probability

[perusal]

A and B play until one has 2 more points than the other. Assuming that each point is independently won by A with probability p, what is the probability they will play a total of 2n points? What is the probability that A wins?

I suggest the answer to the first question is 100% of the time. The lowest number of points that can be played with a winner is 2 then any other victory must be 2n greater than that number.

The second question has me stumped.

From the question:

"Assuming that each point is independently won by A with probability p"

A = $\displaystyle \mathscr{Pr}(p)$

B = $\displaystyle \mathscr{Pr}(p^c)$

The problem I have is that $\displaystyle A^3B$ can be arrived at in 2 ways. But as n gets larger and $\displaystyle A^{n+2}B^n$ the number of ways you can arrive at each result seemingly get endless.

I think I am definitely missing something.

 

[Romsek]

Is the 2nd question what is the probability that A wins at all or that A wins after 2n turns?

 

[perusal]

Is the 2nd question what is the probability that A wins at all or that A wins after 2n turns?

at all I believe.

 

[pka]

A and B play until one has 2 more points than the other. Assuming that each point is independently won by A with probability p, what is the probability they will play a total of 2n points? What is the probability that A wins?
The second question has me stumped.
From the question:
"Assuming that each point is independently won by A with probability p"
A = $\displaystyle \mathscr{Pr}(p)$
B = $\displaystyle \mathscr{Pr}(p^c)$.

A and B play until one has 2 more points than the other. Assuming that each point is independently won by A with probability p, what is the probability they will play a total of 2n points? What is the probability that A wins?

I suggest the answer to the first question is 100% of the time. The lowest number of points that can be played with a winner is 2 then any other victory must be 2n greater than that number.
The second question has me stumped.
From the question:
"Assuming that each point is independently won by A with probability p"
A = $\displaystyle \mathscr{Pr}(p)$
B = $\displaystyle \mathscr{Pr}(p^c)$
The problem I have is that $\displaystyle A^3B$ can be arrived at in 2 ways. But as n gets larger and $\displaystyle A^{n+2}B^n$ the number of ways you can arrive at each result seemingly get endless. I think I am definitely missing something.

I think that this is just a matter of a very poor translation. It seems that each game is a win or loose for either A or B.
So the model $\displaystyle ABABAA$ is six turns where $\displaystyle A$ wins in that $\displaystyle A$ has two more wins than $\displaystyle B$.
Whereas $\displaystyle ABABA$ none wins because neither has two more wins than the other.
Now each of these is a win for $\displaystyle A$: $\displaystyle AA$, $\displaystyle BAA$, $\displaystyle BABAAA$ $\displaystyle ABABAA$.
If that is an incorrect reading, please correct.

 

[JeffM]

Like pka, I am not quite sure what the problem means.

The probability that A and B will play exactly two rounds is

[MATH]p^2 + (1 - p)^2 = 2p^2 - 2p + 1 = 1 - 2p(1 - p).[/MATH]
For example, if p = 0.4, then the probability that A wins in two rounds is 0.16, and the probability B will win in two rounds is 0.36. However, A could win the first round and B the second with a probability 0.4 times 0.6 = 0.24, or vice versa with a probability of 0.6 times 0.4 = 0.24. To summarize, someone will win in two rounds with probability 0.16 + 0.36 = 0.52, which is equal to 1 - (2 * 0.4 * 0.6) = 1 - 0.48. No one wins in two rounds with probability 0.48.

What happens on the third round depends on the rules. If the rule is that whoever wins two rounds first wins, then someone wins in two rounds with probability 0.52 or someone wins in three rounds with probability 0.48. And the probability that someone wins in round 2n where n > 1 is zero. But perhaps the rule is that winning requires winning twice in a row. Things get more interesting.

 

[perusal]

Like pka, I am not quite sure what the problem means.

