Probability that A will win given a condition

[vcsharp2003]

Homework Statement

A and B having equal skill, are playing a game of three points. After A has won 2 points and B has won 1 point, what is the probability that A will win the game?

Above is problem# 22.18 from Schaum's Outline of College Mathematics. It's an unsolved problem in the book with the answer given as $\frac{3} {4}$

Relevant Equations

For an experiment that can have multiple outcomes, $P(X) = \frac {n(X)} {n(S)}$ where $P(X)$ is probability that event X will occur, $n(X)$ is number of ways in which event X can occur, $n(S)$ is total number of ways in which all possible events can occur

I tried to solve this problem using the chart given below. But I get a different answer of $\frac {2}{3}$ rather than $\frac {3}{4}$. Maybe the answer given is incorrect?

I determine from the chart the number of ways in which A could win given that A has already won 2 of first 3 points; from the chart it clearly shows circled A's as these wins. The number of circled A's is 6.

Now I determine the total number of ways in which all possible events can occur given that A has already won 2 of first 3 points. From the chart, it is clearly 9.

$$\therefore P(X) = \frac {6} {9} =\frac{2} {3}$$

 

[BvU]

You are counting the wrong way. A fifth round is only necessary in half of the cases !

$\$

 

[vcsharp2003]

You are counting the wrong way. A fifth round is only necessary in half of the cases !

$\$

I have shown 5th round denoted by row 5 in my chart, only when A or B has not won 3 points till the 4th round. So, I don't get where exactly is the extra counting.

 

[DrClaude]

I have shown 5th round denoted by row 5 in my chart, only when A or B has not won 3 points till the 4th round. So, I don't get where exactly is the extra counting.

You are not taking into account the probabilities. You are counting all event as equally probable, but the probability of A in round 4 is greater than the probability of A or B in round 5.

 

[vcsharp2003]

You are not taking into account the probabilities. You are counting all event as equally probable, but the probability of A in round 4 is greater than the probability of A or B in round 5.

The problem says that A and B have equal skills. Also each time they play a point, probability of A winning is 50% and the probability of B winning is 50%. Right? Each time A and B play for a point, it's like tossing a coin to get heads or tails (either A wins or B wins when they play for each point).

 

[FactChecker]

The problem says that A and B have equal skills. Also each time they play a point, probability of A winning is 50% and the probability of B winning is 50%. Right? Each time A and B play for a point, it's like tossing a coin to get heads or tails (either A wins or B wins when they play for each point).

Correct. With the given condition, A wins immediately half the time. A and B equally split the other half, so B only has 1/2 * 1/2 = 1/4 prob of winning, with A winning an additional 1/2*1/2=1/4 . In all, A wins 1/2+1/4=3/4 and B wins 1/4.

 

[vcsharp2003]

Correct. With the given condition, A wins immediately half the time. A and B equally split the other half, so B only has 1/2 * 1/2 = 1/4 prob of winning, with A winning an additional 1/2*1/2=1/4 . In all, A wins 1/2+1/4=3/4 and B wins 1/4.

With the given condition, according to the chart A can win 6 times, B can win 3 times. Could you please let me know where my chart is wrong? A chart or tree diagram or a table are ways we can count all outcomes, and that's what I tried doing with the chart.

 

[DrClaude]

With the given condition, according to the chart A can win 6 times, B can win 3 times. Could you please let me know where my chart is wrong? A chart or tree diagram or a table are ways we can count all outcomes, and that's what I tried doing with the chart.

Doing it with such a chart, you have to include the 5th round even in cases where A has already won

Round 4:  A    B    A    B    A    B
         / \  / \  / \  / \  / \  / \
Round 5: A B  A B  A B  A B  A B  A B

A wins:  * *  *    * *  *    * *  *

A wins 9 out of 12, so 3/4.

 

[FactChecker]

With the given condition, according to the chart A can win 6 times, B can win 3 times. Could you please let me know where my chart is wrong? A chart or tree diagram or a table are ways we can count all outcomes, and that's what I tried doing with the chart.

An A win on row 4 is worth twice as much as on row 5 because it happens twice as often. As @DrClaude said, you have to pay attention to the probabilities. If you count the row-4 A's twice and row-5 A's &B's once, you get 9 A's out of 12.

 

[vcsharp2003]

Doing it with such a chart, you have to include the 5th round even in cases where A has already won

I see. Could you please tell me why the 5th point has to be included even though the game is over by the 4th point? Once we have decided on the winner, then it seems there is no need to play another point.

 

[vcsharp2003]

An A win on row 4 is worth twice as much as on row 5 because it happens twice as often. As @DrClaude said, you have to pay attention to the probabilities. If you count the row-4 A's twice and row-5 A's &B's once, you get 9 A's out of 12.

Sorry, I don't get it. I thought we just need to count as per the chart, but you're saying to count A wins in 4th row twice. It doesn't make sense to me.

 

[FactChecker]

I see. Could you please tell me why the 5th point has to be included even though the game is over by the 4th point? Once we have decided on the winner, then it seems there is no need to play another point.

I like his explanation. Pretend that they continued to a 5th point even if unnecessary. Then all the 5th point results are equally likely, but an A on row 4 means an A win even if it is followed by a B on row 5. Counting the total A wins gives 9 out of 12.

 

[vcsharp2003]

I like his explanation. Pretend that they continued to a 5th point even if unnecessary. Then all the 5th point results are equally likely, but an A on row 4 means an A win even if it is followed by a B on row 5. Counting the total A wins gives 9 out of 12.

I get that if we include the 5th point for all cases, then A wins 9 out of 12 times. But, still I don't get why the 5th point needs to be played for cases where game is already decided by the 4th point.

 

[FactChecker]

I get that if we include the 5th point for all cases, then A wins 9 out of 12 times. But, still I don't get why the 5th point needs to be played for cases where game is already decided by the 4th point.

It gives you a set of equally likely results so you can just count up the A wins versus the total results. The row-5 results are not as likely as the row-4 results, so simply counting them is not correct. And it doesn't hurt anything to pretend that play always continued to the 5th row before the winner was announced.

 

[vcsharp2003]

Doing it with such a chart, you have to include the 5th round even in cases where A has already won

Round 4:  A    B    A    B    A    B
         / \  / \  / \  / \  / \  / \
Round 5: A B  A B  A B  A B  A B  A B

A wins:  * *  *    * *  *    * *  *

A wins 9 out of 12, so 3/4.

I think the 5th point needs to be included even if A has won by the 4th play since a tree diagram can only be used for counting if all paths are equally likely. I didn't know that a tree diagram needs to be showing equally likely paths when used in a probability problem.

 

[FactChecker]

I didn't know that a tree diagram needs to be showing equally likely paths when used in a probability problem.

Yes. If you are going to use simple counting to determine the probabilities then the things you count must have equal probabilities. That is true, not only for tree problems but for many other approaches.