$P(E|H)$ in a coin-flip problem
I am looking for a simple verification of my thinking:
For the question, "what is the probability of observing a total of $k$ heads when flipping $n$ coins, if one particular of the $n$ coins is known to be heads?" my answer is the probability of observing $k-1$ using $n-1$ coins.
and
For the question, "what is the probability of observing a total of $k$ heads when flipping $n$ coins, if one particular of the $n$ coins is known to be tails?" my answer is the probability of observing $k$ using $n - 1$ coins.
Is my logic correct?
Note: I suppose another way to think of "particular" is to think of a marked coin among the $n$ and considering that this coin is known to be heads (or tail).
probability
asked Nov 28 '18 at 11:04
user120911
233110
add a comment |
I am looking for a simple verification of my thinking:
For the question, "what is the probability of observing a total of $k$ heads when flipping $n$ coins, if one particular of the $n$ coins is known to be heads?" my answer is the probability of observing $k-1$ using $n-1$ coins.
and
For the question, "what is the probability of observing a total of $k$ heads when flipping $n$ coins, if one particular of the $n$ coins is known to be tails?" my answer is the probability of observing $k$ using $n - 1$ coins.
Is my logic correct?
Note: I suppose another way to think of "particular" is to think of a marked coin among the $n$ and considering that this coin is known to be heads (or tail).
probability
asked Nov 28 '18 at 11:04
user120911
233110
Well, let's try $k=1,n=2$. For the first: knowing that one coin is $H$ means we are in one of $HH,TH,HT$ with equal probability. Thus the probability of seeing a total of one $H$ is $frac 23$ which is not what your conjecture gives.
– lulu
Nov 28 '18 at 11:25
You just made me realize my description of the problem is wrong. I didn't realize that writing "one of the coins" as opposed to "one particular coin" matters. Please see my edit.
– user120911
Nov 28 '18 at 13:12
Yes, that makes a big difference. If you say "the first coin is $H$" then your expression is correct.
– lulu
Nov 28 '18 at 13:14
Just to be clear, you believe my stated logic is correct if I consider "the first coin" in the sequence (or, I suppose, any particular or marked coin)?
– user120911
Nov 28 '18 at 13:24
1
Yes. Your reasoning is correct if we specify a single coin.
– lulu
Nov 28 '18 at 13:36
add a comment |
I am looking for a simple verification of my thinking:
For the question, "what is the probability of observing a total of $k$ heads when flipping $n$ coins, if one particular of the $n$ coins is known to be heads?" my answer is the probability of observing $k-1$ using $n-1$ coins.
and
For the question, "what is the probability of observing a total of $k$ heads when flipping $n$ coins, if one particular of the $n$ coins is known to be tails?" my answer is the probability of observing $k$ using $n - 1$ coins.
Is my logic correct?
Note: I suppose another way to think of "particular" is to think of a marked coin among the $n$ and considering that this coin is known to be heads (or tail).
probability
asked Nov 28 '18 at 11:04
user120911
233110
I am looking for a simple verification of my thinking:
For the question, "what is the probability of observing a total of $k$ heads when flipping $n$ coins, if one particular of the $n$ coins is known to be heads?" my answer is the probability of observing $k-1$ using $n-1$ coins.
and
For the question, "what is the probability of observing a total of $k$ heads when flipping $n$ coins, if one particular of the $n$ coins is known to be tails?" my answer is the probability of observing $k$ using $n - 1$ coins.
Is my logic correct?
Note: I suppose another way to think of "particular" is to think of a marked coin among the $n$ and considering that this coin is known to be heads (or tail).
probability
probability
asked Nov 28 '18 at 11:04
user120911
233110
asked Nov 28 '18 at 11:04
user120911
233110
edited Nov 28 '18 at 13:16
asked Nov 28 '18 at 11:04
user120911
233110
asked Nov 28 '18 at 11:04
user120911
233110
asked Nov 28 '18 at 11:04
user120911
233110
233110
Well, let's try $k=1,n=2$. For the first: knowing that one coin is $H$ means we are in one of $HH,TH,HT$ with equal probability. Thus the probability of seeing a total of one $H$ is $frac 23$ which is not what your conjecture gives.
– lulu
Nov 28 '18 at 11:25
You just made me realize my description of the problem is wrong. I didn't realize that writing "one of the coins" as opposed to "one particular coin" matters. Please see my edit.
– user120911
Nov 28 '18 at 13:12
Yes, that makes a big difference. If you say "the first coin is $H$" then your expression is correct.
– lulu
Nov 28 '18 at 13:14
Just to be clear, you believe my stated logic is correct if I consider "the first coin" in the sequence (or, I suppose, any particular or marked coin)?