The probability that A and B will play exactly two rounds is

[MATH]p^2 + (1 - p)^2 = 2p^2 - 2p + 1 = 1 - 2p(1 - p).[/MATH]
For example, if p = 0.4, then the probability that A wins in two rounds is 0.16, and the probability B will win in two rounds is 0.36. However, A could win the first round and B the second with a probability 0.4 times 0.6 = 0.24, or vice versa with a probability of 0.6 times 0.4 = 0.24. To summarize, someone will win in two rounds with probability 0.16 + 0.36 = 0.52, which is equal to 1 - (2 * 0.4 * 0.6) = 1 - 0.48. No one wins in two rounds with probability 0.48.

What happens on the third round depends on the rules. If the rule is that whoever wins two rounds first wins, then someone wins in two rounds with probability 0.52 or someone wins in three rounds with probability 0.48. And the probability that someone wins in round 2n where n > 1 is zero. But perhaps the rule is that winning requires winning twice in a row. Things get more interesting.

It's a bit like a duece in tennis. When both players have 40 points, one player has to then score two points in sucession to win. The sequence goes, 40 points, then advantage and then a win. However, should the player lose a point whilst they have an advantage the score goes back to 40-40 or 40 points each.

 

[perusal]

I think that this is just a matter of a very poor translation. It seems that each game is a win or loose for either A or B.
So the model $\displaystyle ABABAA$ is six turns where $\displaystyle A$ wins in that $\displaystyle A$ has two more wins than $\displaystyle B$.
Whereas $\displaystyle ABABA$ none wins because neither has two more wins than the other.
Now each of these is a win for $\displaystyle A$: $\displaystyle AA$, $\displaystyle BAA$, $\displaystyle BABAAA$ $\displaystyle ABABAA$.
If that is an incorrect reading, please correct.

Yes someone has to win. They keep playing until someone has 2 more points. There is the possibility that they never stop playing as well I suppose.

It's a bit like a duece in tennis. When both players have 40 points, one player has to then score two points in sucession to win. The sequence goes, 40 points, then advantage and then a win. However, should the player lose a point whilst they have an advantage the score goes back to 40-40 or 40 points each.

Like you pointed out AA, BAA, ABABAA etc. are all wins.

However it is worth noting (if I have got this right) that ABAA and BAAA are two ways of obtaining $\displaystyle A^3B$ You can then multiply by AB an infinite amount of times and arrive at a win for A. However there are infinitely further options as the length of the game increases. ABBAABBAABBAABBAABABABABABABABABBAA and so on. So I am uncertain how to arrive at an answer.

 

[pka]

Well now it is a tennis match. Well I give up.
If in the future you do not supply all reverent details evolved in the question, then do not expect me to waste time trying to parse the question. You may want to find a translator who can help you cleanup your efforts.

 

[Steven G]

what is the probability they will play a total of 2n points?

I suggest the answer to the first question is 100% of the time.

What makes you think that there will be an even numbers of games played (2n is even)?
What about BAA or BABAA or BABABAA? Doesn't A win and the number of games played is not even?

 

[perusal]

Well now it is a tennis match. Well I give up.
If in the future you do not supply all reverent details evolved in the question, then do not expect me to waste time trying to parse the question. You may want to find a translator who can help you cleanup your efforts.

A and B play until one has 2 more points than the other. Assuming that each point is independently won by A with probability p, what is the probability they will play a total of 2n points? What is the probability that A wins?

That is exactly how the question is written in the textbook.

I am English so I would be a bit embarrassed getting a translator.

I mentioned tennis because I think that is what the writer had in mind when they came up with the question. I thought it would clarify the question. I guess not.

I don't expect you to waste your time and I appreciate the help you have given me.

 

[perusal]

What makes you think that there will be an even numbers of games played (2n is even)?
What about BAA or BABAA or BABABAA? Doesn't A win and the number of games played is not even?

I agree. The probability that they will play exactly two rounds is.

$\displaystyle p^2 + (1-p)^2$

What makes you think that there will be an even numbers of games played (2n is even)?
What about BAA or BABAA or BABABAA? Doesn't A win and the number of games played is not even?

