Mixing
What is the probability that a player does not have at least 1 card of each suit with a 52-card deck?
$begingroup$
There are 52 cards in a deck, with 4 suits, each with 13 cards. You distribute the 52 cards, at random, to each of four players so that each has 13 in their hand. What is the probability that at least one player does not have at least 1 card of each suit?
I found this post relevant, though it only focuses on the first 13 cards.Standard playing card deck without jokers: Deal 13 cards. What is the probability at least 1 suit is not present?.
My question is after all 52 cards have been dealt. I've used this as a starting point, but am currently stuck as to how to approach this.
combinatorics
asked Jan 14 at 17:09
John MatthessonJohn Matthesson
111
$endgroup$
add a comment |
$begingroup$
There are 52 cards in a deck, with 4 suits, each with 13 cards. You distribute the 52 cards, at random, to each of four players so that each has 13 in their hand. What is the probability that at least one player does not have at least 1 card of each suit?
I found this post relevant, though it only focuses on the first 13 cards.Standard playing card deck without jokers: Deal 13 cards. What is the probability at least 1 suit is not present?.
My question is after all 52 cards have been dealt. I've used this as a starting point, but am currently stuck as to how to approach this.
combinatorics
asked Jan 14 at 17:09
John MatthessonJohn Matthesson
111
$endgroup$
1
$begingroup$
Very often these questions are easier if you look at them the other way round. How would you approach it if the question was "What's the probability that some player has at least one card from each suit?" and can you see how that's relevant to your question?
$endgroup$
– postmortes
Jan 14 at 17:22
$begingroup$
The post I linked to is the answer I'm looking for, I'm dumb. Thanks for your reply, you helped me to figure this out.
$endgroup$
– John Matthesson
Jan 14 at 18:15
$begingroup$
John, I the post you linked does not have the answer you are looking for. Call the players A,B,C,D. The post you linked is the probability that A is missing some suit. Your current question is the probability that at least one of A, B, C or D is missing some suit.
$endgroup$
– Mike Earnest
Jan 14 at 18:20
add a comment |
$begingroup$
There are 52 cards in a deck, with 4 suits, each with 13 cards. You distribute the 52 cards, at random, to each of four players so that each has 13 in their hand. What is the probability that at least one player does not have at least 1 card of each suit?
I found this post relevant, though it only focuses on the first 13 cards.Standard playing card deck without jokers: Deal 13 cards. What is the probability at least 1 suit is not present?.
My question is after all 52 cards have been dealt. I've used this as a starting point, but am currently stuck as to how to approach this.
combinatorics
asked Jan 14 at 17:09
John MatthessonJohn Matthesson
111
$endgroup$
There are 52 cards in a deck, with 4 suits, each with 13 cards. You distribute the 52 cards, at random, to each of four players so that each has 13 in their hand. What is the probability that at least one player does not have at least 1 card of each suit?
I found this post relevant, though it only focuses on the first 13 cards.Standard playing card deck without jokers: Deal 13 cards. What is the probability at least 1 suit is not present?.
My question is after all 52 cards have been dealt. I've used this as a starting point, but am currently stuck as to how to approach this.
combinatorics
combinatorics
asked Jan 14 at 17:09
John MatthessonJohn Matthesson
111
asked Jan 14 at 17:09
John MatthessonJohn Matthesson
111
asked Jan 14 at 17:09
John MatthessonJohn Matthesson
111
asked Jan 14 at 17:09
John MatthessonJohn Matthesson
111
asked Jan 14 at 17:09
John MatthessonJohn Matthesson
111
111
1
$begingroup$
Very often these questions are easier if you look at them the other way round. How would you approach it if the question was "What's the probability that some player has at least one card from each suit?" and can you see how that's relevant to your question?
$endgroup$
– postmortes
Jan 14 at 17:22
$begingroup$
The post I linked to is the answer I'm looking for, I'm dumb. Thanks for your reply, you helped me to figure this out.
