Find the range of p such that team A has the advantage in a best four of seven series.
Two teams, A and B, are playing a series of games. Assume
probability that A won a game is p
result of a game will not affect result of the next game
Find the range of p such that team A has the advantage in a best four of seven series.
Wouldn't the answer simply be p>0.5? Since the probability of winning is more than 0.5, then team A should have an advantage in any series.
Can anyone confirm? Or object?
Here's my attempt using negative binomial.
probability statistics negative-binomial
asked 2 days ago
WadeWade
179211
add a comment |
Two teams, A and B, are playing a series of games. Assume
probability that A won a game is p
result of a game will not affect result of the next game
Find the range of p such that team A has the advantage in a best four of seven series.
Wouldn't the answer simply be p>0.5? Since the probability of winning is more than 0.5, then team A should have an advantage in any series.
Can anyone confirm? Or object?
Here's my attempt using negative binomial.
probability statistics negative-binomial
asked 2 days ago
WadeWade
179211
Well, I agree with the answer but you don't really present an argument for it. What you write amounts to saying "it's true because it is true." You could, for instance, compute the probability of $A$ winning explicitly. Probably a good exercise anyway. Or you could just argue directly (looking at winning and losing paths for $A$), though if you are new to probability you might find it hard to write such an argument out clearly and completely.
– lulu
2 days ago
Ok, since this problem looks like a binomial type of question then the probability of team A winning 4 games out of the 7 would be [7!/(4!*3!)]*(p^4)*(1-p)^3. How could the range be found in this? Would. that probability I. just calculated be more than 0.5 and I can find the range of p from that? ex. [7!/(4!*3!)]*(p^4)*(1-p)^3 >0.5
– Wade
2 days ago
That's a start, but of course $A$ might win more than $4$ games out of $7$. As a first step, convince yourself that the situation is unchanged if you imagine that all $7$ games are played (even though in reality the series usually stops once one team has won $4$ games). Next compute the probability that $A$ wins $4,5,6,7$ games out of the seven.
– lulu
2 days ago
In this question, it's assumed that the game stops when a team reaches 4 wins. In this case I tried to solve for the range of p by adding the probability (using negative binomial) that A wins 4 games after player 4, 5, 6 and 7 games. However, my answer came with the summation of p raised to various powers and it became difficult to solve.
– Wade
yesterday
With my interpretation of the rules, which your notes confirm, it doesn't matter if the series stops or not. You get the same winner regardless. Thus you can just sum the probabilities that $A$ gets $4,5,6,7$ wins. My posted solution, though, avoids the binomial calculation entirely.
– lulu
yesterday
add a comment |
Two teams, A and B, are playing a series of games. Assume
probability that A won a game is p
result of a game will not affect result of the next game
Find the range of p such that team A has the advantage in a best four of seven series.
Wouldn't the answer simply be p>0.5? Since the probability of winning is more than 0.5, then team A should have an advantage in any series.
Can anyone confirm? Or object?
Here's my attempt using negative binomial.
probability statistics negative-binomial
asked 2 days ago
WadeWade
179211
Two teams, A and B, are playing a series of games. Assume
probability that A won a game is p
result of a game will not affect result of the next game
Find the range of p such that team A has the advantage in a best four of seven series.
Wouldn't the answer simply be p>0.5? Since the probability of winning is more than 0.5, then team A should have an advantage in any series.
Can anyone confirm? Or object?
Here's my attempt using negative binomial.
probability statistics negative-binomial
probability statistics negative-binomial
asked 2 days ago
WadeWade
179211
asked 2 days ago
WadeWade
179211
edited yesterday
Wade
asked 2 days ago
WadeWade
179211
asked 2 days ago
WadeWade
179211
asked 2 days ago
WadeWade
179211
179211
Well, I agree with the answer but you don't really present an argument for it. What you write amounts to saying "it's true because it is true." You could, for instance, compute the probability of $A$ winning explicitly. Probably a good exercise anyway. Or you could just argue directly (looking at winning and losing paths for $A$), though if you are new to probability you might find it hard to write such an argument out clearly and completely.
– lulu
2 days ago
Ok, since this problem looks like a binomial type of question then the probability of team A winning 4 games out of the 7 would be [7!/(4!*3!)]*(p^4)*(1-p)^3. How could the range be found in this? Would. that probability I. just calculated be more than 0.5 and I can find the range of p from that? ex. [7!/(4!*3!)]*(p^4)*(1-p)^3 >0.5
– Wade
2 days ago
That's a start, but of course $A$ might win more than $4$ games out of $7$. As a first step, convince yourself that the situation is unchanged if you imagine that all $7$ games are played (even though in reality the series usually stops once one team has won $4$ games). Next compute the probability that $A$ wins $4,5,6,7$ games out of the seven.