I checked again. A player must have 2 more points than the other player to win. So in you examples BAA A=2 B=1, BABAA A=3 B=2, BABABAA A = 4 B =3 A only has one more point than B in each of these cases.

So A would not win in this case. I think there has to be an even amount of points played for there to be a winner.

$\displaystyle p^{n+2} * (1-p)^n$ or $\displaystyle (1-p)^{n+2} * p^n$

Are the only possible wins I think for all natural numbers n including 0

 

[perusal]

I think that this is just a matter of a very poor translation. It seems that each game is a win or loose for either A or B.
So the model $\displaystyle ABABAA$ is six turns where $\displaystyle A$ wins in that $\displaystyle A$ has two more wins than $\displaystyle B$.
Whereas $\displaystyle ABABA$ none wins because neither has two more wins than the other.
Now each of these is a win for $\displaystyle A$: $\displaystyle AA$, $\displaystyle BAA$, $\displaystyle BABAAA$ $\displaystyle ABABAA$.
If that is an incorrect reading, please correct.

BAA is not a win, because one player must have two more points than the opposing player to win. So if B has 2 points then A must have 4

The other examples you gave would be wins.

I think the answer to the first part of the question is that there is a 100% chance that a game resulting in a winning player has a total of 2n points played where n is a natural number not including 0.

$\displaystyle p^{n+2} * (1-p)^n$ or $\displaystyle (1-p)^{n+2} * p^n$

Are the only possible wins I think for all natural numbers n including 0

But "what is the probability that A wins?". I cannot answer.

 

[JeffM]

I am trying to get this.

Let a be the number of rounds won by player A when the game is over. Let b be the number of rounds won by player B when the game is over. Let n be the number of rounds that have been played when the game is over. By definition then

a + b = n.

The game is over when

a + b = n and |a - b| = 2. If a - b = 2, then A wins; but if b - a = 2, then B wins.

Do I have it now? By the way, that is not tennis.

[MATH]|a - b| = 2 \implies (a - b)^2 = 4 \implies a^2 - 2ab + b^2 = 4 \implies a^2 + b^2 = 2(ab + 4, \text { an even number.}[/MATH]
Thus, either both a and b are even, or else both are odd. In either case, n is even.

So I can see the 2n formulation, but I am not sure it is helpful, particularly not if we want to use mathematical induction.

But, assuming I have a clue, I agree that the probability that n is even is 100%.

The probability that A wins in 2 rounds means we had AA and so is [MATH]p^2.[/MATH]
The probability that A wins in 4 rounds means we had ABAA or BAAA and so is

[MATH]p(1 - p)pp + (1 - p)ppp = 2p^3(1- p)[/MATH].

The probability that A wins in 6 rounds means we had ABABAA, ABBAAA, BAABAA, or BABAAA and so is

[MATH]p(1 -p)p(1 - p)pp + p(1 - p)(1 - p)ppp + (1 - p)pp(1 - p)pp + (1 - p)p(1 - p)ppp = 4p^4(1 - p)^2.[/MATH]
Now that is not a proof, but I see a pattern, namely the probability that A will win in exactly 2k rounds is

[MATH]2kp^{(k+1)}(1 - p)^{(k-1)}.[/MATH]
See if you can prove that.

 

[Steven G]

Oops, my examples did not have a winner.

 

[perusal]

I am trying to get this.

Let a be the number of rounds won by player A when the game is over. Let b be the number of rounds won by player B when the game is over. Let n be the number of rounds that have been played when the game is over. By definition then

a + b = n.

The game is over when

a + b = n and |a - b| = 2. If a - b = 2, then A wins; but if b - a = 2, then B wins.

Do I have it now? By the way, that is not tennis.

[MATH]|a - b| = 2 \implies (a - b)^2 = 4 \implies a^2 - 2ab + b^2 = 4 \implies a^2 + b^2 = 2(ab + 4, \text { an even number.}[/MATH]
Thus, either both a and b are even, or else both are odd. In either case, n is even.

