Mixing
Soccer penalty shots
Clash Royale CLAN TAG#URR8PPP
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Imagine a scenario where we have a soccer player who is shooting penalty shots. He makes his first shot with probability of failure $1/3$, then his second with probability of failure $1/5$ and so on, and at the $n$-th throw the probability of failure is $1/(2n+1)$.
During practice, he shoots $19$ consecutive kicks. What is the probability he succeeds in an even number of kicks?
For each throw, the respective probability of success is $$1-frac1(2n+1)= frac2n(2n+1)$$
To find the total probability of success in an even number of throws, we must add the probability of $0$ successful shots, $2, 4, cdots, 18$.
But the $2$ successful throws can be, for example, the $1$st and the $2$nd, or the $4$th and the $19$th. Then, obviously, we will multiply the $2$.
So how do we calculate the individual probabilities?
Any help will be highly appreciated!
probability combinatorics
edited Jul 18 at 10:02
TheSimpliFire
9,67361951
asked Jul 18 at 9:42
Creton Laplace
1085
add a comment |Â
up vote
10
down vote
favorite
Imagine a scenario where we have a soccer player who is shooting penalty shots. He makes his first shot with probability of failure $1/3$, then his second with probability of failure $1/5$ and so on, and at the $n$-th throw the probability of failure is $1/(2n+1)$.
During practice, he shoots $19$ consecutive kicks. What is the probability he succeeds in an even number of kicks?
For each throw, the respective probability of success is $$1-frac1(2n+1)= frac2n(2n+1)$$
To find the total probability of success in an even number of throws, we must add the probability of $0$ successful shots, $2, 4, cdots, 18$.
But the $2$ successful throws can be, for example, the $1$st and the $2$nd, or the $4$th and the $19$th. Then, obviously, we will multiply the $2$.
So how do we calculate the individual probabilities?
Any help will be highly appreciated!
probability combinatorics
edited Jul 18 at 10:02
TheSimpliFire
9,67361951
asked Jul 18 at 9:42
Creton Laplace
1085
add a comment |Â
up vote
10
down vote
favorite
up vote
10
down vote
favorite
Imagine a scenario where we have a soccer player who is shooting penalty shots. He makes his first shot with probability of failure $1/3$, then his second with probability of failure $1/5$ and so on, and at the $n$-th throw the probability of failure is $1/(2n+1)$.
During practice, he shoots $19$ consecutive kicks. What is the probability he succeeds in an even number of kicks?
For each throw, the respective probability of success is $$1-frac1(2n+1)= frac2n(2n+1)$$
To find the total probability of success in an even number of throws, we must add the probability of $0$ successful shots, $2, 4, cdots, 18$.
But the $2$ successful throws can be, for example, the $1$st and the $2$nd, or the $4$th and the $19$th. Then, obviously, we will multiply the $2$.
So how do we calculate the individual probabilities?
Any help will be highly appreciated!
probability combinatorics
edited Jul 18 at 10:02
TheSimpliFire
9,67361951
asked Jul 18 at 9:42
Creton Laplace
1085
Imagine a scenario where we have a soccer player who is shooting penalty shots. He makes his first shot with probability of failure $1/3$, then his second with probability of failure $1/5$ and so on, and at the $n$-th throw the probability of failure is $1/(2n+1)$.
During practice, he shoots $19$ consecutive kicks. What is the probability he succeeds in an even number of kicks?
For each throw, the respective probability of success is $$1-frac1(2n+1)= frac2n(2n+1)$$
To find the total probability of success in an even number of throws, we must add the probability of $0$ successful shots, $2, 4, cdots, 18$.
But the $2$ successful throws can be, for example, the $1$st and the $2$nd, or the $4$th and the $19$th. Then, obviously, we will multiply the $2$.
So how do we calculate the individual probabilities?
