How many different ways are there to get a 21 in Black Jack?
1
$begingroup$
Using one deck of cards.
Taking into account different suites. However, the way they are arranged would not be accounted for. Ace of Spades and Queen of Diamonds would be the same as Queen of Diamonds and Ace of Spades.
Pictures (Jacks, Queens, Kings) = 10
Aces = 11, or 1
The way I see it, 21 can be split into:
20 + 1, 19 + 2, 18 + 3 ... 1 + 20. Each of those numbers can further be split down. It appears to be kind of recursive?
Then we have 2-card ways, 3-card ways, 4-card ways and so on ...
It seems to me a very, very tough problem and I have no idea how to approach it. Any ideas?
permutations combinations card-games
asked Aug 19 '16 at 3:39
KosKos
12517
$endgroup$
add a comment |
1
$begingroup$
Using one deck of cards.
Taking into account different suites. However, the way they are arranged would not be accounted for. Ace of Spades and Queen of Diamonds would be the same as Queen of Diamonds and Ace of Spades.
Pictures (Jacks, Queens, Kings) = 10
Aces = 11, or 1
The way I see it, 21 can be split into:
20 + 1, 19 + 2, 18 + 3 ... 1 + 20. Each of those numbers can further be split down. It appears to be kind of recursive?
Then we have 2-card ways, 3-card ways, 4-card ways and so on ...
It seems to me a very, very tough problem and I have no idea how to approach it. Any ideas?
permutations combinations card-games
asked Aug 19 '16 at 3:39
KosKos
12517
$endgroup$
$begingroup$
How many decks are in use, or can we assume there are an unlimited number of cards?
$endgroup$
– Parcly Taxel
Aug 19 '16 at 3:58
$begingroup$
Let us assume that there is one deck ... no repetitions of a card
$endgroup$
– Kos
Aug 19 '16 at 4:01
$begingroup$
You could if you really wanted to break it into cases: $2$ cards used, $3$ cards used, $4$ cards used... and in each break into cases further based on number of repeated values used... break into cases even further based on which numbers they actually are. There are for example $4cdot binom{16}{2}$ ways to have the points add up as $1+10+10$ while there are $4cdot 4cdot 4$ ways to have the points add up $6+7+8$. It seems to me that there would be a huge amount of brute force required to continue this way, but with the condition that no card can be repeated it might be unavoidable.
$endgroup$
– JMoravitz
Aug 19 '16 at 4:41
$begingroup$
math.stackexchange.com/questions/1461118/…
$endgroup$
– Matt Watkins
Aug 19 '16 at 6:08
$begingroup$
Perhaps we can come up with a computer algorithm to calculate this?
$endgroup$
– Kos
Aug 19 '16 at 14:51
add a comment |
1
1
1
$begingroup$
Using one deck of cards.
Taking into account different suites. However, the way they are arranged would not be accounted for. Ace of Spades and Queen of Diamonds would be the same as Queen of Diamonds and Ace of Spades.
Pictures (Jacks, Queens, Kings) = 10
Aces = 11, or 1
The way I see it, 21 can be split into:
20 + 1, 19 + 2, 18 + 3 ... 1 + 20. Each of those numbers can further be split down. It appears to be kind of recursive?
Then we have 2-card ways, 3-card ways, 4-card ways and so on ...
It seems to me a very, very tough problem and I have no idea how to approach it. Any ideas?
permutations combinations card-games
asked Aug 19 '16 at 3:39
KosKos
12517
$endgroup$
Using one deck of cards.
Taking into account different suites. However, the way they are arranged would not be accounted for. Ace of Spades and Queen of Diamonds would be the same as Queen of Diamonds and Ace of Spades.
Pictures (Jacks, Queens, Kings) = 10
Aces = 11, or 1
The way I see it, 21 can be split into:
20 + 1, 19 + 2, 18 + 3 ... 1 + 20. Each of those numbers can further be split down. It appears to be kind of recursive?
Then we have 2-card ways, 3-card ways, 4-card ways and so on ...
It seems to me a very, very tough problem and I have no idea how to approach it. Any ideas?
permutations combinations card-games
permutations combinations card-games
asked Aug 19 '16 at 3:39
KosKos
12517
asked Aug 19 '16 at 3:39
KosKos
12517
edited Aug 19 '16 at 4:02
Kos
asked Aug 19 '16 at 3:39
KosKos
12517
asked Aug 19 '16 at 3:39
KosKos
12517
asked Aug 19 '16 at 3:39
KosKos
12517
12517
$begingroup$
How many decks are in use, or can we assume there are an unlimited number of cards?