– user120911
Nov 28 '18 at 13:24
1
Yes. Your reasoning is correct if we specify a single coin.
– lulu
Nov 28 '18 at 13:36
add a comment |
Well, let's try $k=1,n=2$. For the first: knowing that one coin is $H$ means we are in one of $HH,TH,HT$ with equal probability. Thus the probability of seeing a total of one $H$ is $frac 23$ which is not what your conjecture gives.
– lulu
Nov 28 '18 at 11:25
You just made me realize my description of the problem is wrong. I didn't realize that writing "one of the coins" as opposed to "one particular coin" matters. Please see my edit.
– user120911
Nov 28 '18 at 13:12
Yes, that makes a big difference. If you say "the first coin is $H$" then your expression is correct.
– lulu
Nov 28 '18 at 13:14
Just to be clear, you believe my stated logic is correct if I consider "the first coin" in the sequence (or, I suppose, any particular or marked coin)?
– user120911
Nov 28 '18 at 13:24
1
Yes. Your reasoning is correct if we specify a single coin.
– lulu
Nov 28 '18 at 13:36
Well, let's try $k=1,n=2$. For the first: knowing that one coin is $H$ means we are in one of $HH,TH,HT$ with equal probability. Thus the probability of seeing a total of one $H$ is $frac 23$ which is not what your conjecture gives.
– lulu
Nov 28 '18 at 11:25
Well, let's try $k=1,n=2$. For the first: knowing that one coin is $H$ means we are in one of $HH,TH,HT$ with equal probability. Thus the probability of seeing a total of one $H$ is $frac 23$ which is not what your conjecture gives.
– lulu
Nov 28 '18 at 11:25
You just made me realize my description of the problem is wrong. I didn't realize that writing "one of the coins" as opposed to "one particular coin" matters. Please see my edit.
– user120911
Nov 28 '18 at 13:12
You just made me realize my description of the problem is wrong. I didn't realize that writing "one of the coins" as opposed to "one particular coin" matters. Please see my edit.
– user120911
Nov 28 '18 at 13:12
Yes, that makes a big difference. If you say "the first coin is $H$" then your expression is correct.
– lulu
Nov 28 '18 at 13:14
Yes, that makes a big difference. If you say "the first coin is $H$" then your expression is correct.
– lulu
Nov 28 '18 at 13:14
Just to be clear, you believe my stated logic is correct if I consider "the first coin" in the sequence (or, I suppose, any particular or marked coin)?
– user120911
Nov 28 '18 at 13:24
Just to be clear, you believe my stated logic is correct if I consider "the first coin" in the sequence (or, I suppose, any particular or marked coin)?
– user120911
Nov 28 '18 at 13:24
1
1
Yes. Your reasoning is correct if we specify a single coin.
– lulu
Nov 28 '18 at 13:36
Yes. Your reasoning is correct if we specify a single coin.
– lulu
Nov 28 '18 at 13:36
add a comment |
1 Answer
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I will use 1, 0 instead of H (head) and T (tails) below, i feel it is simpler to type. Let us consider the particular case of $n=3$ coins, and we let $k$ be one of $0,1,2,3$.
(1) The conditional probability w.r.t. the fact that there is at least one H (1).
All possibilities, the "big probability space" for modeling the problem is $Omega={0,1}^{times3}$, the cartesian product of three copies of the space modeling one coin flip. It has $2^3=8$ elements. Now the "restricted space" is the subspace $Omega'=Omega-{000}$. (I will write please the elements as words instead of tuples, so $000$ is short for $(0,0,0)$.) Then $k$ heads occur respectively
$$
binom 30 ,
binom 31 ,
binom 32 ,
binom 33
$$
many times
in $Omega$, thus with probability $binom 3k/2^3$ in $Omega$. In the "restricted space", we count one less element for the first case, thus respectively for $k=0,1,2,3$
$$
binom 30-1 ,
binom 31 ,
binom 32 ,
binom 33
$$
many times, and the probability is obtained by dividing with $|Omega'|=|Omega-1|=2^3-1$.
So the conditional probability asked for is $0$ for $k=0$, and $binom 3k/(2^3-1)$.
Try to count considering explicitly the elements of the space $Omega$:
000 - not in Omega'
001
010
011
100
101
110
111
For a general $n$ we have for $k=0$ the conditional probability $0$, else, for $k>0$ the corresponding $binom nk/(2^n-1)$.
(2) The conditional probability w.r.t. the fact that there is at least one T (0).