Any help will be highly appreciated!
probability combinatorics
edited Jul 18 at 10:02
TheSimpliFire
9,67361951
asked Jul 18 at 9:42
Creton Laplace
1085
edited Jul 18 at 10:02
TheSimpliFire
9,67361951
edited Jul 18 at 10:02
TheSimpliFire
9,67361951
edited Jul 18 at 10:02
TheSimpliFire
9,67361951
9,67361951
asked Jul 18 at 9:42
Creton Laplace
1085
asked Jul 18 at 9:42
Creton Laplace
1085
asked Jul 18 at 9:42
Creton Laplace
1085
1085
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
The probability generating function for the number of failures is
$$
f(x)=prod_n=1^19frac2n+x2n+1;.
$$
The probability of getting an even number of failures is
begineqnarray*
fracf(1)+f(-1)2=frac12left(prod_n=1^19frac2n+12n+1+prod_n=1^19frac2n-12n+1right)=frac12left(1+frac139right)=frac2039;.
endeqnarray*
Since the total number of kicks is odd, the probability of getting an even number of successes is
$$1-frac2039=frac1939;.$$
answered Jul 18 at 9:59
joriki
164k10180328
1
@joriki: Many thanks! I thought it would be something simpler... Could you possibly explain a bit further about the generating function (what is x???) and also why the probability of getting an even number of successes is 1/2f(1)+f(-1)? Apologies but I am not very familiar with these topics!!
– Creton Laplace
Jul 18 at 10:24
@CretonLaplace: Nothing to apologize for! It just seemed more efficient to first see whether you are and then explain if not, than to explain first, possibly without need :-). In a probability generating function $f(x)=sum_kp_kx^k$ for the number of successes, the coefficient $p_k$ of the power $x^k$ is the probability for obtaining $k$ successes. $x$ is just a marker variable for the bookkeeping. The $n$-th kick contributes a factor $frac2n+x2n+1$, since the probability of success is $frac12n+1$. (Multiply it out up to $n=2$ or $3$ and you'll see why this works.)
– joriki
Jul 18 at 10:32
@CretonLaplace: The probability of getting an even number of successes is the sum of the coefficients $p_k$ for even $k$. This is the the average of the sum of the coefficients and the alternating sum of the coefficients. The sum of the coefficients is $f(1)$, and the alternating sum of the coefficients is $f(-1)$.
– joriki
Jul 18 at 10:33
And, of course, see en.wikipedia.org/wiki/Probability-generating_function. About it being something simpler -- I was surprised by the simple result! :-) There may well be a simpler way to derive it; this was just the first one that occurred to me.
– joriki
Jul 18 at 10:35
It seems to me something must be wrong here. If $n=1$ then he will succeed in an even number of shots iff he fails at his first attempt. So $p_1=frac13$. This does not agree with $frac12(1+frac13)$. I come to $p_2n=frac2n+14n+1$ and $p_2n+1=frac2n+14n+3$, so that $p_19=frac1939$
– drhab
Jul 18 at 11:58
 |Â
show 2 more comments
up vote
2
down vote
Hint
Try to work out the recursive equality:
$$left(2n+1right)p_n=p_n-1+2nleft(1-p_n-1right)$$
Where the probability that he succeeds in an even number of kicks by
taking $n$ penalties be denoted by $p_n$.
This is based on the thought that an even number can be achieved in two ways:
- an even number in the first $n-1$ efforts followed by a failure
- an odd number in the first $n-1$ efforts followed by a success.
Here $p_0=1$.
answered Jul 18 at 9:59
drhab
86.5k541118
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
The probability generating function for the number of failures is
$$
f(x)=prod_n=1^19frac2n+x2n+1;.
$$
The probability of getting an even number of failures is
begineqnarray*
fracf(1)+f(-1)2=frac12left(prod_n=1^19frac2n+12n+1+prod_n=1^19frac2n-12n+1right)=frac12left(1+frac139right)=frac2039;.
endeqnarray*
Since the total number of kicks is odd, the probability of getting an even number of successes is
$$1-frac2039=frac1939;.$$
answered Jul 18 at 9:59
joriki
164k10180328
1
@joriki: Many thanks! I thought it would be something simpler... Could you possibly explain a bit further about the generating function (what is x???) and also why the probability of getting an even number of successes is 1/2f(1)+f(-1)? Apologies but I am not very familiar with these topics!!
