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statistics and odds question

Assume that player A has 70% chance to win any single match and player B has a 65% chance to win any single match.

What are the % chance that player A will win the match between player A and B? 

What are the % chance that player B will win the match between player A and B?

Anyone know a general math to do the above?

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My feeling is that the ratio would be 0.7 × 0.35 to 0.65 × 0.3 or 49:39; 55.7% for A, 44.3% for B.

That is, it sounds like a binomial tree problem.

Simplify the complicated side; don't complify the simplicated side.

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Another problem, the last one:

If player A has a 75% chance of scoring 1 goal and 25% chance of scoring 2 goals

And player B has a 55% chance of scoring 1 goal and 45% chance of scoring 2 goals:

What are the odds of the score 1-1

What are the chance of the score 1-2

What are the chances of 2-1

What are the chances 2-2

I thought it would be for the 1-1 instance , .75*.5 + .55 * .5 = 65%

And I thought 1-2 would be .75 * .55 + .45 * .5 = 60%

but the math doesnt seem right because that is already overt 100% There’s something I’m missing in the math.

Actually I think I just figured it out.

The combined chances of 1-1 and 1-2 is 75%, therefore, 1-1 is 41.25% while 1-2 is 33.75%

And the odds of 2-1 and 2-2, combined is 25%.. therefore 2-1 is 13.75% and 2-2 is 11.25%

What u think, is that right?

S2000magician wrote:

My feeling is that the ratio would be 0.7 × 0.35 to 0.65 × 0.3 or 49:39; 55.7% for A, 44.3% for B.

That is, it sounds like a binomial tree problem.

How would you do that tree given 3 ways:

For any game, player A wins 60%, Loses 20% and Draws 20%

For any game, player B Wins 30%, Loses 25%, Draws 45%

What are the probability player A will win, lose or draw if playing player B.

What are athe probability player B will win, lose or draw if playing player A.

125mph wrote:
Another problem, the last one:

If player A has a 75% chance of scoring 1 goal and 25% chance of scoring 2 goals

And player B has a 55% chance of scoring 1 goal and 45% chance of scoring 2 goals:

What are the odds of the score 1-1

What are the chance of the score 1-2

What are the chances of 2-1

What are the chances 2-2

I thought it would be for the 1-1 instance , .75*.5 + .55 * .5 = 65%

And I thought 1-2 would be .75 * .55 + .45 * .5 = 60%

but the math doesnt seem right because that is already overt 100% There’s something I’m missing in the math.

Assuming the goal-scoring is statistically independent, this is clearly a binomial tree problem:

  • A scores 1: 75%
    • B scores 1: 55%
    • B scores 2: 45%
  • A scores 2: 25%
    • B scores 1: 55%
    • B scores 2: 45%

P(1–1) = 75% × 55% = 41.25%

P(1–2) = 75% × 45% = 33.75%

P(2–1) = 25% × 55% = 13.75%

P(2–2) = 25% × 45% = 11.25%

Simplify the complicated side; don't complify the simplicated side.

Financial Exam Help 123: The place to get help for the CFA® exams
http://financialexamhelp123.com/

125mph wrote:
S2000magician wrote:
My feeling is that the ratio would be 0.7 × 0.35 to 0.65 × 0.3 or 49:39; 55.7% for A, 44.3% for B.

That is, it sounds like a binomial tree problem.

How would you do that tree given 3 ways:

For any game, player A wins 60%, Loses 20% and Draws 20%

For any game, player B Wins 30%, Loses 25%, Draws 45%

What are the probability player A will win, lose or draw if playing player B.

What are athe probability player B will win, lose or draw if playing player A.

This sounds like a trinomial tree problem:

  • A wins: 60%
    • B wins: 30%
    • B draws: 45%
    • B loses: 25%
  • A draws: 20%
    • B wins: 30%
    • B draws: 45%
    • B loses: 25%
  • A loses: 20%
    • B wins: 30%
    • B draws: 45%
    • B loses: 25%

P(A wins & B loses) = (60% × 25%) / ((60% × 25%) + (20% × 45%) + (20% × 30%)) = 50%

P(A draws & B draws) = (20% × 45%) / ((60% × 25%) + (20% × 45%) + (20% × 30%)) = 30%

P(A loses & B wins) = (20% × 30%) / ((60% × 25%) + (20% × 45%) + (20% × 30%)) = 20%

Simplify the complicated side; don't complify the simplicated side.

Financial Exam Help 123: The place to get help for the CFA® exams
http://financialexamhelp123.com/

S2000magician wrote:

P(A wins & B loses) = (60% × 25%) / ((60% × 25%) + (20% × 45%) + (20% × 30%)) = 50%

P(A draws & B draws) = (20% × 45%) / ((60% × 25%) + (20% × 45%) + (20% × 30%)) = 30%

P(A loses & B wins) = (20% × 30%) / ((60% × 25%) + (20% × 45%) + (20% × 30%)) = 20%

Nice one! This explanation was a lot better than the original binomial with the 49:39! lol.

125mph wrote:
S2000magician wrote:
P(A wins & B loses) = (60% × 25%) / ((60% × 25%) + (20% × 45%) + (20% × 30%)) = 50%

P(A draws & B draws) = (20% × 45%) / ((60% × 25%) + (20% × 45%) + (20% × 30%)) = 30%

P(A loses & B wins) = (20% × 30%) / ((60% × 25%) + (20% × 45%) + (20% × 30%)) = 20%

Nice one! This explanation was a lot better than the original binomial with the 49:39! lol.

I just hope that it’s correct!

Simplify the complicated side; don't complify the simplicated side.

Financial Exam Help 123: The place to get help for the CFA® exams
http://financialexamhelp123.com/