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Find the probability that the thirteenth spade will appear before the thirteenth diamond?
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The $52$ cards of an ordinary deck of cards are placed successively one after the other from left to right. Find the probability that the thirteenth spade will appear before the thirteenth diamond?
This appears to be a hard problem since counting the favorable cases are hard to count. Clearly, the position of the $13$th diamond must be between $26$ and $N$. Now, I tried to count for specific values, but I find this difficult, perhaps there is a better way?
Note:
I checked the answer and the favorable cases are supposed to be:
$$sum_k=26^52(n-1)!(52-n)!binom26n-26cdot 13.$$
So I guess this ugly sum divided by $(52)!$ is equal to $1/2,$ right?
PS:
Question:
Answer:
probability combinatorics
asked 32 mins ago
Hello_World
3,43321429
add a comment |Â
up vote
3
down vote
favorite
The $52$ cards of an ordinary deck of cards are placed successively one after the other from left to right. Find the probability that the thirteenth spade will appear before the thirteenth diamond?
This appears to be a hard problem since counting the favorable cases are hard to count. Clearly, the position of the $13$th diamond must be between $26$ and $N$. Now, I tried to count for specific values, but I find this difficult, perhaps there is a better way?
Note:
I checked the answer and the favorable cases are supposed to be:
$$sum_k=26^52(n-1)!(52-n)!binom26n-26cdot 13.$$
So I guess this ugly sum divided by $(52)!$ is equal to $1/2,$ right?
PS:
Question:
Answer:
probability combinatorics
asked 32 mins ago
Hello_World
3,43321429
Your sum involves products of positive integers so will be more than $frac12$ and is indeed more than $8 times 10^52$
– Henry
21 mins ago
I have made an edit.
– Hello_World
19 mins ago
Divided by $52!$ is it about $10^-15$
– Henry
18 mins ago
@Henry Actually, I get $frac12$ when I divide the sum by $52!$, assuming that where $n$ appears it is supposed to be $k$.
– Misha Lavrov
16 mins ago
I made another edit @Henry.
– Hello_World
15 mins ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
The $52$ cards of an ordinary deck of cards are placed successively one after the other from left to right. Find the probability that the thirteenth spade will appear before the thirteenth diamond?
This appears to be a hard problem since counting the favorable cases are hard to count. Clearly, the position of the $13$th diamond must be between $26$ and $N$. Now, I tried to count for specific values, but I find this difficult, perhaps there is a better way?
Note:
I checked the answer and the favorable cases are supposed to be:
$$sum_k=26^52(n-1)!(52-n)!binom26n-26cdot 13.$$
So I guess this ugly sum divided by $(52)!$ is equal to $1/2,$ right?
PS:
Question:
Answer:
probability combinatorics
asked 32 mins ago
Hello_World
3,43321429
The $52$ cards of an ordinary deck of cards are placed successively one after the other from left to right. Find the probability that the thirteenth spade will appear before the thirteenth diamond?
This appears to be a hard problem since counting the favorable cases are hard to count. Clearly, the position of the $13$th diamond must be between $26$ and $N$. Now, I tried to count for specific values, but I find this difficult, perhaps there is a better way?
Note:
I checked the answer and the favorable cases are supposed to be:
$$sum_k=26^52(n-1)!(52-n)!binom26n-26cdot 13.$$
So I guess this ugly sum divided by $(52)!$ is equal to $1/2,$ right?
PS:
Question:
Answer:
probability combinatorics
probability combinatorics
asked 32 mins ago
Hello_World
3,43321429
asked 32 mins ago
Hello_World
3,43321429
edited 15 mins ago
asked 32 mins ago
Hello_World
3,43321429
asked 32 mins ago
Hello_World
3,43321429
asked 32 mins ago
Hello_World
3,43321429
3,43321429
Your sum involves products of positive integers so will be more than $frac12$ and is indeed more than $8 times 10^52$
– Henry
21 mins ago
I have made an edit.
– Hello_World
19 mins ago
Divided by $52!$ is it about $10^-15$
– Henry
18 mins ago
@Henry Actually, I get $frac12$ when I divide the sum by $52!$, assuming that where $n$ appears it is supposed to be $k$.
– Misha Lavrov
16 mins ago
I made another edit @Henry.
– Hello_World
15 mins ago
add a comment |Â
Your sum involves products of positive integers so will be more than $frac12$ and is indeed more than $8 times 10^52$
– Henry
21 mins ago
I have made an edit.