So I can see the 2n formulation, but I am not sure it is helpful, particularly not if we want to use mathematical induction.

But, assuming I have a clue, I agree that the probability that n is even is 100%.

The probability that A wins in 2 rounds means we had AA and so is [MATH]p^2.[/MATH]
The probability that A wins in 4 rounds means we had ABAA or BAAA and so is

[MATH]p(1 - p)pp + (1 - p)ppp = 2p^3(1- p)[/MATH].

The probability that A wins in 6 rounds means we had ABABAA, ABBAAA, BAABAA, or BABAAA and so is

[MATH]p(1 -p)p(1 - p)pp + p(1 - p)(1 - p)ppp + (1 - p)pp(1 - p)pp + (1 - p)p(1 - p)ppp = 4p^4(1 - p)^2.[/MATH]
Now that is not a proof, but I see a pattern, namely the probability that A will win in exactly 2k rounds is

[MATH]2kp^{(k+1)}(1 - p)^{(k-1)}.[/MATH]
See if you can prove that.

Thank you.

I followed your reasoning and it leads me to the following.

For the probability that A wins in 8 rounds we have 8 possible sequence of events: ABABABAA, ABBAABAA, ABABBAAA, ABBABAAA, BABABAAA, BAABBAAA, BABAABAA and BAABABAA

For 10 rounds played there are 16 winning outcomes for A.

With each additional 2 points played we have $\displaystyle 2^n$ as the possible number of winning outcomes for A in that specific round when n is equal to $\displaystyle \frac{points played}{2} - 1$

Below is the pattern you gave.

$\displaystyle 2kp^{k+1}(1−p)^{k−1}$

I think this is flawed because when k is equal to 1 you get $\displaystyle 2p^2 * (1-p)$
I suggest the following.

Probability that A wins is equal to $\displaystyle \sum_{n=0}^{\infty} 2^n * p^{n+2} * (1-n)^{n}$

I think this is a geometric with common ratio $\displaystyle 2*p*(1-p)$ and starting value $\displaystyle p^2$.

Therefore:

Probability that A wins is equal to: $\displaystyle \frac{p^2}{1+2p^2-2p}$

What do you think? I'm not 100% I got this right.

However, when p = 1/2 is plugged into the equation, you get a result of 1/2 which is promising.

 

[perusal]

Thank you.

I followed your reasoning and it leads me to the following.

For the probability that A wins in 8 rounds we have 8 possible sequence of events: ABABABAA, ABBAABAA, ABABBAAA, ABBABAAA, BABABAAA, BAABBAAA, BABAABAA and BAABABAA

For 10 rounds played there are 16 winning outcomes for A.

With each additional 2 points played we have $\displaystyle 2^n$ as the possible number of winning outcomes for A in that specific round when n is equal to $\displaystyle \frac{points played}{2} - 1$

Below is the pattern you gave.

$\displaystyle 2kp^{k+1}(1−p)^{k−1}$

I think this is flawed because when k is equal to 1 you get $\displaystyle 2p^2 * (1-p)$
I suggest the following.

Probability that A wins is equal to $\displaystyle \sum_{n=0}^{\infty} 2^n * p^{n+2} * (1-n)^{n}$

I think this is a geometric with common ratio $\displaystyle 2*p*(1-p)$ and starting value $\displaystyle p^2$.

Therefore:

Probability that A wins is equal to: $\displaystyle \frac{p^2}{1+2p^2-2p}$

What do you think? I'm not 100% I got this right.

However, when p = 1/2 is plugged into the equation, you get a result of 1/2 which is promising.

Sorry just to make a correction.

$\displaystyle 2kp^{k+1}(1−p)^{k−1}$

This actually equals $\displaystyle p^2$ when k = 1

However, I still believe it to be flawed as I have hopefully made clear in my previous post.

 

[perusal]

I am trying to get this.