– Creton Laplace
Jul 18 at 10:24
@CretonLaplace: Nothing to apologize for! It just seemed more efficient to first see whether you are and then explain if not, than to explain first, possibly without need :-). In a probability generating function $f(x)=sum_kp_kx^k$ for the number of successes, the coefficient $p_k$ of the power $x^k$ is the probability for obtaining $k$ successes. $x$ is just a marker variable for the bookkeeping. The $n$-th kick contributes a factor $frac2n+x2n+1$, since the probability of success is $frac12n+1$. (Multiply it out up to $n=2$ or $3$ and you'll see why this works.)
– joriki
Jul 18 at 10:32
@CretonLaplace: The probability of getting an even number of successes is the sum of the coefficients $p_k$ for even $k$. This is the the average of the sum of the coefficients and the alternating sum of the coefficients. The sum of the coefficients is $f(1)$, and the alternating sum of the coefficients is $f(-1)$.
– joriki
Jul 18 at 10:33
And, of course, see en.wikipedia.org/wiki/Probability-generating_function. About it being something simpler -- I was surprised by the simple result! :-) There may well be a simpler way to derive it; this was just the first one that occurred to me.
– joriki
Jul 18 at 10:35
It seems to me something must be wrong here. If $n=1$ then he will succeed in an even number of shots iff he fails at his first attempt. So $p_1=frac13$. This does not agree with $frac12(1+frac13)$. I come to $p_2n=frac2n+14n+1$ and $p_2n+1=frac2n+14n+3$, so that $p_19=frac1939$
– drhab
Jul 18 at 11:58
 |Â
show 2 more comments
up vote
3
down vote
The probability generating function for the number of failures is
$$
f(x)=prod_n=1^19frac2n+x2n+1;.
$$
The probability of getting an even number of failures is
begineqnarray*
fracf(1)+f(-1)2=frac12left(prod_n=1^19frac2n+12n+1+prod_n=1^19frac2n-12n+1right)=frac12left(1+frac139right)=frac2039;.
endeqnarray*
Since the total number of kicks is odd, the probability of getting an even number of successes is
$$1-frac2039=frac1939;.$$
answered Jul 18 at 9:59
joriki
164k10180328
1
@joriki: Many thanks! I thought it would be something simpler... Could you possibly explain a bit further about the generating function (what is x???) and also why the probability of getting an even number of successes is 1/2f(1)+f(-1)? Apologies but I am not very familiar with these topics!!
– Creton Laplace
Jul 18 at 10:24
@CretonLaplace: Nothing to apologize for! It just seemed more efficient to first see whether you are and then explain if not, than to explain first, possibly without need :-). In a probability generating function $f(x)=sum_kp_kx^k$ for the number of successes, the coefficient $p_k$ of the power $x^k$ is the probability for obtaining $k$ successes. $x$ is just a marker variable for the bookkeeping. The $n$-th kick contributes a factor $frac2n+x2n+1$, since the probability of success is $frac12n+1$. (Multiply it out up to $n=2$ or $3$ and you'll see why this works.)
– joriki
Jul 18 at 10:32
@CretonLaplace: The probability of getting an even number of successes is the sum of the coefficients $p_k$ for even $k$. This is the the average of the sum of the coefficients and the alternating sum of the coefficients. The sum of the coefficients is $f(1)$, and the alternating sum of the coefficients is $f(-1)$.
