Mixing
How many ways are there to draw a four card hand, with replacement, where at least one card appears multiple...
$begingroup$
You draw four cards from a standard deck, with replacement. How many possible hands are there where at least one card appears multiple times?
Here's what I have so far: to count the ways using the Fundamental Counting Principle it looks like 52*1*52*52 because you have 52 options for the first draw, only one option for the second draw so it matches the first, and then the third and fourth don't matter. But, now I think I need to divide because the order of the draws don't matter. Should I divide by 4 since the 1 above could be in 4 different positions? Should I divide by 4! since any 4 card hand can be arranged in 4! ways? But, what about the hands where the same card is repeated 3 times? Are those hands included in how I'm counting?
Is there a simple and organized way to keep track and count the possibilities?
combinatorics
asked Jan 25 at 13:54
jontailjontail
32
$endgroup$
add a comment |
$begingroup$
You draw four cards from a standard deck, with replacement. How many possible hands are there where at least one card appears multiple times?
Here's what I have so far: to count the ways using the Fundamental Counting Principle it looks like 52*1*52*52 because you have 52 options for the first draw, only one option for the second draw so it matches the first, and then the third and fourth don't matter. But, now I think I need to divide because the order of the draws don't matter. Should I divide by 4 since the 1 above could be in 4 different positions? Should I divide by 4! since any 4 card hand can be arranged in 4! ways? But, what about the hands where the same card is repeated 3 times? Are those hands included in how I'm counting?
Is there a simple and organized way to keep track and count the possibilities?
combinatorics
asked Jan 25 at 13:54
jontailjontail
32
$endgroup$
1
$begingroup$
It's easier to count the ways to draw $4$ distinct cards
$endgroup$
– lulu
Jan 25 at 13:59
add a comment |
$begingroup$
You draw four cards from a standard deck, with replacement. How many possible hands are there where at least one card appears multiple times?
Here's what I have so far: to count the ways using the Fundamental Counting Principle it looks like 52*1*52*52 because you have 52 options for the first draw, only one option for the second draw so it matches the first, and then the third and fourth don't matter. But, now I think I need to divide because the order of the draws don't matter. Should I divide by 4 since the 1 above could be in 4 different positions? Should I divide by 4! since any 4 card hand can be arranged in 4! ways? But, what about the hands where the same card is repeated 3 times? Are those hands included in how I'm counting?
Is there a simple and organized way to keep track and count the possibilities?
combinatorics
asked Jan 25 at 13:54
jontailjontail
32
$endgroup$
You draw four cards from a standard deck, with replacement. How many possible hands are there where at least one card appears multiple times?
Here's what I have so far: to count the ways using the Fundamental Counting Principle it looks like 52*1*52*52 because you have 52 options for the first draw, only one option for the second draw so it matches the first, and then the third and fourth don't matter. But, now I think I need to divide because the order of the draws don't matter. Should I divide by 4 since the 1 above could be in 4 different positions? Should I divide by 4! since any 4 card hand can be arranged in 4! ways? But, what about the hands where the same card is repeated 3 times? Are those hands included in how I'm counting?
Is there a simple and organized way to keep track and count the possibilities?
combinatorics
combinatorics
asked Jan 25 at 13:54
jontailjontail
32
asked Jan 25 at 13:54
jontailjontail
32
asked Jan 25 at 13:54
jontailjontail
32
asked Jan 25 at 13:54
jontailjontail
32
asked Jan 25 at 13:54
jontailjontail
32
32
1
$begingroup$
It's easier to count the ways to draw $4$ distinct cards
$endgroup$
– lulu
Jan 25 at 13:59
add a comment |
1
$begingroup$
It's easier to count the ways to draw $4$ distinct cards
$endgroup$
– lulu
Jan 25 at 13:59
1
1
$begingroup$
It's easier to count the ways to draw $4$ distinct cards
$endgroup$
– lulu
Jan 25 at 13:59
$begingroup$
It's easier to count the ways to draw $4$ distinct cards
$endgroup$
– lulu
Jan 25 at 13:59
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I assume you are disregarding the order of the cards in the hand.
The difficulty here is that the total number of possible hands isn't clear (trusting that, as usual, the order of the cards in a hand is not important).
Stars and Bars gives us a handy way to do that count. Here we are looking for the number of $52-$ tuples of non-negative integers that sum to $4$. (a coordinate in such a $52-$ tuple tells us how many of that card appears in the hand). The standard result shows that this number is $binom {52+4-1}{4}=binom {55}4$
As there are clearly $binom {52}4$ ways to choose a hand of $4$ distinct cards, the answer is the difference: $$binom {55}{4}-binom {52}4=boxed {70,330}$$
answered Jan 25 at 14:18
lulululu
43k25080
$endgroup$
$begingroup$
This is an elegant solution. Thank you.
$endgroup$
– jontail
Jan 28 at 15:15
add a comment |
$begingroup$
I would approach this from the other persepctive - look at how many possible hands there are and remove those which have no repeated cards.