$endgroup$
– John Matthesson
Jan 14 at 18:15
$begingroup$
John, I the post you linked does not have the answer you are looking for. Call the players A,B,C,D. The post you linked is the probability that A is missing some suit. Your current question is the probability that at least one of A, B, C or D is missing some suit.
$endgroup$
– Mike Earnest
Jan 14 at 18:20
add a comment |
1
$begingroup$
Very often these questions are easier if you look at them the other way round. How would you approach it if the question was "What's the probability that some player has at least one card from each suit?" and can you see how that's relevant to your question?
$endgroup$
– postmortes
Jan 14 at 17:22
$begingroup$
The post I linked to is the answer I'm looking for, I'm dumb. Thanks for your reply, you helped me to figure this out.
$endgroup$
– John Matthesson
Jan 14 at 18:15
$begingroup$
John, I the post you linked does not have the answer you are looking for. Call the players A,B,C,D. The post you linked is the probability that A is missing some suit. Your current question is the probability that at least one of A, B, C or D is missing some suit.
$endgroup$
– Mike Earnest
Jan 14 at 18:20
1
1
$begingroup$
Very often these questions are easier if you look at them the other way round. How would you approach it if the question was "What's the probability that some player has at least one card from each suit?" and can you see how that's relevant to your question?
$endgroup$
– postmortes
Jan 14 at 17:22
$begingroup$
Very often these questions are easier if you look at them the other way round. How would you approach it if the question was "What's the probability that some player has at least one card from each suit?" and can you see how that's relevant to your question?
$endgroup$
– postmortes
Jan 14 at 17:22
$begingroup$
The post I linked to is the answer I'm looking for, I'm dumb. Thanks for your reply, you helped me to figure this out.
$endgroup$
– John Matthesson
Jan 14 at 18:15
$begingroup$
The post I linked to is the answer I'm looking for, I'm dumb. Thanks for your reply, you helped me to figure this out.
$endgroup$
– John Matthesson
Jan 14 at 18:15
$begingroup$
John, I the post you linked does not have the answer you are looking for. Call the players A,B,C,D. The post you linked is the probability that A is missing some suit. Your current question is the probability that at least one of A, B, C or D is missing some suit.
$endgroup$
– Mike Earnest
Jan 14 at 18:20
$begingroup$
John, I the post you linked does not have the answer you are looking for. Call the players A,B,C,D. The post you linked is the probability that A is missing some suit. Your current question is the probability that at least one of A, B, C or D is missing some suit.
$endgroup$
– Mike Earnest
Jan 14 at 18:20
add a comment |
1 Answer
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$begingroup$
Perhaps there's some smart way to count the number of permutations of the 52 cards such that each player gets at least one card of each suite, which is the complement of what is asked for. But sometimes we do not need the exact solution and a simulation can give a pretty good approximation. In this case, with $10^6$ samples, we can estimate the probability being asked for is about $0.184171$. I bet this is pretty close the exact result.
Here's the code in Mathematica
In[2]:= n = 52
Out[2]= 52
In[13]:= n0 = n/4
Out[13]= 13
In[3]:= l = Range[n]
Out[3]= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17,
18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34,
35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52}
In[14]:= ll = Partition[l, n0]
Out[14]= {{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}, {14, 15, 16,
17, 18, 19, 20, 21, 22, 23, 24, 25, 26}, {27, 28, 29, 30, 31, 32,
33, 34, 35, 36, 37, 38, 39}, {40, 41, 42, 43, 44, 45, 46, 47, 48,
49, 50, 51, 52}}
In[39]:= containAll[sample_] := Module[{i, sl, crossl},
sl = Partition[sample, n0];
crossl = Tuples[{sl, ll}];
If[And @@ Map[ContainsAny[#[[1]], #[[2]]] &, crossl], 1, 0]
]
In[53]:= s0 =
Total[Parallelize[
Map[(Permute[l, #] // containAll) &,
RandomPermutation[n, 1000000]]]] // AbsoluteTiming
Out[53]= {191.22, 815829}
In[59]:= 1 - 815829/1000000 // N
Out[59]= 0.184171
answered Jan 14 at 18:39
ablmfablmf
2,54342452
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Perhaps there's some smart way to count the number of permutations of the 52 cards such that each player gets at least one card of each suite, which is the complement of what is asked for. But sometimes we do not need the exact solution and a simulation can give a pretty good approximation. In this case, with $10^6$ samples, we can estimate the probability being asked for is about $0.184171$. I bet this is pretty close the exact result.
