Find the range of p such that team A has the advantage in a best four of seven series.
$begingroup$
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 Jan 5 at 13:46
WadeWade
305311
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 13:46
WadeWade
305311
$endgroup$
$begingroup$
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.
$endgroup$
– lulu
Jan 5 at 13:53
$begingroup$
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
$endgroup$
– Wade
Jan 5 at 14:00
$begingroup$
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.
$endgroup$
– lulu
Jan 5 at 14:24
$begingroup$
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.
$endgroup$
– Wade
Jan 6 at 1:34
$begingroup$
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.
$endgroup$
– lulu
Jan 6 at 11:49
add a comment |
$begingroup$
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 Jan 5 at 13:46
WadeWade
305311
$endgroup$
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 Jan 5 at 13:46
WadeWade
305311
asked Jan 5 at 13:46
WadeWade
305311
edited Jan 6 at 2:03
Wade
asked Jan 5 at 13:46
WadeWade
305311
asked Jan 5 at 13:46
WadeWade
305311
asked Jan 5 at 13:46
WadeWade
305311
305311
$begingroup$
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.
$endgroup$
– lulu
Jan 5 at 13:53
$begingroup$
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
$endgroup$
– Wade
Jan 5 at 14:00
$begingroup$
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.
$endgroup$
– lulu
Jan 5 at 14:24
$begingroup$
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.
$endgroup$
– Wade
Jan 6 at 1:34
$begingroup$
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.
$endgroup$
– lulu
Jan 6 at 11:49
add a comment |
$begingroup$
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.
$endgroup$
– lulu
Jan 5 at 13:53
$begingroup$
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
$endgroup$
– Wade
Jan 5 at 14:00
$begingroup$
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.
$endgroup$
– lulu
Jan 5 at 14:24
$begingroup$
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.
$endgroup$
– Wade
Jan 6 at 1:34
$begingroup$
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.
$endgroup$
– lulu
Jan 6 at 11:49
$begingroup$
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.
$endgroup$
– lulu
Jan 5 at 13:53
$begingroup$
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.
$endgroup$
– lulu
Jan 5 at 13:53
$begingroup$
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
$endgroup$
– Wade
Jan 5 at 14:00
$begingroup$
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
$endgroup$
– Wade
Jan 5 at 14:00
$begingroup$
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.
$endgroup$
– lulu
Jan 5 at 14:24
$begingroup$
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.
$endgroup$
– lulu
Jan 5 at 14:24
$begingroup$
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.
$endgroup$
– Wade
Jan 6 at 1:34
$begingroup$
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.
$endgroup$
– Wade
Jan 6 at 1:34
$begingroup$
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.
$endgroup$
– lulu
Jan 6 at 11:49
$begingroup$
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.
$endgroup$
– lulu
Jan 6 at 11:49
add a comment |
1 Answer
1
active
oldest
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$begingroup$
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 Jan 5 at 15:04
lulululu
43.5k25081
$endgroup$
add a comment |
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$begingroup$
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 Jan 5 at 15:04
lulululu
43.5k25081
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 15:04
lulululu
43.5k25081
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 15:04
lulululu
43.5k25081
$endgroup$
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 Jan 5 at 15:04
lulululu
43.5k25081
edited Jan 5 at 15:23
answered Jan 5 at 15:04
lulululu
43.5k25081
answered Jan 5 at 15:04
lulululu
43.5k25081
answered Jan 5 at 15:04
lulululu
43.5k25081
43.5k25081
add a comment |
add a comment |
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$begingroup$
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.
$endgroup$
– lulu
Jan 5 at 13:53
$begingroup$
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
$endgroup$
– Wade
Jan 5 at 14:00
$begingroup$
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.
$endgroup$
– lulu
Jan 5 at 14:24
$begingroup$
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.
$endgroup$
– Wade
Jan 6 at 1:34
$begingroup$
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.
$endgroup$
– lulu
Jan 6 at 11:49