We have $52times 52times 52times 52 = 7,311,616$ possible combinations of all the cards (provided we care about the ordering).
We now need to see how many of those hands had no repeated cards in, which we know is equal to $52times 51times 50times 49 = 6,497,400$ (again, provided we care about the ordering)
Therefore, there are $$7,311,616-6,497,400=814,206$$ hands where at least one card appears multiple times.
answered Jan 25 at 14:06
lioness99alioness99a
3,8612727
$endgroup$
$begingroup$
Unfortunately, it isn't true that "If ordering does not matter then we have counted each hand $4$ times." You only count ${Aspadesuit, Aspadesuit,Aspadesuit,Aspadesuit}$ once, for example.
$endgroup$
– lulu
Jan 25 at 14:10
$begingroup$
@lulu That's is true I have removed that half of my answer
$endgroup$
– lioness99a
Jan 25 at 14:15
add a comment |
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2 Answers
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2 Answers
2
active
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$begingroup$
I assume you are disregarding the order of the cards in the hand.
The difficulty here is that the total number of possible hands isn't clear (trusting that, as usual, the order of the cards in a hand is not important).
Stars and Bars gives us a handy way to do that count. Here we are looking for the number of $52-$ tuples of non-negative integers that sum to $4$. (a coordinate in such a $52-$ tuple tells us how many of that card appears in the hand). The standard result shows that this number is $binom {52+4-1}{4}=binom {55}4$
As there are clearly $binom {52}4$ ways to choose a hand of $4$ distinct cards, the answer is the difference: $$binom {55}{4}-binom {52}4=boxed {70,330}$$
answered Jan 25 at 14:18
lulululu
43k25080
$endgroup$
$begingroup$
This is an elegant solution. Thank you.
$endgroup$
– jontail
Jan 28 at 15:15
add a comment |
$begingroup$
I assume you are disregarding the order of the cards in the hand.
The difficulty here is that the total number of possible hands isn't clear (trusting that, as usual, the order of the cards in a hand is not important).
Stars and Bars gives us a handy way to do that count. Here we are looking for the number of $52-$ tuples of non-negative integers that sum to $4$. (a coordinate in such a $52-$ tuple tells us how many of that card appears in the hand). The standard result shows that this number is $binom {52+4-1}{4}=binom {55}4$
As there are clearly $binom {52}4$ ways to choose a hand of $4$ distinct cards, the answer is the difference: $$binom {55}{4}-binom {52}4=boxed {70,330}$$
answered Jan 25 at 14:18
lulululu
43k25080
$endgroup$
$begingroup$
This is an elegant solution. Thank you.
$endgroup$
– jontail
Jan 28 at 15:15
add a comment |
$begingroup$
I assume you are disregarding the order of the cards in the hand.
The difficulty here is that the total number of possible hands isn't clear (trusting that, as usual, the order of the cards in a hand is not important).
Stars and Bars gives us a handy way to do that count. Here we are looking for the number of $52-$ tuples of non-negative integers that sum to $4$. (a coordinate in such a $52-$ tuple tells us how many of that card appears in the hand). The standard result shows that this number is $binom {52+4-1}{4}=binom {55}4$
As there are clearly $binom {52}4$ ways to choose a hand of $4$ distinct cards, the answer is the difference: $$binom {55}{4}-binom {52}4=boxed {70,330}$$
answered Jan 25 at 14:18
lulululu
43k25080
$endgroup$
I assume you are disregarding the order of the cards in the hand.
The difficulty here is that the total number of possible hands isn't clear (trusting that, as usual, the order of the cards in a hand is not important).
Stars and Bars gives us a handy way to do that count. Here we are looking for the number of $52-$ tuples of non-negative integers that sum to $4$. (a coordinate in such a $52-$ tuple tells us how many of that card appears in the hand). The standard result shows that this number is $binom {52+4-1}{4}=binom {55}4$
As there are clearly $binom {52}4$ ways to choose a hand of $4$ distinct cards, the answer is the difference: $$binom {55}{4}-binom {52}4=boxed {70,330}$$
answered Jan 25 at 14:18
lulululu
43k25080
answered Jan 25 at 14:18
lulululu
43k25080
answered Jan 25 at 14:18
lulululu
43k25080
answered Jan 25 at 14:18
lulululu
43k25080
43k25080
$begingroup$
This is an elegant solution. Thank you.
$endgroup$
– jontail
Jan 28 at 15:15
add a comment |
$begingroup$
This is an elegant solution. Thank you.
$endgroup$
– jontail
Jan 28 at 15:15
$begingroup$
This is an elegant solution. Thank you.
$endgroup$
– jontail
Jan 28 at 15:15
$begingroup$
This is an elegant solution. Thank you.
$endgroup$
– jontail
Jan 28 at 15:15
add a comment |
$begingroup$
I would approach this from the other persepctive - look at how many possible hands there are and remove those which have no repeated cards.
We have $52times 52times 52times 52 = 7,311,616$ possible combinations of all the cards (provided we care about the ordering).