Here's the code in Mathematica
In[2]:= n = 52
Out[2]= 52
In[13]:= n0 = n/4
Out[13]= 13
In[3]:= l = Range[n]
Out[3]= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17,
18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34,
35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52}
In[14]:= ll = Partition[l, n0]
Out[14]= {{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}, {14, 15, 16,
17, 18, 19, 20, 21, 22, 23, 24, 25, 26}, {27, 28, 29, 30, 31, 32,
33, 34, 35, 36, 37, 38, 39}, {40, 41, 42, 43, 44, 45, 46, 47, 48,
49, 50, 51, 52}}
In[39]:= containAll[sample_] := Module[{i, sl, crossl},
sl = Partition[sample, n0];
crossl = Tuples[{sl, ll}];
If[And @@ Map[ContainsAny[#[[1]], #[[2]]] &, crossl], 1, 0]
]
In[53]:= s0 =
Total[Parallelize[
Map[(Permute[l, #] // containAll) &,
RandomPermutation[n, 1000000]]]] // AbsoluteTiming
Out[53]= {191.22, 815829}
In[59]:= 1 - 815829/1000000 // N
Out[59]= 0.184171
answered Jan 14 at 18:39
ablmfablmf
2,54342452
$endgroup$
add a comment |
$begingroup$
Perhaps there's some smart way to count the number of permutations of the 52 cards such that each player gets at least one card of each suite, which is the complement of what is asked for. But sometimes we do not need the exact solution and a simulation can give a pretty good approximation. In this case, with $10^6$ samples, we can estimate the probability being asked for is about $0.184171$. I bet this is pretty close the exact result.
Here's the code in Mathematica
In[2]:= n = 52
Out[2]= 52
In[13]:= n0 = n/4
Out[13]= 13
In[3]:= l = Range[n]
Out[3]= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17,
18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34,
35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52}
In[14]:= ll = Partition[l, n0]
Out[14]= {{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}, {14, 15, 16,
17, 18, 19, 20, 21, 22, 23, 24, 25, 26}, {27, 28, 29, 30, 31, 32,
33, 34, 35, 36, 37, 38, 39}, {40, 41, 42, 43, 44, 45, 46, 47, 48,
49, 50, 51, 52}}
In[39]:= containAll[sample_] := Module[{i, sl, crossl},
sl = Partition[sample, n0];
crossl = Tuples[{sl, ll}];
If[And @@ Map[ContainsAny[#[[1]], #[[2]]] &, crossl], 1, 0]
]
In[53]:= s0 =
Total[Parallelize[
Map[(Permute[l, #] // containAll) &,
RandomPermutation[n, 1000000]]]] // AbsoluteTiming
Out[53]= {191.22, 815829}
In[59]:= 1 - 815829/1000000 // N
Out[59]= 0.184171
answered Jan 14 at 18:39
ablmfablmf
2,54342452
$endgroup$
add a comment |
$begingroup$
Perhaps there's some smart way to count the number of permutations of the 52 cards such that each player gets at least one card of each suite, which is the complement of what is asked for. But sometimes we do not need the exact solution and a simulation can give a pretty good approximation. In this case, with $10^6$ samples, we can estimate the probability being asked for is about $0.184171$. I bet this is pretty close the exact result.