Let a be the number of rounds won by player A when the game is over. Let b be the number of rounds won by player B when the game is over. Let n be the number of rounds that have been played when the game is over. By definition then

a + b = n.

The game is over when

a + b = n and |a - b| = 2. If a - b = 2, then A wins; but if b - a = 2, then B wins.

Do I have it now? By the way, that is not tennis.

[MATH]|a - b| = 2 \implies (a - b)^2 = 4 \implies a^2 - 2ab + b^2 = 4 \implies a^2 + b^2 = 2(ab + 4, \text { an even number.}[/MATH]
Thus, either both a and b are even, or else both are odd. In either case, n is even.

So I can see the 2n formulation, but I am not sure it is helpful, particularly not if we want to use mathematical induction.

But, assuming I have a clue, I agree that the probability that n is even is 100%.

The probability that A wins in 2 rounds means we had AA and so is [MATH]p^2.[/MATH]
The probability that A wins in 4 rounds means we had ABAA or BAAA and so is

[MATH]p(1 - p)pp + (1 - p)ppp = 2p^3(1- p)[/MATH].

The probability that A wins in 6 rounds means we had ABABAA, ABBAAA, BAABAA, or BABAAA and so is

[MATH]p(1 -p)p(1 - p)pp + p(1 - p)(1 - p)ppp + (1 - p)pp(1 - p)pp + (1 - p)p(1 - p)ppp = 4p^4(1 - p)^2.[/MATH]
Now that is not a proof, but I see a pattern, namely the probability that A will win in exactly 2k rounds is

[MATH]2kp^{(k+1)}(1 - p)^{(k-1)}.[/MATH]
See if you can prove that.

I misread your definition of k sorry.

you are quite right.

$\displaystyle 2kp^{n+1}*(1-p)^{n-1}$

I just changed the meanings of the variables slightly to make it easier for me to form a geometric series.

 

[JeffM]

Yes, I think I made an error.

[MATH]2^{(k-1)}p^{(k+1)}(1 - p)^{(k-1>}[/MATH]
looks better for the probability of A winning in 2k rounds.

[MATH]2^{(k-1)}p^{(k-1)}(1 - p)^{(k+1)}[/MATH]
is then the probability of B winning in 2k rounds.

But that is not a proof.

If k = 1, the probability that A wins in exactly 2k rounds is

[MATH]p^2 = 1 * p^2 * 1 = 2^0 * p^2 * (1 - p)^0 = 2^{(k-1)}p^{(k+1)}(1 - p)^{(k-1)}.[/MATH]
If k = 1, the probability that B wins in exactly 2k rounds is

[MATH](1 - p)^2 = 2^{(k-1)}p^{(k-1)}(1 - p)^{(k+1)}.[/MATH]
If k = 1, the probability that no one wins in a 2k rounds is

[MATH]1 - p^2 - (1 - p)^2 = 2p - 2p^2 = 2p(1 - p) = \sum_{j=1}^k \{2p(1 - p)\}^j.[/MATH]
This is what must be generalized.

If k = 2, the probability that A wins in exactly 2k rounds is

[MATH]2p(1 - p) * p^2 = 2^{(k-1)}p^{(k+1)}(1 - p)^{(k-1)}.[/MATH]
If k = 2, the probability that B wins in exactly 2k rounds is

[MATH]2p(1 - p) * (1 - p)^2 = 2^{(k-1)}p^{(k - 1)}(1 - p)^{(k+1)}.[/MATH]
If k = 2, the probability that no one wins in 2k rounds is

[MATH]2p(1 - p) + 2p(1 - p)\{1 - p^2 - (1 - p)^2\} = \sum_{j=1}^k\{2p(1 - p)\}^j.[/MATH]
You could try induction now.

Alternatively, you could try to show that, for finite k, the probability that A wins is less than p, and the probability that B wins is less than (1 - p), but that, in the limit, the probability that no one wins is zero. If I knew any analysis, I could probably show that, in the limit, the probability that A wins approaches p and the probability that B wins approaches (1 - p). But I don't know analysis.