$endgroup$
– Parcly Taxel
Aug 19 '16 at 3:58
$begingroup$
Let us assume that there is one deck ... no repetitions of a card
$endgroup$
– Kos
Aug 19 '16 at 4:01
$begingroup$
You could if you really wanted to break it into cases: $2$ cards used, $3$ cards used, $4$ cards used... and in each break into cases further based on number of repeated values used... break into cases even further based on which numbers they actually are. There are for example $4cdot binom{16}{2}$ ways to have the points add up as $1+10+10$ while there are $4cdot 4cdot 4$ ways to have the points add up $6+7+8$. It seems to me that there would be a huge amount of brute force required to continue this way, but with the condition that no card can be repeated it might be unavoidable.
$endgroup$
– JMoravitz
Aug 19 '16 at 4:41
$begingroup$
math.stackexchange.com/questions/1461118/…
$endgroup$
– Matt Watkins
Aug 19 '16 at 6:08
$begingroup$
Perhaps we can come up with a computer algorithm to calculate this?
$endgroup$
– Kos
Aug 19 '16 at 14:51
add a comment |
$begingroup$
How many decks are in use, or can we assume there are an unlimited number of cards?
$endgroup$
– Parcly Taxel
Aug 19 '16 at 3:58
$begingroup$
Let us assume that there is one deck ... no repetitions of a card
$endgroup$
– Kos
Aug 19 '16 at 4:01
$begingroup$
You could if you really wanted to break it into cases: $2$ cards used, $3$ cards used, $4$ cards used... and in each break into cases further based on number of repeated values used... break into cases even further based on which numbers they actually are. There are for example $4cdot binom{16}{2}$ ways to have the points add up as $1+10+10$ while there are $4cdot 4cdot 4$ ways to have the points add up $6+7+8$. It seems to me that there would be a huge amount of brute force required to continue this way, but with the condition that no card can be repeated it might be unavoidable.
$endgroup$
– JMoravitz
Aug 19 '16 at 4:41
$begingroup$
math.stackexchange.com/questions/1461118/…
$endgroup$
– Matt Watkins
Aug 19 '16 at 6:08
$begingroup$
Perhaps we can come up with a computer algorithm to calculate this?
$endgroup$
– Kos
Aug 19 '16 at 14:51
$begingroup$
How many decks are in use, or can we assume there are an unlimited number of cards?
$endgroup$
– Parcly Taxel
Aug 19 '16 at 3:58
$begingroup$
How many decks are in use, or can we assume there are an unlimited number of cards?
$endgroup$
– Parcly Taxel
Aug 19 '16 at 3:58
$begingroup$
Let us assume that there is one deck ... no repetitions of a card
$endgroup$
– Kos
Aug 19 '16 at 4:01
$begingroup$
Let us assume that there is one deck ... no repetitions of a card
$endgroup$
– Kos
Aug 19 '16 at 4:01
$begingroup$
You could if you really wanted to break it into cases: $2$ cards used, $3$ cards used, $4$ cards used... and in each break into cases further based on number of repeated values used... break into cases even further based on which numbers they actually are. There are for example $4cdot binom{16}{2}$ ways to have the points add up as $1+10+10$ while there are $4cdot 4cdot 4$ ways to have the points add up $6+7+8$. It seems to me that there would be a huge amount of brute force required to continue this way, but with the condition that no card can be repeated it might be unavoidable.
$endgroup$
– JMoravitz
Aug 19 '16 at 4:41
$begingroup$
You could if you really wanted to break it into cases: $2$ cards used, $3$ cards used, $4$ cards used... and in each break into cases further based on number of repeated values used... break into cases even further based on which numbers they actually are. There are for example $4cdot binom{16}{2}$ ways to have the points add up as $1+10+10$ while there are $4cdot 4cdot 4$ ways to have the points add up $6+7+8$. It seems to me that there would be a huge amount of brute force required to continue this way, but with the condition that no card can be repeated it might be unavoidable.
$endgroup$
– JMoravitz
Aug 19 '16 at 4:41
$begingroup$
math.stackexchange.com/questions/1461118/…
$endgroup$
– Matt Watkins
Aug 19 '16 at 6:08
$begingroup$
math.stackexchange.com/questions/1461118/…
$endgroup$
– Matt Watkins
Aug 19 '16 at 6:08
$begingroup$
Perhaps we can come up with a computer algorithm to calculate this?
$endgroup$
– Kos
Aug 19 '16 at 14:51
$begingroup$
Perhaps we can come up with a computer algorithm to calculate this?
