- #1
chwala
Gold Member
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- 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;
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 said: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?
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 said: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.
Ok, I will...talk later cheers mate.Orodruin said: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;Orodruin said: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?
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 said: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##
This is a different approach to the problem...looks like I did not understand your question...I will re look at it later...Orodruin said: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?
I used this approach, just to be clear;chwala said: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 want to get it right, If the first game is a draw then the possibilities of Andy Winning are;Orodruin said: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?
@pasmith eeeèeish! Looks tough there...cheers matepasmith said: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 [tex]\begin{split}
P(\mbox{Andy wins the final game}) &=
P(\mbox{Andy wins game}\,|\,\mbox{Andy or Beth wins game}) \\
&=
\frac{P(\mbox{Andy wins game})}{1 - P(\mbox{Neither player wins game})}.\end{split}[/tex] The probability that the final game is played is [tex]
P(\mbox{The final game is played}) = 1 - \lim_{n \to \infty} P(\mbox{Neither player wins game})^n = 1.[/tex] Putting these together: [tex]
\begin{split}
P(\mbox{Andy wins match}) &= P(\mbox{Andy wins the final game})P(\mbox{The final game is played}) \\
&= \frac{P(\mbox{Andy wins game})}{1 - P(\mbox{Neither player wins game})}.\end{split}[/tex] As expected, this is equal to[tex]
P(\mbox{Andy wins game}) \sum_{n=0}^\infty P(\mbox{Neither player wins game})^n
.[/tex]
^^ What he said.PeroK said: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.
Not sure. The game may end up in a draw with nonzero probability , so your top equation won't hold up.chwala said: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##
This gives you;PeroK said: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.
For sure i am not getting you...you mean this alternative method is wrong? Kindly be specific.WWGD said:Not sure. The game may end up in a draw with nonzero probability , so your top equation won't hold up.
I think you are missing the point and confusing "wins the match" with "wins the game".chwala said:This gives you;
##0.256+(0.24×0.256)=0.31744≠0.337## unless i am missing the point...
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.chwala said: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.
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.chwala said:For sure i am not getting you...you mean this alternative method is wrong? Kindly be specific.
There is a more advanced example of this technique here (see eventually post #30):PeroK said:I think you are missing the point and confusing "wins the match" with "wins the game".
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.WWGD said: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.
Looks interesting...i will look at it over the weekend...PeroK said: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/
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 noticeOrodruin said: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.
@WWGD Thanks a lot ...well noted mate. I learn a a lot from the team here. We are family!...I used;WWGD said: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
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 said:@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.
Ok, noted...I guess I need to let that sync in...because of the possibility of the draw? Cheers mate.Orodruin said:What is being pointed out in the last few messages is that ”Andy does not win game” is not equivalent to ”Bev wins game”.
Probability is a measure of the likelihood that a certain event will occur. It is expressed as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty.
Probability is calculated by dividing the number of favorable outcomes by the total number of possible outcomes. For example, if there are 5 red marbles and 10 total marbles in a bag, the probability of picking a red marble is 5/10 or 0.5.
There are many factors that can affect the probability of Andy winning the match, including his skill level, physical condition, mental state, and the skill level of his opponent.
No, probability cannot guarantee a win for Andy. It only provides a measure of the likelihood of him winning based on the available information.
Probability can be useful in predicting the outcome of a match by providing insight into the likelihood of a certain result. It can also help in making informed decisions and strategies based on the probabilities of different outcomes.