# 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?

You’ve made it this far, and you know what it takes to pass. Don’t be fooled by false promises and unrealistic claims. Schweser’s study packages give you the proven study tools and expert instruction you need to finish the job.

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.

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

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/