– lulu
2 days ago
In this question, it's assumed that the game stops when a team reaches 4 wins. In this case I tried to solve for the range of p by adding the probability (using negative binomial) that A wins 4 games after player 4, 5, 6 and 7 games. However, my answer came with the summation of p raised to various powers and it became difficult to solve.
– Wade
yesterday
With my interpretation of the rules, which your notes confirm, it doesn't matter if the series stops or not. You get the same winner regardless. Thus you can just sum the probabilities that $A$ gets $4,5,6,7$ wins. My posted solution, though, avoids the binomial calculation entirely.
– lulu
yesterday
add a comment |
Well, I agree with the answer but you don't really present an argument for it. What you write amounts to saying "it's true because it is true." You could, for instance, compute the probability of $A$ winning explicitly. Probably a good exercise anyway. Or you could just argue directly (looking at winning and losing paths for $A$), though if you are new to probability you might find it hard to write such an argument out clearly and completely.
– lulu
2 days ago
Ok, since this problem looks like a binomial type of question then the probability of team A winning 4 games out of the 7 would be [7!/(4!*3!)]*(p^4)*(1-p)^3. How could the range be found in this? Would. that probability I. just calculated be more than 0.5 and I can find the range of p from that? ex. [7!/(4!*3!)]*(p^4)*(1-p)^3 >0.5
– Wade
2 days ago
That's a start, but of course $A$ might win more than $4$ games out of $7$. As a first step, convince yourself that the situation is unchanged if you imagine that all $7$ games are played (even though in reality the series usually stops once one team has won $4$ games). Next compute the probability that $A$ wins $4,5,6,7$ games out of the seven.
– lulu
2 days ago
In this question, it's assumed that the game stops when a team reaches 4 wins. In this case I tried to solve for the range of p by adding the probability (using negative binomial) that A wins 4 games after player 4, 5, 6 and 7 games. However, my answer came with the summation of p raised to various powers and it became difficult to solve.
– Wade
yesterday
With my interpretation of the rules, which your notes confirm, it doesn't matter if the series stops or not. You get the same winner regardless. Thus you can just sum the probabilities that $A$ gets $4,5,6,7$ wins. My posted solution, though, avoids the binomial calculation entirely.
– lulu
yesterday
Well, I agree with the answer but you don't really present an argument for it. What you write amounts to saying "it's true because it is true." You could, for instance, compute the probability of $A$ winning explicitly. Probably a good exercise anyway. Or you could just argue directly (looking at winning and losing paths for $A$), though if you are new to probability you might find it hard to write such an argument out clearly and completely.
– lulu
2 days ago
Well, I agree with the answer but you don't really present an argument for it. What you write amounts to saying "it's true because it is true." You could, for instance, compute the probability of $A$ winning explicitly. Probably a good exercise anyway. Or you could just argue directly (looking at winning and losing paths for $A$), though if you are new to probability you might find it hard to write such an argument out clearly and completely.
– lulu
2 days ago
Ok, since this problem looks like a binomial type of question then the probability of team A winning 4 games out of the 7 would be [7!/(4!*3!)]*(p^4)*(1-p)^3. How could the range be found in this? Would. that probability I. just calculated be more than 0.5 and I can find the range of p from that? ex. [7!/(4!*3!)]*(p^4)*(1-p)^3 >0.5
– Wade
2 days ago
Ok, since this problem looks like a binomial type of question then the probability of team A winning 4 games out of the 7 would be [7!/(4!*3!)]*(p^4)*(1-p)^3. How could the range be found in this? Would. that probability I. just calculated be more than 0.5 and I can find the range of p from that? ex. [7!/(4!*3!)]*(p^4)*(1-p)^3 >0.5
– Wade
2 days ago
That's a start, but of course $A$ might win more than $4$ games out of $7$. As a first step, convince yourself that the situation is unchanged if you imagine that all $7$ games are played (even though in reality the series usually stops once one team has won $4$ games). Next compute the probability that $A$ wins $4,5,6,7$ games out of the seven.
– lulu
2 days ago
That's a start, but of course $A$ might win more than $4$ games out of $7$. As a first step, convince yourself that the situation is unchanged if you imagine that all $7$ games are played (even though in reality the series usually stops once one team has won $4$ games). Next compute the probability that $A$ wins $4,5,6,7$ games out of the seven.
– lulu
2 days ago
In this question, it's assumed that the game stops when a team reaches 4 wins. In this case I tried to solve for the range of p by adding the probability (using negative binomial) that A wins 4 games after player 4, 5, 6 and 7 games. However, my answer came with the summation of p raised to various powers and it became difficult to solve.
– Wade
yesterday
In this question, it's assumed that the game stops when a team reaches 4 wins. In this case I tried to solve for the range of p by adding the probability (using negative binomial) that A wins 4 games after player 4, 5, 6 and 7 games. However, my answer came with the summation of p raised to various powers and it became difficult to solve.