– joriki
Jul 18 at 10:33
And, of course, see en.wikipedia.org/wiki/Probability-generating_function. About it being something simpler -- I was surprised by the simple result! :-) There may well be a simpler way to derive it; this was just the first one that occurred to me.
– joriki
Jul 18 at 10:35
It seems to me something must be wrong here. If $n=1$ then he will succeed in an even number of shots iff he fails at his first attempt. So $p_1=frac13$. This does not agree with $frac12(1+frac13)$. I come to $p_2n=frac2n+14n+1$ and $p_2n+1=frac2n+14n+3$, so that $p_19=frac1939$
– drhab
Jul 18 at 11:58
 |Â
show 2 more comments
up vote
3
down vote
up vote
3
down vote
The probability generating function for the number of failures is
$$
f(x)=prod_n=1^19frac2n+x2n+1;.
$$
The probability of getting an even number of failures is
begineqnarray*
fracf(1)+f(-1)2=frac12left(prod_n=1^19frac2n+12n+1+prod_n=1^19frac2n-12n+1right)=frac12left(1+frac139right)=frac2039;.
endeqnarray*
Since the total number of kicks is odd, the probability of getting an even number of successes is
$$1-frac2039=frac1939;.$$
answered Jul 18 at 9:59
joriki
164k10180328
The probability generating function for the number of failures is
$$
f(x)=prod_n=1^19frac2n+x2n+1;.
$$
The probability of getting an even number of failures is
begineqnarray*
fracf(1)+f(-1)2=frac12left(prod_n=1^19frac2n+12n+1+prod_n=1^19frac2n-12n+1right)=frac12left(1+frac139right)=frac2039;.
endeqnarray*
Since the total number of kicks is odd, the probability of getting an even number of successes is
$$1-frac2039=frac1939;.$$
answered Jul 18 at 9:59
joriki
164k10180328
edited Jul 18 at 12:04
answered Jul 18 at 9:59
joriki
164k10180328
answered Jul 18 at 9:59
joriki
164k10180328
answered Jul 18 at 9:59
joriki
164k10180328
164k10180328
1
@joriki: Many thanks! I thought it would be something simpler... Could you possibly explain a bit further about the generating function (what is x???) and also why the probability of getting an even number of successes is 1/2f(1)+f(-1)? Apologies but I am not very familiar with these topics!!
– Creton Laplace
Jul 18 at 10:24
@CretonLaplace: Nothing to apologize for! It just seemed more efficient to first see whether you are and then explain if not, than to explain first, possibly without need :-). In a probability generating function $f(x)=sum_kp_kx^k$ for the number of successes, the coefficient $p_k$ of the power $x^k$ is the probability for obtaining $k$ successes. $x$ is just a marker variable for the bookkeeping. The $n$-th kick contributes a factor $frac2n+x2n+1$, since the probability of success is $frac12n+1$. (Multiply it out up to $n=2$ or $3$ and you'll see why this works.)
– joriki
Jul 18 at 10:32
@CretonLaplace: The probability of getting an even number of successes is the sum of the coefficients $p_k$ for even $k$. This is the the average of the sum of the coefficients and the alternating sum of the coefficients. The sum of the coefficients is $f(1)$, and the alternating sum of the coefficients is $f(-1)$.
– joriki
Jul 18 at 10:33
And, of course, see en.wikipedia.org/wiki/Probability-generating_function. About it being something simpler -- I was surprised by the simple result! :-) There may well be a simpler way to derive it; this was just the first one that occurred to me.
– joriki
Jul 18 at 10:35
It seems to me something must be wrong here. If $n=1$ then he will succeed in an even number of shots iff he fails at his first attempt. So $p_1=frac13$. This does not agree with $frac12(1+frac13)$. I come to $p_2n=frac2n+14n+1$ and $p_2n+1=frac2n+14n+3$, so that $p_19=frac1939$
– drhab
Jul 18 at 11:58
 |Â
show 2 more comments
1
@joriki: Many thanks! I thought it would be something simpler... Could you possibly explain a bit further about the generating function (what is x???) and also why the probability of getting an even number of successes is 1/2f(1)+f(-1)? Apologies but I am not very familiar with these topics!!
