Determine the probability that Andy wins the match

[chwala]

Homework Statement

See attached

Relevant Equations

probability

My interest is on the highlighted part only...the other questions are well understood. Find ms solution here;

Even this is well understood...they made use of sum to infinity to arrive at the solution. I am interested on an alternative approach. Cheers guys.

 

[Orodruin]

Consider the probability that Andy wins. How can this be expressed in terms of the probabilities for the outcomes of the next game?

In particular, if the next game is a draw, what is the probability that Andy will win?

 

[chwala]

Consider the probability that Andy wins. How can this be expressed in terms of the probabilities for the outcomes of the next game?

In particular, if the next game is a draw, what is the probability that Andy will win?

I think ms addressed that...it looked at all possibilities! i.e possibilities of Andy winning on the first game... then not winning first game but winning on second game, ...not winning on second game but winning on third game ...I understand that...looks like its the only approach on the problem.

 

[Orodruin]

I think ms addressed that...it looked at all possibilities! i.e possibilities of Andy winning on the first game... then not winning first game but winning on second game, ...not winning on second game but winning on third game ...I understand that...looks like its the only approach on the problem.

No, that is not what I am suggesting. Just look at the first game. What is the probability that Andy wins if the first game is a draw? Can you relate it to the probability that Andy wins?

 

[chwala]

No, that is not what I am suggesting. Just look at the first game. What is the probability that Andy wins if the first game is a draw? Can you relate it to the probability that Andy wins?

Ok, I will...talk later cheers mate.

 

[pasmith]

The winner of the match is the winner of the final game.
A game is the final game if and only if it is not a draw.

The probabilities of these events are independent of the number of previously drawn games.

 

[chwala]

No, that is not what I am suggesting. Just look at the first game. What is the probability that Andy wins if the first game is a draw? Can you relate it to the probability that Andy wins?

We can also have;

$P_{Andy win} = 1-P_{Bev win}$

$P_{Bev win} = 0.36+0.4×0.6×0.6=0.504$

$P_{Andy win}=1-\dfrac{0.504}{1-0.24}$

$P_{Andy win}=1-\dfrac{0.504}{0.76}$

$P_{Andy win}=1-0.663157$

$P_{Andy win}=0.336843≅0.337$

 

[Orodruin]

We can also have;

$P_{Andy win} = 1-P_{Bev win}$

$P_{Bev win} = 0.36+0.4×0.6×0.6=0.504$

$P_{Andy win}=1-\dfrac{0.504}{1-0.24}$

$P_{Andy win}=1-\dfrac{0.504}{0.76}$

$P_{Andy win}=1-0.663157$

$P_{Andy win}=0.336843≅0.337$

It is not clear to me what exactly you are trying to do here. Why don't you try to answer the questions I already posed?

 

[chwala]

It is not clear to me what exactly you are trying to do here. Why don't you try to answer the questions I already posed?

This is a different approach to the problem...looks like I did not understand your question...I will re look at it later...

 

[chwala]

We can also have;

$P_{Andy win} = 1-P_{Bev win}$

$P_{Bev win} = 0.36+0.4×0.6×0.6=0.504$

$P_{Andy win}=1-\dfrac{0.504}{1-0.24}$

$P_{Andy win}=1-\dfrac{0.504}{0.76}$

$P_{Andy win}=1-0.663157$

$P_{Andy win}=0.336843≅0.337$

I used this approach, just to be clear;

The P[Andy winning] = [1 - P(Bev Winning)]

In that case we shall have,

$P_{ Andy Win} =1-[0.504 + 0.24 ×0.504+ 0.24^2×0.504+0.24^3×0.504+...]$ that will eventually bring you up to the working that i indicated as attached.

 

[chwala]

No, that is not what I am suggesting. Just look at the first game. What is the probability that Andy wins if the first game is a draw? Can you relate it to the probability that Andy wins?

I want to get it right, If the first game is a draw then the possibilities of Andy Winning are;

[Draw or Draw Win or Draw Draw Win or Draw Draw Draw Win...]

On a side note,

$P_{Andy win}= 0.256 + 0.24 ×0.256+ 0.24^2×0.256+0.24^3×0.256+...]$ I hope you did not mean that i show this... as i had earlier stated that i fully understand this.

I was looking for an alternative approach which seemingly has been done by me on post $7$ and $10.$

 

[pasmith]

Andy wins the match by winning the final game. By definition the final game is won either by Andy or by Beth, so the probability of Andy winning this game is \begin{split}<br /> P(\mbox{Andy wins the final game}) &amp;=<br /> P(\mbox{Andy wins game}\,|\,\mbox{Andy or Beth wins game}) \\<br /> &amp;= <br /> \frac{P(\mbox{Andy wins game})}{1 - P(\mbox{Neither player wins game})}.\end{split} The probability that the final game is played is <br /> P(\mbox{The final game is played}) = 1 - \lim_{n \to \infty} P(\mbox{Neither player wins game})^n = 1. Putting these together: <br /> \begin{split}<br /> P(\mbox{Andy wins match}) &amp;= P(\mbox{Andy wins the final game})P(\mbox{The final game is played}) \\<br /> &amp;= \frac{P(\mbox{Andy wins game})}{1 - P(\mbox{Neither player wins game})}.\end{split} As expected, this is equal to<br /> P(\mbox{Andy wins game}) \sum_{n=0}^\infty P(\mbox{Neither player wins game})^n<br /> .