We now need to see how many of those hands had no repeated cards in, which we know is equal to $52times 51times 50times 49 = 6,497,400$ (again, provided we care about the ordering)
Therefore, there are $$7,311,616-6,497,400=814,206$$ hands where at least one card appears multiple times.
answered Jan 25 at 14:06
lioness99alioness99a
3,8612727
$endgroup$
$begingroup$
Unfortunately, it isn't true that "If ordering does not matter then we have counted each hand $4$ times." You only count ${Aspadesuit, Aspadesuit,Aspadesuit,Aspadesuit}$ once, for example.
$endgroup$
– lulu
Jan 25 at 14:10
$begingroup$
@lulu That's is true I have removed that half of my answer
$endgroup$
– lioness99a
Jan 25 at 14:15
add a comment |
$begingroup$
I would approach this from the other persepctive - look at how many possible hands there are and remove those which have no repeated cards.
We have $52times 52times 52times 52 = 7,311,616$ possible combinations of all the cards (provided we care about the ordering).
We now need to see how many of those hands had no repeated cards in, which we know is equal to $52times 51times 50times 49 = 6,497,400$ (again, provided we care about the ordering)
Therefore, there are $$7,311,616-6,497,400=814,206$$ hands where at least one card appears multiple times.
answered Jan 25 at 14:06
lioness99alioness99a
3,8612727
$endgroup$
$begingroup$
Unfortunately, it isn't true that "If ordering does not matter then we have counted each hand $4$ times." You only count ${Aspadesuit, Aspadesuit,Aspadesuit,Aspadesuit}$ once, for example.
$endgroup$
– lulu
Jan 25 at 14:10
$begingroup$
@lulu That's is true I have removed that half of my answer
$endgroup$
– lioness99a
Jan 25 at 14:15
add a comment |
$begingroup$
I would approach this from the other persepctive - look at how many possible hands there are and remove those which have no repeated cards.
We have $52times 52times 52times 52 = 7,311,616$ possible combinations of all the cards (provided we care about the ordering).
We now need to see how many of those hands had no repeated cards in, which we know is equal to $52times 51times 50times 49 = 6,497,400$ (again, provided we care about the ordering)
Therefore, there are $$7,311,616-6,497,400=814,206$$ hands where at least one card appears multiple times.
answered Jan 25 at 14:06
lioness99alioness99a
3,8612727
$endgroup$
I would approach this from the other persepctive - look at how many possible hands there are and remove those which have no repeated cards.
We have $52times 52times 52times 52 = 7,311,616$ possible combinations of all the cards (provided we care about the ordering).
We now need to see how many of those hands had no repeated cards in, which we know is equal to $52times 51times 50times 49 = 6,497,400$ (again, provided we care about the ordering)
Therefore, there are $$7,311,616-6,497,400=814,206$$ hands where at least one card appears multiple times.
answered Jan 25 at 14:06
lioness99alioness99a
3,8612727
edited Jan 25 at 14:15
answered Jan 25 at 14:06
lioness99alioness99a
3,8612727
answered Jan 25 at 14:06
lioness99alioness99a
3,8612727
answered Jan 25 at 14:06
lioness99alioness99a
3,8612727
3,8612727
$begingroup$
Unfortunately, it isn't true that "If ordering does not matter then we have counted each hand $4$ times." You only count ${Aspadesuit, Aspadesuit,Aspadesuit,Aspadesuit}$ once, for example.
$endgroup$
– lulu
Jan 25 at 14:10
$begingroup$
@lulu That's is true I have removed that half of my answer
$endgroup$
– lioness99a
Jan 25 at 14:15
add a comment |
$begingroup$
Unfortunately, it isn't true that "If ordering does not matter then we have counted each hand $4$ times." You only count ${Aspadesuit, Aspadesuit,Aspadesuit,Aspadesuit}$ once, for example.
$endgroup$
– lulu
Jan 25 at 14:10
$begingroup$
@lulu That's is true I have removed that half of my answer
$endgroup$
– lioness99a
Jan 25 at 14:15
$begingroup$
Unfortunately, it isn't true that "If ordering does not matter then we have counted each hand $4$ times." You only count ${Aspadesuit, Aspadesuit,Aspadesuit,Aspadesuit}$ once, for example.
$endgroup$
– lulu
Jan 25 at 14:10
$begingroup$
Unfortunately, it isn't true that "If ordering does not matter then we have counted each hand $4$ times." You only count ${Aspadesuit, Aspadesuit,Aspadesuit,Aspadesuit}$ once, for example.
$endgroup$
– lulu
Jan 25 at 14:10
$begingroup$
@lulu That's is true I have removed that half of my answer
$endgroup$
– lioness99a
Jan 25 at 14:15
$begingroup$
@lulu That's is true I have removed that half of my answer
$endgroup$
– lioness99a
Jan 25 at 14:15
add a comment |
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1
$begingroup$
It's easier to count the ways to draw $4$ distinct cards
$endgroup$
– lulu
Jan 25 at 13:59