– Creton Laplace
Jul 18 at 10:24
@CretonLaplace: Nothing to apologize for! It just seemed more efficient to first see whether you are and then explain if not, than to explain first, possibly without need :-). In a probability generating function $f(x)=sum_kp_kx^k$ for the number of successes, the coefficient $p_k$ of the power $x^k$ is the probability for obtaining $k$ successes. $x$ is just a marker variable for the bookkeeping. The $n$-th kick contributes a factor $frac2n+x2n+1$, since the probability of success is $frac12n+1$. (Multiply it out up to $n=2$ or $3$ and you'll see why this works.)
– joriki
Jul 18 at 10:32
@CretonLaplace: The probability of getting an even number of successes is the sum of the coefficients $p_k$ for even $k$. This is the the average of the sum of the coefficients and the alternating sum of the coefficients. The sum of the coefficients is $f(1)$, and the alternating sum of the coefficients is $f(-1)$.
– joriki
Jul 18 at 10:33
And, of course, see en.wikipedia.org/wiki/Probability-generating_function. About it being something simpler -- I was surprised by the simple result! :-) There may well be a simpler way to derive it; this was just the first one that occurred to me.
– joriki
Jul 18 at 10:35
It seems to me something must be wrong here. If $n=1$ then he will succeed in an even number of shots iff he fails at his first attempt. So $p_1=frac13$. This does not agree with $frac12(1+frac13)$. I come to $p_2n=frac2n+14n+1$ and $p_2n+1=frac2n+14n+3$, so that $p_19=frac1939$
– drhab
Jul 18 at 11:58
1
1
@joriki: Many thanks! I thought it would be something simpler... Could you possibly explain a bit further about the generating function (what is x???) and also why the probability of getting an even number of successes is 1/2f(1)+f(-1)? Apologies but I am not very familiar with these topics!!
– Creton Laplace
Jul 18 at 10:24
@joriki: Many thanks! I thought it would be something simpler... Could you possibly explain a bit further about the generating function (what is x???) and also why the probability of getting an even number of successes is 1/2f(1)+f(-1)? Apologies but I am not very familiar with these topics!!
– Creton Laplace
Jul 18 at 10:24
@CretonLaplace: Nothing to apologize for! It just seemed more efficient to first see whether you are and then explain if not, than to explain first, possibly without need :-). In a probability generating function $f(x)=sum_kp_kx^k$ for the number of successes, the coefficient $p_k$ of the power $x^k$ is the probability for obtaining $k$ successes. $x$ is just a marker variable for the bookkeeping. The $n$-th kick contributes a factor $frac2n+x2n+1$, since the probability of success is $frac12n+1$. (Multiply it out up to $n=2$ or $3$ and you'll see why this works.)
– joriki
Jul 18 at 10:32
@CretonLaplace: Nothing to apologize for! It just seemed more efficient to first see whether you are and then explain if not, than to explain first, possibly without need :-). In a probability generating function $f(x)=sum_kp_kx^k$ for the number of successes, the coefficient $p_k$ of the power $x^k$ is the probability for obtaining $k$ successes. $x$ is just a marker variable for the bookkeeping. The $n$-th kick contributes a factor $frac2n+x2n+1$, since the probability of success is $frac12n+1$. (Multiply it out up to $n=2$ or $3$ and you'll see why this works.)
– joriki
Jul 18 at 10:32
@CretonLaplace: The probability of getting an even number of successes is the sum of the coefficients $p_k$ for even $k$. This is the the average of the sum of the coefficients and the alternating sum of the coefficients. The sum of the coefficients is $f(1)$, and the alternating sum of the coefficients is $f(-1)$.