 

[JeffM]

My alternative approach requires some new definitions

[MATH]q = 1 - p.[/MATH]
[MATH]r_k = \text {P(A wins in no more than 2k rounds).}[/MATH]
[MATH]s_k = \text {P(B wins in no more than 2k rounds).}[/MATH]
[MATH]u_k = \text {P(no one wins in 2k rounds).}[/MATH]
[MATH]r_k + s_k + u_k = 1.[/MATH]
[MATH]r_1 = p^2,\ s_1 = q^2 = (1 - p)^2 = 1 - 2p + p^2,\ u_1 = 1 - (r_1 + s_1) = 2p(1 - p) = 2pq.[/MATH]
[MATH]0 < p < 1 \implies 0 < p < 1 \text { and } 0 < q < 1 \implies[/MATH]
[MATH]0 < p^2 < p \text { and } 0 < q^2 < q \implies r_1 < p \text { and } s_1 < q.[/MATH]
[MATH]\delta = p - 0.5 \implies -\ 0.5 < \delta < 0.5, p = 0.5 + \delta, \text { and } q = 1 - p = 0.5 - \delta.[/MATH]
[MATH]pq = (0.5 + \delta)(0.5 - \delta) = 0.25 - \delta^2.[/MATH]
[[MATH]0 \le \delta^2 < 0.25 \implies -\ 0.25 \le \delta^2 - 0.25 < 0 \implies 0 < pq \le 0.25 \implies[/MATH]
[MATH]0 < 2pq \le 0.5 \implies 0 < u_1 \le 0.5[/MATH]
[MATH]r_{j+1} = r_j + u_jp^2, \ s_{j+1} = s_j + u_jq^2, \text { and }[/MATH]
[MATH]u_{j+ 1} = u_j + u_j(1 - p^2 - q^2) = u_j(u_j + u_1).[/MATH]
This is all I have time for now. I'll check it later. What I am hoping to show is that r_k < p, s_k < q, and the limit of u_k is zero, which implies, I think, that the limit of r_k is p and the limit of s_k is q.[/MATH]

 

[JeffM]

Still busy on other things, but there is an error above.

[MATH]u_{j+1} = 1 - r_{j+1} - s_{j+1} = 1 - r_j - u_jp^2 - s_j - u_jq^2 =[/MATH]
[MATH]u_{j+1} = u_j + r_j + s_j - r_j - s_j - u_jp^2 - u_jq^2 =[/MATH]
[MATH]u_j - u_jp^2 - u_jq^2 = ]u_j(1 - p^2 - q^2).[/MATH]
Let's check.

[MATH]r_{j+1} + s_{j+1} + u_{j + 1} =[/MATH]
[MATH]r_j + u_jp^2 + s_j + u_jq^2 + u_j(1 - p^2 - s^2) =[/MATH]
[MATH]r_j + s_j + u_j + u_jp^2 - u_jp^2 + u_jq^2 -u_jq^2 =[/MATH]
[MATH]r_j + s_j + u_j = 1.[/MATH]
[MATH]\text {But } (1 - p^2 - q^2) = 2pq \implies u_{j+1} = 2pqu_j.[/MATH]
Now we can do induction on u_k.

[MATH]k = 1 \implies u_1 = 2pq = (2pq)^1.[/MATH]
[MATH]\therefore \exists \ m \in \mathbb Z^+ \text { such that } u_m = (2pq)^m.[/MATH]
[MATH]\text {But } u_{m+1} = 2pqu_m = 2pq * (2pq)^m = (2pq)^{(m+1)}.[/MATH]
Thus by induction we have

[MATH]u_k = (2pq)^k \text { for all } k \in \mathbb Z^+[/MATH]
[MATH]\therefore \lim_{k \rightarrow \infty} u_k = 0 \ \because \ 0 < 2pq \le 0.5.[/MATH]