– Hello_World
19 mins ago
Divided by $52!$ is it about $10^-15$
– Henry
18 mins ago
@Henry Actually, I get $frac12$ when I divide the sum by $52!$, assuming that where $n$ appears it is supposed to be $k$.
– Misha Lavrov
16 mins ago
I made another edit @Henry.
– Hello_World
15 mins ago
Your sum involves products of positive integers so will be more than $frac12$ and is indeed more than $8 times 10^52$
– Henry
21 mins ago
Your sum involves products of positive integers so will be more than $frac12$ and is indeed more than $8 times 10^52$
– Henry
21 mins ago
I have made an edit.
– Hello_World
19 mins ago
I have made an edit.
– Hello_World
19 mins ago
Divided by $52!$ is it about $10^-15$
– Henry
18 mins ago
Divided by $52!$ is it about $10^-15$
– Henry
18 mins ago
@Henry Actually, I get $frac12$ when I divide the sum by $52!$, assuming that where $n$ appears it is supposed to be $k$.
– Misha Lavrov
16 mins ago
@Henry Actually, I get $frac12$ when I divide the sum by $52!$, assuming that where $n$ appears it is supposed to be $k$.
– Misha Lavrov
16 mins ago
I made another edit @Henry.
– Hello_World
15 mins ago
I made another edit @Henry.
– Hello_World
15 mins ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
4
down vote
Use symmetry.
Let us write $P_1$ as the probability that the thirteenth spade comes before the thirteenth diamond, and let $P_2$ be the probability that the thirteenth diamond comes before the thirteenth spade.
Clearly, one of these must occur, and they can't both occur. Thus, we have that
$$P_1+P_2=1$$
Furthermore, since suits are arbitrary, we may exploit symmetry to say that $P_1=P_2$.
Thus, $P_1=P_2=1/2$.
Equivalently, the number of deck arrangements in which the last spade comes before the last diamond is $52!/2$.
answered 27 mins ago
Frpzzd
17.6k63491
add a comment |Â
up vote
3
down vote
By symmetry, the probability is $frac 12$
answered 27 mins ago
Ross Millikan
282k23191359
add a comment |Â
up vote
1
down vote
By symmetry, the probability is $frac12$.
Formally, we may biject the deck orderings where the last spade comes out first, with the deck orderings where the last diamond comes out first, by the involution that switches $Aspadesuit$ with $Adiamondsuit$, $2spadesuit$ with $2diamondsuit$, ..., $Kspadesuit$ with $Kdiamondsuit$. So there must be $frac12 cdot 52!$ orderings of each kind, which makes the probability $frac12$.
answered 27 mins ago
Misha Lavrov
38.7k55195
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Use symmetry.
Let us write $P_1$ as the probability that the thirteenth spade comes before the thirteenth diamond, and let $P_2$ be the probability that the thirteenth diamond comes before the thirteenth spade.
Clearly, one of these must occur, and they can't both occur. Thus, we have that
$$P_1+P_2=1$$
Furthermore, since suits are arbitrary, we may exploit symmetry to say that $P_1=P_2$.
Thus, $P_1=P_2=1/2$.
Equivalently, the number of deck arrangements in which the last spade comes before the last diamond is $52!/2$.
answered 27 mins ago
Frpzzd
17.6k63491
add a comment |Â
up vote
4
down vote
Use symmetry.
Let us write $P_1$ as the probability that the thirteenth spade comes before the thirteenth diamond, and let $P_2$ be the probability that the thirteenth diamond comes before the thirteenth spade.
Clearly, one of these must occur, and they can't both occur. Thus, we have that
$$P_1+P_2=1$$
Furthermore, since suits are arbitrary, we may exploit symmetry to say that $P_1=P_2$.
Thus, $P_1=P_2=1/2$.
Equivalently, the number of deck arrangements in which the last spade comes before the last diamond is $52!/2$.
answered 27 mins ago
Frpzzd
17.6k63491
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Use symmetry.
Let us write $P_1$ as the probability that the thirteenth spade comes before the thirteenth diamond, and let $P_2$ be the probability that the thirteenth diamond comes before the thirteenth spade.
Clearly, one of these must occur, and they can't both occur. Thus, we have that
$$P_1+P_2=1$$
Furthermore, since suits are arbitrary, we may exploit symmetry to say that $P_1=P_2$.
Thus, $P_1=P_2=1/2$.
Equivalently, the number of deck arrangements in which the last spade comes before the last diamond is $52!/2$.
answered 27 mins ago
Frpzzd
17.6k63491
Use symmetry.