Same argument. Answer:
For a general $n$ we have for $k=n$ the conditional probability $0$, else, for $k<n$ the corresponding $binom nk/(2^n-1)$.
answered Nov 28 '18 at 11:28
dan_fulea
6,2151312
I am sorry to have made you venture this answer when my question was imprecise.
– user120911
Nov 28 '18 at 13:21
add a comment |
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I will use 1, 0 instead of H (head) and T (tails) below, i feel it is simpler to type. Let us consider the particular case of $n=3$ coins, and we let $k$ be one of $0,1,2,3$.
(1) The conditional probability w.r.t. the fact that there is at least one H (1).
All possibilities, the "big probability space" for modeling the problem is $Omega={0,1}^{times3}$, the cartesian product of three copies of the space modeling one coin flip. It has $2^3=8$ elements. Now the "restricted space" is the subspace $Omega'=Omega-{000}$. (I will write please the elements as words instead of tuples, so $000$ is short for $(0,0,0)$.) Then $k$ heads occur respectively
$$
binom 30 ,
binom 31 ,
binom 32 ,
binom 33
$$
many times
in $Omega$, thus with probability $binom 3k/2^3$ in $Omega$. In the "restricted space", we count one less element for the first case, thus respectively for $k=0,1,2,3$
$$
binom 30-1 ,
binom 31 ,
binom 32 ,
binom 33
$$
many times, and the probability is obtained by dividing with $|Omega'|=|Omega-1|=2^3-1$.
So the conditional probability asked for is $0$ for $k=0$, and $binom 3k/(2^3-1)$.
Try to count considering explicitly the elements of the space $Omega$:
000 - not in Omega'
001
010
011
100
101
110
111
For a general $n$ we have for $k=0$ the conditional probability $0$, else, for $k>0$ the corresponding $binom nk/(2^n-1)$.
(2) The conditional probability w.r.t. the fact that there is at least one T (0).
Same argument. Answer:
For a general $n$ we have for $k=n$ the conditional probability $0$, else, for $k<n$ the corresponding $binom nk/(2^n-1)$.
answered Nov 28 '18 at 11:28
dan_fulea
6,2151312
I am sorry to have made you venture this answer when my question was imprecise.
– user120911
Nov 28 '18 at 13:21
add a comment |
I will use 1, 0 instead of H (head) and T (tails) below, i feel it is simpler to type. Let us consider the particular case of $n=3$ coins, and we let $k$ be one of $0,1,2,3$.
(1) The conditional probability w.r.t. the fact that there is at least one H (1).
All possibilities, the "big probability space" for modeling the problem is $Omega={0,1}^{times3}$, the cartesian product of three copies of the space modeling one coin flip. It has $2^3=8$ elements. Now the "restricted space" is the subspace $Omega'=Omega-{000}$. (I will write please the elements as words instead of tuples, so $000$ is short for $(0,0,0)$.) Then $k$ heads occur respectively
$$
binom 30 ,
binom 31 ,
binom 32 ,
binom 33
$$
many times
in $Omega$, thus with probability $binom 3k/2^3$ in $Omega$. In the "restricted space", we count one less element for the first case, thus respectively for $k=0,1,2,3$
$$
binom 30-1 ,
binom 31 ,
binom 32 ,
binom 33
$$
many times, and the probability is obtained by dividing with $|Omega'|=|Omega-1|=2^3-1$.
So the conditional probability asked for is $0$ for $k=0$, and $binom 3k/(2^3-1)$.
Try to count considering explicitly the elements of the space $Omega$:
000 - not in Omega'
001
010
011
100
101
110
111
For a general $n$ we have for $k=0$ the conditional probability $0$, else, for $k>0$ the corresponding $binom nk/(2^n-1)$.
(2) The conditional probability w.r.t. the fact that there is at least one T (0).
Same argument. Answer:
For a general $n$ we have for $k=n$ the conditional probability $0$, else, for $k<n$ the corresponding $binom nk/(2^n-1)$.
answered Nov 28 '18 at 11:28
dan_fulea
6,2151312
I am sorry to have made you venture this answer when my question was imprecise.
– user120911
Nov 28 '18 at 13:21
add a comment |
I will use 1, 0 instead of H (head) and T (tails) below, i feel it is simpler to type. Let us consider the particular case of $n=3$ coins, and we let $k$ be one of $0,1,2,3$.
(1) The conditional probability w.r.t. the fact that there is at least one H (1).