$endgroup$
– Kos
Aug 19 '16 at 14:51
add a comment |
1 Answer
1
active
oldest
votes
0
$begingroup$
This Python code by @niemmi on Stack Overflow says that there are 186,184 ways.
import operator
from math import factorial
def bcoef(n, k):
return factorial(n) / (factorial(k) * factorial(n - k))
def combinations(limit, start, used):
if limit == 0:
# For each face value figure out how many combinations can be used
combs = (bcoef(4 if i != 10 else 16, x) for i, x in enumerate(used))
res = reduce(operator.mul, combs, 1)
return res
res = 0
for i in range(start, 12):
if i > limit:
break
index = i if i != 11 else 1
if used[index] < 4:
used[index] += 1
res += combinations(limit - i, i, used)
used[index] -= 1
return res
print combinations(21, 1, [0] * 11) # 186184
answered Aug 26 '16 at 0:51
KosKos
12517
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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0
$begingroup$
This Python code by @niemmi on Stack Overflow says that there are 186,184 ways.
import operator
from math import factorial
def bcoef(n, k):
return factorial(n) / (factorial(k) * factorial(n - k))
def combinations(limit, start, used):
if limit == 0:
# For each face value figure out how many combinations can be used
combs = (bcoef(4 if i != 10 else 16, x) for i, x in enumerate(used))
res = reduce(operator.mul, combs, 1)
return res
res = 0
for i in range(start, 12):
if i > limit:
break
index = i if i != 11 else 1
if used[index] < 4:
used[index] += 1
res += combinations(limit - i, i, used)
used[index] -= 1
return res
print combinations(21, 1, [0] * 11) # 186184
answered Aug 26 '16 at 0:51
KosKos
12517
$endgroup$
add a comment |
0
$begingroup$
This Python code by @niemmi on Stack Overflow says that there are 186,184 ways.
import operator
from math import factorial
def bcoef(n, k):
return factorial(n) / (factorial(k) * factorial(n - k))
def combinations(limit, start, used):
if limit == 0:
# For each face value figure out how many combinations can be used
combs = (bcoef(4 if i != 10 else 16, x) for i, x in enumerate(used))
res = reduce(operator.mul, combs, 1)
return res
res = 0
for i in range(start, 12):
if i > limit:
break
index = i if i != 11 else 1
if used[index] < 4:
used[index] += 1
res += combinations(limit - i, i, used)
used[index] -= 1
return res
print combinations(21, 1, [0] * 11) # 186184
answered Aug 26 '16 at 0:51
KosKos
12517
$endgroup$
add a comment |
0
0
0
$begingroup$
This Python code by @niemmi on Stack Overflow says that there are 186,184 ways.
import operator
from math import factorial
def bcoef(n, k):
return factorial(n) / (factorial(k) * factorial(n - k))
def combinations(limit, start, used):
if limit == 0:
# For each face value figure out how many combinations can be used
combs = (bcoef(4 if i != 10 else 16, x) for i, x in enumerate(used))
res = reduce(operator.mul, combs, 1)
return res
res = 0
for i in range(start, 12):
if i > limit:
break
index = i if i != 11 else 1
if used[index] < 4:
used[index] += 1
res += combinations(limit - i, i, used)
used[index] -= 1
return res
print combinations(21, 1, [0] * 11) # 186184
answered Aug 26 '16 at 0:51
KosKos
12517
$endgroup$
This Python code by @niemmi on Stack Overflow says that there are 186,184 ways.
import operator
from math import factorial
def bcoef(n, k):
return factorial(n) / (factorial(k) * factorial(n - k))
def combinations(limit, start, used):
if limit == 0:
# For each face value figure out how many combinations can be used
combs = (bcoef(4 if i != 10 else 16, x) for i, x in enumerate(used))
res = reduce(operator.mul, combs, 1)
return res
res = 0
for i in range(start, 12):
if i > limit:
break
index = i if i != 11 else 1
if used[index] < 4:
used[index] += 1
res += combinations(limit - i, i, used)
used[index] -= 1
return res
print combinations(21, 1, [0] * 11) # 186184
answered Aug 26 '16 at 0:51
KosKos
12517
answered Aug 26 '16 at 0:51
KosKos
12517
answered Aug 26 '16 at 0:51
KosKos
12517
answered Aug 26 '16 at 0:51
KosKos
12517
12517
add a comment |
add a comment |
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$begingroup$
How many decks are in use, or can we assume there are an unlimited number of cards?
$endgroup$
– Parcly Taxel
Aug 19 '16 at 3:58
$begingroup$
Let us assume that there is one deck ... no repetitions of a card
$endgroup$
– Kos
Aug 19 '16 at 4:01
$begingroup$
You could if you really wanted to break it into cases: $2$ cards used, $3$ cards used, $4$ cards used... and in each break into cases further based on number of repeated values used... break into cases even further based on which numbers they actually are. There are for example $4cdot binom{16}{2}$ ways to have the points add up as $1+10+10$ while there are $4cdot 4cdot 4$ ways to have the points add up $6+7+8$. It seems to me that there would be a huge amount of brute force required to continue this way, but with the condition that no card can be repeated it might be unavoidable.
$endgroup$
– JMoravitz
Aug 19 '16 at 4:41
$begingroup$
math.stackexchange.com/questions/1461118/…
$endgroup$
– Matt Watkins
Aug 19 '16 at 6:08
$begingroup$
Perhaps we can come up with a computer algorithm to calculate this?
$endgroup$
– Kos
Aug 19 '16 at 14:51