Here's the code in Mathematica
In[2]:= n = 52
Out[2]= 52
In[13]:= n0 = n/4
Out[13]= 13
In[3]:= l = Range[n]
Out[3]= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17,
18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34,
35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52}
In[14]:= ll = Partition[l, n0]
Out[14]= {{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}, {14, 15, 16,
17, 18, 19, 20, 21, 22, 23, 24, 25, 26}, {27, 28, 29, 30, 31, 32,
33, 34, 35, 36, 37, 38, 39}, {40, 41, 42, 43, 44, 45, 46, 47, 48,
49, 50, 51, 52}}
In[39]:= containAll[sample_] := Module[{i, sl, crossl},
sl = Partition[sample, n0];
crossl = Tuples[{sl, ll}];
If[And @@ Map[ContainsAny[#[[1]], #[[2]]] &, crossl], 1, 0]
]
In[53]:= s0 =
Total[Parallelize[
Map[(Permute[l, #] // containAll) &,
RandomPermutation[n, 1000000]]]] // AbsoluteTiming
Out[53]= {191.22, 815829}
In[59]:= 1 - 815829/1000000 // N
Out[59]= 0.184171
answered Jan 14 at 18:39
ablmfablmf
2,54342452
$endgroup$
Perhaps there's some smart way to count the number of permutations of the 52 cards such that each player gets at least one card of each suite, which is the complement of what is asked for. But sometimes we do not need the exact solution and a simulation can give a pretty good approximation. In this case, with $10^6$ samples, we can estimate the probability being asked for is about $0.184171$. I bet this is pretty close the exact result.
Here's the code in Mathematica
In[2]:= n = 52
Out[2]= 52
In[13]:= n0 = n/4
Out[13]= 13
In[3]:= l = Range[n]
Out[3]= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17,
18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34,
35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52}
In[14]:= ll = Partition[l, n0]
Out[14]= {{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}, {14, 15, 16,
17, 18, 19, 20, 21, 22, 23, 24, 25, 26}, {27, 28, 29, 30, 31, 32,
33, 34, 35, 36, 37, 38, 39}, {40, 41, 42, 43, 44, 45, 46, 47, 48,
49, 50, 51, 52}}
In[39]:= containAll[sample_] := Module[{i, sl, crossl},
sl = Partition[sample, n0];
crossl = Tuples[{sl, ll}];
If[And @@ Map[ContainsAny[#[[1]], #[[2]]] &, crossl], 1, 0]
]
In[53]:= s0 =
Total[Parallelize[
Map[(Permute[l, #] // containAll) &,
RandomPermutation[n, 1000000]]]] // AbsoluteTiming
Out[53]= {191.22, 815829}
In[59]:= 1 - 815829/1000000 // N
Out[59]= 0.184171
answered Jan 14 at 18:39
ablmfablmf
2,54342452
answered Jan 14 at 18:39
ablmfablmf
2,54342452
answered Jan 14 at 18:39
ablmfablmf
2,54342452
answered Jan 14 at 18:39
ablmfablmf
2,54342452
2,54342452
add a comment |
add a comment |
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1
$begingroup$
Very often these questions are easier if you look at them the other way round. How would you approach it if the question was "What's the probability that some player has at least one card from each suit?" and can you see how that's relevant to your question?
$endgroup$
– postmortes
Jan 14 at 17:22
$begingroup$
The post I linked to is the answer I'm looking for, I'm dumb. Thanks for your reply, you helped me to figure this out.
$endgroup$
– John Matthesson
Jan 14 at 18:15
$begingroup$
John, I the post you linked does not have the answer you are looking for. Call the players A,B,C,D. The post you linked is the probability that A is missing some suit. Your current question is the probability that at least one of A, B, C or D is missing some suit.
$endgroup$
– Mike Earnest
Jan 14 at 18:20