All possibilities, the "big probability space" for modeling the problem is $Omega={0,1}^{times3}$, the cartesian product of three copies of the space modeling one coin flip. It has $2^3=8$ elements. Now the "restricted space" is the subspace $Omega'=Omega-{000}$. (I will write please the elements as words instead of tuples, so $000$ is short for $(0,0,0)$.) Then $k$ heads occur respectively
$$
binom 30 ,
binom 31 ,
binom 32 ,
binom 33
$$
many times
in $Omega$, thus with probability $binom 3k/2^3$ in $Omega$. In the "restricted space", we count one less element for the first case, thus respectively for $k=0,1,2,3$
$$
binom 30-1 ,
binom 31 ,
binom 32 ,
binom 33
$$
many times, and the probability is obtained by dividing with $|Omega'|=|Omega-1|=2^3-1$.
So the conditional probability asked for is $0$ for $k=0$, and $binom 3k/(2^3-1)$.
Try to count considering explicitly the elements of the space $Omega$:
000 - not in Omega'
001
010
011
100
101
110
111
For a general $n$ we have for $k=0$ the conditional probability $0$, else, for $k>0$ the corresponding $binom nk/(2^n-1)$.
(2) The conditional probability w.r.t. the fact that there is at least one T (0).
Same argument. Answer:
For a general $n$ we have for $k=n$ the conditional probability $0$, else, for $k<n$ the corresponding $binom nk/(2^n-1)$.
answered Nov 28 '18 at 11:28
dan_fulea
6,2151312
I will use 1, 0 instead of H (head) and T (tails) below, i feel it is simpler to type. Let us consider the particular case of $n=3$ coins, and we let $k$ be one of $0,1,2,3$.
(1) The conditional probability w.r.t. the fact that there is at least one H (1).
All possibilities, the "big probability space" for modeling the problem is $Omega={0,1}^{times3}$, the cartesian product of three copies of the space modeling one coin flip. It has $2^3=8$ elements. Now the "restricted space" is the subspace $Omega'=Omega-{000}$. (I will write please the elements as words instead of tuples, so $000$ is short for $(0,0,0)$.) Then $k$ heads occur respectively
$$
binom 30 ,
binom 31 ,
binom 32 ,
binom 33
$$
many times
in $Omega$, thus with probability $binom 3k/2^3$ in $Omega$. In the "restricted space", we count one less element for the first case, thus respectively for $k=0,1,2,3$
$$
binom 30-1 ,
binom 31 ,
binom 32 ,
binom 33
$$
many times, and the probability is obtained by dividing with $|Omega'|=|Omega-1|=2^3-1$.
So the conditional probability asked for is $0$ for $k=0$, and $binom 3k/(2^3-1)$.
Try to count considering explicitly the elements of the space $Omega$:
000 - not in Omega'
001
010
011
100
101
110
111
For a general $n$ we have for $k=0$ the conditional probability $0$, else, for $k>0$ the corresponding $binom nk/(2^n-1)$.
(2) The conditional probability w.r.t. the fact that there is at least one T (0).
Same argument. Answer:
For a general $n$ we have for $k=n$ the conditional probability $0$, else, for $k<n$ the corresponding $binom nk/(2^n-1)$.
answered Nov 28 '18 at 11:28
dan_fulea
6,2151312
answered Nov 28 '18 at 11:28
dan_fulea
6,2151312
answered Nov 28 '18 at 11:28
dan_fulea
6,2151312
answered Nov 28 '18 at 11:28
dan_fulea
6,2151312
6,2151312
I am sorry to have made you venture this answer when my question was imprecise.
– user120911
Nov 28 '18 at 13:21
add a comment |
I am sorry to have made you venture this answer when my question was imprecise.
– user120911
Nov 28 '18 at 13:21
I am sorry to have made you venture this answer when my question was imprecise.
– user120911
Nov 28 '18 at 13:21
I am sorry to have made you venture this answer when my question was imprecise.
– user120911
Nov 28 '18 at 13:21
add a comment |
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Well, let's try $k=1,n=2$. For the first: knowing that one coin is $H$ means we are in one of $HH,TH,HT$ with equal probability. Thus the probability of seeing a total of one $H$ is $frac 23$ which is not what your conjecture gives.
– lulu
Nov 28 '18 at 11:25
You just made me realize my description of the problem is wrong. I didn't realize that writing "one of the coins" as opposed to "one particular coin" matters. Please see my edit.
– user120911
Nov 28 '18 at 13:12
Yes, that makes a big difference. If you say "the first coin is $H$" then your expression is correct.
– lulu
Nov 28 '18 at 13:14
Just to be clear, you believe my stated logic is correct if I consider "the first coin" in the sequence (or, I suppose, any particular or marked coin)?
– user120911
Nov 28 '18 at 13:24
1
Yes. Your reasoning is correct if we specify a single coin.
– lulu
Nov 28 '18 at 13:36