Let us write $P_1$ as the probability that the thirteenth spade comes before the thirteenth diamond, and let $P_2$ be the probability that the thirteenth diamond comes before the thirteenth spade.
Clearly, one of these must occur, and they can't both occur. Thus, we have that
$$P_1+P_2=1$$
Furthermore, since suits are arbitrary, we may exploit symmetry to say that $P_1=P_2$.
Thus, $P_1=P_2=1/2$.
Equivalently, the number of deck arrangements in which the last spade comes before the last diamond is $52!/2$.
answered 27 mins ago
Frpzzd
17.6k63491
edited 17 mins ago
answered 27 mins ago
Frpzzd
17.6k63491
answered 27 mins ago
Frpzzd
17.6k63491
answered 27 mins ago
Frpzzd
17.6k63491
17.6k63491
add a comment |Â
add a comment |Â
up vote
3
down vote
By symmetry, the probability is $frac 12$
answered 27 mins ago
Ross Millikan
282k23191359
add a comment |Â
up vote
3
down vote
By symmetry, the probability is $frac 12$
answered 27 mins ago
Ross Millikan
282k23191359
add a comment |Â
up vote
3
down vote
up vote
3
down vote
By symmetry, the probability is $frac 12$
answered 27 mins ago
Ross Millikan
282k23191359
By symmetry, the probability is $frac 12$
answered 27 mins ago
Ross Millikan
282k23191359
answered 27 mins ago
Ross Millikan
282k23191359
answered 27 mins ago
Ross Millikan
282k23191359
answered 27 mins ago
Ross Millikan
282k23191359
282k23191359
add a comment |Â
add a comment |Â
up vote
1
down vote
By symmetry, the probability is $frac12$.
Formally, we may biject the deck orderings where the last spade comes out first, with the deck orderings where the last diamond comes out first, by the involution that switches $Aspadesuit$ with $Adiamondsuit$, $2spadesuit$ with $2diamondsuit$, ..., $Kspadesuit$ with $Kdiamondsuit$. So there must be $frac12 cdot 52!$ orderings of each kind, which makes the probability $frac12$.
answered 27 mins ago
Misha Lavrov
38.7k55195
add a comment |Â
up vote
1
down vote
By symmetry, the probability is $frac12$.
Formally, we may biject the deck orderings where the last spade comes out first, with the deck orderings where the last diamond comes out first, by the involution that switches $Aspadesuit$ with $Adiamondsuit$, $2spadesuit$ with $2diamondsuit$, ..., $Kspadesuit$ with $Kdiamondsuit$. So there must be $frac12 cdot 52!$ orderings of each kind, which makes the probability $frac12$.
answered 27 mins ago
Misha Lavrov
38.7k55195
add a comment |Â
up vote
1
down vote
up vote
1
down vote
By symmetry, the probability is $frac12$.
Formally, we may biject the deck orderings where the last spade comes out first, with the deck orderings where the last diamond comes out first, by the involution that switches $Aspadesuit$ with $Adiamondsuit$, $2spadesuit$ with $2diamondsuit$, ..., $Kspadesuit$ with $Kdiamondsuit$. So there must be $frac12 cdot 52!$ orderings of each kind, which makes the probability $frac12$.
answered 27 mins ago
Misha Lavrov
38.7k55195
By symmetry, the probability is $frac12$.
Formally, we may biject the deck orderings where the last spade comes out first, with the deck orderings where the last diamond comes out first, by the involution that switches $Aspadesuit$ with $Adiamondsuit$, $2spadesuit$ with $2diamondsuit$, ..., $Kspadesuit$ with $Kdiamondsuit$. So there must be $frac12 cdot 52!$ orderings of each kind, which makes the probability $frac12$.
answered 27 mins ago
Misha Lavrov
38.7k55195
answered 27 mins ago
Misha Lavrov
38.7k55195
answered 27 mins ago
Misha Lavrov
38.7k55195
answered 27 mins ago
Misha Lavrov
38.7k55195
38.7k55195
add a comment |Â
add a comment |Â
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Your sum involves products of positive integers so will be more than $frac12$ and is indeed more than $8 times 10^52$
– Henry
21 mins ago
I have made an edit.
– Hello_World
19 mins ago
Divided by $52!$ is it about $10^-15$
– Henry
18 mins ago
@Henry Actually, I get $frac12$ when I divide the sum by $52!$, assuming that where $n$ appears it is supposed to be $k$.
– Misha Lavrov
16 mins ago
I made another edit @Henry.
– Hello_World
15 mins ago