– Wade
yesterday
With my interpretation of the rules, which your notes confirm, it doesn't matter if the series stops or not. You get the same winner regardless. Thus you can just sum the probabilities that $A$ gets $4,5,6,7$ wins. My posted solution, though, avoids the binomial calculation entirely.
– lulu
yesterday
With my interpretation of the rules, which your notes confirm, it doesn't matter if the series stops or not. You get the same winner regardless. Thus you can just sum the probabilities that $A$ gets $4,5,6,7$ wins. My posted solution, though, avoids the binomial calculation entirely.
– lulu
yesterday
add a comment |
1 Answer
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Note: I am interpreting the question as "assume that $A$ wins (or loses) any given game with probability $p$, independent of all other games. Find the range of $p$ such that team A has the advantage in a best four of seven series." If some other interpretation was intended, please clarify.
As a way to do the problem without significant computation, look at the various paths. Here a "path" means a sequence of game winners until one side or the other appears $4$ times. Thus possible paths would include $AAAA$, $ABAAA$,$BBABAAA$ and so on (read left to right).
We divide the set of paths in two according to whether $A$ or $B$ comes up $4$ times. Let's call the paths along which $A$ wins an $A-$path.
We note that the map which switches $A,B$ defines a bijection between the $A-$ paths and the $B-$paths. If $P$ is a path we'll let $overline P$ be the same path with $A,B$ interchanged.
Consider an $A-$path $P$. We claim that $p>.5$ implies that the probability of $P$ is greater than the probability of $overline P$. Indeed, we know that $P$ contains exactly $4$ $A's$ and $i$ $B's$ where $iin {0,1,2,3}$. Thus $$Prob(P)=p^4(1-p)^iquad Prob(overline P)=p^i(1-p)^4$$ dividing we see that $$frac {Prob(P)}{Prob(overline P)}=left(frac p{1-p}right)^{4-i}$$ and $p>.5$ implies that this is greater than $1$, so we are done.
answered 2 days ago
lulululu
39.2k24677
add a comment |
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Note: I am interpreting the question as "assume that $A$ wins (or loses) any given game with probability $p$, independent of all other games. Find the range of $p$ such that team A has the advantage in a best four of seven series." If some other interpretation was intended, please clarify.
As a way to do the problem without significant computation, look at the various paths. Here a "path" means a sequence of game winners until one side or the other appears $4$ times. Thus possible paths would include $AAAA$, $ABAAA$,$BBABAAA$ and so on (read left to right).
We divide the set of paths in two according to whether $A$ or $B$ comes up $4$ times. Let's call the paths along which $A$ wins an $A-$path.
We note that the map which switches $A,B$ defines a bijection between the $A-$ paths and the $B-$paths. If $P$ is a path we'll let $overline P$ be the same path with $A,B$ interchanged.
Consider an $A-$path $P$. We claim that $p>.5$ implies that the probability of $P$ is greater than the probability of $overline P$. Indeed, we know that $P$ contains exactly $4$ $A's$ and $i$ $B's$ where $iin {0,1,2,3}$. Thus $$Prob(P)=p^4(1-p)^iquad Prob(overline P)=p^i(1-p)^4$$ dividing we see that $$frac {Prob(P)}{Prob(overline P)}=left(frac p{1-p}right)^{4-i}$$ and $p>.5$ implies that this is greater than $1$, so we are done.
answered 2 days ago
lulululu
39.2k24677
add a comment |
Note: I am interpreting the question as "assume that $A$ wins (or loses) any given game with probability $p$, independent of all other games. Find the range of $p$ such that team A has the advantage in a best four of seven series." If some other interpretation was intended, please clarify.
As a way to do the problem without significant computation, look at the various paths. Here a "path" means a sequence of game winners until one side or the other appears $4$ times. Thus possible paths would include $AAAA$, $ABAAA$,$BBABAAA$ and so on (read left to right).
We divide the set of paths in two according to whether $A$ or $B$ comes up $4$ times. Let's call the paths along which $A$ wins an $A-$path.
We note that the map which switches $A,B$ defines a bijection between the $A-$ paths and the $B-$paths. If $P$ is a path we'll let $overline P$ be the same path with $A,B$ interchanged.
Consider an $A-$path $P$. We claim that $p>.5$ implies that the probability of $P$ is greater than the probability of $overline P$. Indeed, we know that $P$ contains exactly $4$ $A's$ and $i$ $B's$ where $iin {0,1,2,3}$. Thus $$Prob(P)=p^4(1-p)^iquad Prob(overline P)=p^i(1-p)^4$$ dividing we see that $$frac {Prob(P)}{Prob(overline P)}=left(frac p{1-p}right)^{4-i}$$ and $p>.5$ implies that this is greater than $1$, so we are done.
answered 2 days ago
lulululu
39.2k24677
add a comment |
Note: I am interpreting the question as "assume that $A$ wins (or loses) any given game with probability $p$, independent of all other games. Find the range of $p$ such that team A has the advantage in a best four of seven series." If some other interpretation was intended, please clarify.
