with coefficients of 1 and see if it's the correct answer. Probability isn't my strong suit

anyway I calculated the above as:

0.0056667911

or half of 1%.

Yeah that's got to be wrong. I bet the binomial coefficients have to be put on (but not the coeffiecients for .5 but the coefficients for .5545 I think). Oh well I tried

This approach of randomly drawing a game from the 11 and calculating its chance of success (.55) is a novel one. I didn't consider it. If we assume that we can do repeat random draws and then use the binomial theorem as unreality suggested, the chance of winning the tournament for A is about 64%. It's a bit off from the true percentage, but it's way closer than previous attempts.

The reason the answer is off because this approach isn't a true binomial experiment. While the first random draw from the pool of 11 games has a chance of success = .55, subsequent repetition of this exercise draw from an increasingly smaller pool of games, and thus the chance of success changes.

I haven't figured out an exact solution to this problem. But I have an approximate solution that is accurate to within 2 decimal digits, and increases in accuracy as the number of game increases. Hint = normal approximation.

[Guest]

I get around 80% now. Is that anywhere near what you get?

[Guest]

...approximately 68.64%.

I wish I had an interesting story on how I figured this out, but basically I spent a half-hour figuring the odds for every W/L scenario in Excel.

Spoiler for As a bonus, the odds of A winning exactly n number of games::

0: <0.01%

1: 0.01%

2: 0.19%

3: 1.69%

4: 8.10%

5: 21.37%

6: 31.05%

7: 24.58%

8: 10.40%

9: 2.33%

10: .26%

11: .01%

M. Dale

Edited May 18, 2009 by M. Dale

[bushindo]

I wish I had an interesting story on how I figured this out, but basically I spent a half-hour figuring the odds for every W/L scenario in Excel.

n number of games::

0: <0.01%

1: 0.01%

2: 0.19%

3: 1.69%

4: 8.10%

5: 21.37%

6: 31.05%

7: 24.58%

8: 10.40%

9: 2.33%

10: .26%

11: .01%

...approximately 68.64%.