– joriki
Jul 18 at 10:33
@CretonLaplace: The probability of getting an even number of successes is the sum of the coefficients $p_k$ for even $k$. This is the the average of the sum of the coefficients and the alternating sum of the coefficients. The sum of the coefficients is $f(1)$, and the alternating sum of the coefficients is $f(-1)$.
– joriki
Jul 18 at 10:33
And, of course, see en.wikipedia.org/wiki/Probability-generating_function. About it being something simpler -- I was surprised by the simple result! :-) There may well be a simpler way to derive it; this was just the first one that occurred to me.
– joriki
Jul 18 at 10:35
And, of course, see en.wikipedia.org/wiki/Probability-generating_function. About it being something simpler -- I was surprised by the simple result! :-) There may well be a simpler way to derive it; this was just the first one that occurred to me.
– joriki
Jul 18 at 10:35
It seems to me something must be wrong here. If $n=1$ then he will succeed in an even number of shots iff he fails at his first attempt. So $p_1=frac13$. This does not agree with $frac12(1+frac13)$. I come to $p_2n=frac2n+14n+1$ and $p_2n+1=frac2n+14n+3$, so that $p_19=frac1939$
– drhab
Jul 18 at 11:58
It seems to me something must be wrong here. If $n=1$ then he will succeed in an even number of shots iff he fails at his first attempt. So $p_1=frac13$. This does not agree with $frac12(1+frac13)$. I come to $p_2n=frac2n+14n+1$ and $p_2n+1=frac2n+14n+3$, so that $p_19=frac1939$
– drhab
Jul 18 at 11:58
 |Â
show 2 more comments
up vote
2
down vote
Hint
Try to work out the recursive equality:
$$left(2n+1right)p_n=p_n-1+2nleft(1-p_n-1right)$$
Where the probability that he succeeds in an even number of kicks by
taking $n$ penalties be denoted by $p_n$.
This is based on the thought that an even number can be achieved in two ways:
- an even number in the first $n-1$ efforts followed by a failure
- an odd number in the first $n-1$ efforts followed by a success.
Here $p_0=1$.
answered Jul 18 at 9:59
drhab
86.5k541118
add a comment |Â
up vote
2
down vote
Hint
Try to work out the recursive equality:
$$left(2n+1right)p_n=p_n-1+2nleft(1-p_n-1right)$$
Where the probability that he succeeds in an even number of kicks by
taking $n$ penalties be denoted by $p_n$.
This is based on the thought that an even number can be achieved in two ways:
- an even number in the first $n-1$ efforts followed by a failure
- an odd number in the first $n-1$ efforts followed by a success.
Here $p_0=1$.
answered Jul 18 at 9:59
drhab
86.5k541118
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Hint
Try to work out the recursive equality:
$$left(2n+1right)p_n=p_n-1+2nleft(1-p_n-1right)$$
Where the probability that he succeeds in an even number of kicks by
taking $n$ penalties be denoted by $p_n$.
This is based on the thought that an even number can be achieved in two ways:
- an even number in the first $n-1$ efforts followed by a failure
- an odd number in the first $n-1$ efforts followed by a success.
Here $p_0=1$.
answered Jul 18 at 9:59
drhab
86.5k541118
Hint
Try to work out the recursive equality:
$$left(2n+1right)p_n=p_n-1+2nleft(1-p_n-1right)$$
Where the probability that he succeeds in an even number of kicks by
taking $n$ penalties be denoted by $p_n$.
This is based on the thought that an even number can be achieved in two ways:
- an even number in the first $n-1$ efforts followed by a failure
- an odd number in the first $n-1$ efforts followed by a success.
Here $p_0=1$.
answered Jul 18 at 9:59
drhab
86.5k541118
answered Jul 18 at 9:59
drhab
86.5k541118
answered Jul 18 at 9:59
drhab
86.5k541118
answered Jul 18 at 9:59
drhab
86.5k541118
86.5k541118
add a comment |Â
add a comment |Â
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