As a way to do the problem without significant computation, look at the various paths. Here a "path" means a sequence of game winners until one side or the other appears $4$ times. Thus possible paths would include $AAAA$, $ABAAA$,$BBABAAA$ and so on (read left to right).
We divide the set of paths in two according to whether $A$ or $B$ comes up $4$ times. Let's call the paths along which $A$ wins an $A-$path.
We note that the map which switches $A,B$ defines a bijection between the $A-$ paths and the $B-$paths. If $P$ is a path we'll let $overline P$ be the same path with $A,B$ interchanged.
Consider an $A-$path $P$. We claim that $p>.5$ implies that the probability of $P$ is greater than the probability of $overline P$. Indeed, we know that $P$ contains exactly $4$ $A's$ and $i$ $B's$ where $iin {0,1,2,3}$. Thus $$Prob(P)=p^4(1-p)^iquad Prob(overline P)=p^i(1-p)^4$$ dividing we see that $$frac {Prob(P)}{Prob(overline P)}=left(frac p{1-p}right)^{4-i}$$ and $p>.5$ implies that this is greater than $1$, so we are done.
answered 2 days ago
lulululu
39.2k24677
Note: I am interpreting the question as "assume that $A$ wins (or loses) any given game with probability $p$, independent of all other games. Find the range of $p$ such that team A has the advantage in a best four of seven series." If some other interpretation was intended, please clarify.
As a way to do the problem without significant computation, look at the various paths. Here a "path" means a sequence of game winners until one side or the other appears $4$ times. Thus possible paths would include $AAAA$, $ABAAA$,$BBABAAA$ and so on (read left to right).
We divide the set of paths in two according to whether $A$ or $B$ comes up $4$ times. Let's call the paths along which $A$ wins an $A-$path.
We note that the map which switches $A,B$ defines a bijection between the $A-$ paths and the $B-$paths. If $P$ is a path we'll let $overline P$ be the same path with $A,B$ interchanged.
Consider an $A-$path $P$. We claim that $p>.5$ implies that the probability of $P$ is greater than the probability of $overline P$. Indeed, we know that $P$ contains exactly $4$ $A's$ and $i$ $B's$ where $iin {0,1,2,3}$. Thus $$Prob(P)=p^4(1-p)^iquad Prob(overline P)=p^i(1-p)^4$$ dividing we see that $$frac {Prob(P)}{Prob(overline P)}=left(frac p{1-p}right)^{4-i}$$ and $p>.5$ implies that this is greater than $1$, so we are done.
answered 2 days ago
lulululu
39.2k24677
edited 2 days ago
answered 2 days ago
lulululu
39.2k24677
answered 2 days ago
lulululu
39.2k24677
answered 2 days ago
lulululu
39.2k24677
39.2k24677
add a comment |
add a comment |
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Well, I agree with the answer but you don't really present an argument for it. What you write amounts to saying "it's true because it is true." You could, for instance, compute the probability of $A$ winning explicitly. Probably a good exercise anyway. Or you could just argue directly (looking at winning and losing paths for $A$), though if you are new to probability you might find it hard to write such an argument out clearly and completely.
– lulu
2 days ago
Ok, since this problem looks like a binomial type of question then the probability of team A winning 4 games out of the 7 would be [7!/(4!*3!)]*(p^4)*(1-p)^3. How could the range be found in this? Would. that probability I. just calculated be more than 0.5 and I can find the range of p from that? ex. [7!/(4!*3!)]*(p^4)*(1-p)^3 >0.5
– Wade
2 days ago
That's a start, but of course $A$ might win more than $4$ games out of $7$. As a first step, convince yourself that the situation is unchanged if you imagine that all $7$ games are played (even though in reality the series usually stops once one team has won $4$ games). Next compute the probability that $A$ wins $4,5,6,7$ games out of the seven.
– lulu
2 days ago
In this question, it's assumed that the game stops when a team reaches 4 wins. In this case I tried to solve for the range of p by adding the probability (using negative binomial) that A wins 4 games after player 4, 5, 6 and 7 games. However, my answer came with the summation of p raised to various powers and it became difficult to solve.
– Wade
yesterday
With my interpretation of the rules, which your notes confirm, it doesn't matter if the series stops or not. You get the same winner regardless. Thus you can just sum the probabilities that $A$ gets $4,5,6,7$ wins. My posted solution, though, avoids the binomial calculation entirely.
– lulu
yesterday