 

[chwala]

Andy wins the match by winning the final game. By definition the final game is won either by Andy or by Beth, so the probability of Andy winning this game is \begin{split}<br /> P(\mbox{Andy wins the final game}) &amp;=<br /> P(\mbox{Andy wins game}\,|\,\mbox{Andy or Beth wins game}) \\<br /> &amp;=<br /> \frac{P(\mbox{Andy wins game})}{1 - P(\mbox{Neither player wins game})}.\end{split} The probability that the final game is played is <br /> P(\mbox{The final game is played}) = 1 - \lim_{n \to \infty} P(\mbox{Neither player wins game})^n = 1. Putting these together: <br /> \begin{split}<br /> P(\mbox{Andy wins match}) &amp;= P(\mbox{Andy wins the final game})P(\mbox{The final game is played}) \\<br /> &amp;= \frac{P(\mbox{Andy wins game})}{1 - P(\mbox{Neither player wins game})}.\end{split} As expected, this is equal to<br /> P(\mbox{Andy wins game}) \sum_{n=0}^\infty P(\mbox{Neither player wins game})^n<br /> .

@pasmith eeeèeish! Looks tough there...cheers mate

 

[PeroK]

The idea that @Orodruin was trying to get you to consider is that the state of the match after a draw is equivalent to the state of the match at the beginning. So that Andy has the same probability of winning after a draw as he did at the start. Hence:
$$P(\text{Andy wins match}) = P(\text{Andy wins game}) + P(\text{Draw})P(\text{Andy wins match})$$And we can use this simple equation rather than an infinite sequence.

 

[Orodruin]

The idea that @Orodruin was trying to get you to consider is that the state of the match after a draw is equivalent to the state of the match at the beginning. So that Andy has the same probability of winning after a draw as he did at the start. Hence:
$$P(\text{Andy wins match}) = P(\text{Andy wins game}) + P(\text{Draw})P(\text{Andy wins match})$$And we can use this simple equation rather than an infinite sequence.

^^ What he said.

 

[WWGD]

We can also have;

$P_{Andy win} = 1-P_{Bev win}$

$P_{Bev win} = 0.36+0.4×0.6×0.6=0.504$

$P_{Andy win}=1-\dfrac{0.504}{1-0.24}$

$P_{Andy win}=1-\dfrac{0.504}{0.76}$

$P_{Andy win}=1-0.663157$

$P_{Andy win}=0.336843≅0.337$

Not sure. The game may end up in a draw with nonzero probability , so your top equation won't hold up.

 

[chwala]

The idea that @Orodruin was trying to get you to consider is that the state of the match after a draw is equivalent to the state of the match at the beginning. So that Andy has the same probability of winning after a draw as he did at the start. Hence:
$$P(\text{Andy wins match}) = P(\text{Andy wins game}) + P(\text{Draw})P(\text{Andy wins match})$$And we can use this simple equation rather than an infinite sequence.

This gives you;
$0.256+(0.24×0.256)=0.31744≠0.337$ unless i am missing the point...we do require to find sum to infinity as shown in the ms guide.

 

[chwala]

Not sure. The game may end up in a draw with nonzero probability , so your top equation won't hold up.

For sure i am not getting you...you mean this alternative method is wrong? Kindly be specific.

 

[PeroK]

This gives you;
$0.256+(0.24×0.256)=0.31744≠0.337$ unless i am missing the point...

I think you are missing the point and confusing "wins the match" with "wins the game".

 

[Orodruin]

This gives you;
$0.256+(0.24×0.256)=0.31744≠0.337$ unless i am missing the point...we do require to find sum to infinity as shown in the ms guide.

No, as @PeroK said you are confusing "wins game" with "wins match" in the second term of the RHS. You have P(wins match) on both sides so you must solve for it algebraically. Only after doing that can you start inserting numbers.

 

[WWGD]

For sure i am not getting you...you mean this alternative method is wrong? Kindly be specific.

When you write Pa=1-Pb ( Andy, Bevin), you assume , as I can tell, that those are the only two options: That either of them will win.

 

[PeroK]

I think you are missing the point and confusing "wins the match" with "wins the game".

There is a more advanced example of this technique here (see eventually post #30):

https://www.physicsforums.com/threads/how-many-tosses-to-flip-a-coin-hhth.1014702/

 

[Orodruin]

When you write Pa=1-Pb ( Andy, Bevin), you assume , as I can tell, that those are the only two options: That either of them will win.

Will win the game to be specific. One of them will eventually win the match as the probability of the match continuing goes exponentially to zero with the number of games played. The important thing to separate here is the probabilities of either one winning a particular game and either one winning the match.

 

[chwala]

There is a more advanced example of this technique here (see eventually post #30):

https://www.physicsforums.com/threads/how-many-tosses-to-flip-a-coin-hhth.1014702/

Looks interesting...i will look at it over the weekend...

 

[WWGD]

Will win the game to be specific. One of them will eventually win the match as the probability of the match continuing goes exponentially to zero with the number of games played. The important thing to separate here is the probabilities of either one winning a particular game and either one winning the match.

I understand, just thought Chwala may want to address and prove it, given this is new material for him. Apologies if he already did and I didn't notice

 

[chwala]

I understand, just thought Chwala may want to address and prove it, given this is new material for him. Apologies if he already did and I didn't notice

@WWGD Thanks a lot ...well noted mate. I learn a a lot from the team here. We are family!...I used;
$P_{Event happening} = 1- P_{Event NOT happening}$ for Mutually Exclusive events.

 

[Orodruin]

@WWGD Thanks a lot ...well noted mate. I learn a a lot from the team here. We are family!...I used;
$P_{Event happening} = 1- P_{Event NOT happening}$ for Mutually Exclusive events.

What is being pointed out in the last few messages is that ”Andy does not win game” is not equivalent to ”Bev wins game”.

 

[chwala]

What is being pointed out in the last few messages is that ”Andy does not win game” is not equivalent to ”Bev wins game”.

Ok, noted...I guess I need to let that sync in...because of the possibility of the draw? Cheers mate.