Probability of full house?
1
$begingroup$
I have seen there are other answers on this, but they all follow the pattern
- first, choose which card you have $3$ of
- then, choose $3$ of the $4$ available cards
- then, the card you have $2$ of
- then, $2$ of $4$ cards
and they get to
$$binom{13}{1} binom{4}{3} binom{12}{1} binom{4}{2}$$
for the numerator.
My question is: doesn't this imply some kind of ordering?
By following the same logic, I would think that for the denominator
- pick any of $52$ cards
- then, pick any of $51$ remaining cards
and so on, getting $52 cdot 51 cdot 50 cdot 49 cdot 48$. But instead, the correct value of the denominator is $binom{52}{5}$.
Why are those cases different?
probability combinatorics combinations
asked Nov 25 '18 at 13:17
blue_noteblue_note
45948
$endgroup$
|
show 1 more comment
1
$begingroup$
I have seen there are other answers on this, but they all follow the pattern
- first, choose which card you have $3$ of
- then, choose $3$ of the $4$ available cards
- then, the card you have $2$ of
- then, $2$ of $4$ cards
and they get to
$$binom{13}{1} binom{4}{3} binom{12}{1} binom{4}{2}$$
for the numerator.
My question is: doesn't this imply some kind of ordering?
By following the same logic, I would think that for the denominator
- pick any of $52$ cards
- then, pick any of $51$ remaining cards
and so on, getting $52 cdot 51 cdot 50 cdot 49 cdot 48$. But instead, the correct value of the denominator is $binom{52}{5}$.
Why are those cases different?
probability combinatorics combinations
asked Nov 25 '18 at 13:17
blue_noteblue_note
45948
$endgroup$
$begingroup$
There is no order imposed by the construction. A full house is determined by the rank of the triple, the suits within that rank, the rank of the pair, the suits within that rank. No order involved.
$endgroup$
– lulu
Nov 25 '18 at 13:18
$begingroup$
By contrast, your second calculation does impose an order...since you choose the first card, then the second, and so on.
$endgroup$
– lulu
Nov 25 '18 at 13:19
$begingroup$
@lulu: i realise that for the second calculation, but I don't see why the first one does not. how could I visualize it?
$endgroup$
– blue_note
Nov 25 '18 at 13:22
1
$begingroup$
Work with a smaller hand. Say you want to count the two card hands consisting of an ace and a queen. There are four ways to choose the ace and four to choose the queen, so $4times 4=16$. If you order them, there are $32$.
$endgroup$
– lulu
Nov 25 '18 at 13:25
2
$begingroup$
Build up from two cards. Say I want to count the unordered hands with two spades and a heart. Convince yourself that this is $binom {13}2times binom {13}1$.
$endgroup$
– lulu
Nov 25 '18 at 13:27
|
show 1 more comment
1
1
1
1
$begingroup$
I have seen there are other answers on this, but they all follow the pattern
- first, choose which card you have $3$ of
- then, choose $3$ of the $4$ available cards
- then, the card you have $2$ of
- then, $2$ of $4$ cards
and they get to
$$binom{13}{1} binom{4}{3} binom{12}{1} binom{4}{2}$$
for the numerator.
My question is: doesn't this imply some kind of ordering?
By following the same logic, I would think that for the denominator
- pick any of $52$ cards
- then, pick any of $51$ remaining cards
and so on, getting $52 cdot 51 cdot 50 cdot 49 cdot 48$. But instead, the correct value of the denominator is $binom{52}{5}$.
Why are those cases different?
probability combinatorics combinations
asked Nov 25 '18 at 13:17
blue_noteblue_note
45948
$endgroup$
I have seen there are other answers on this, but they all follow the pattern
- first, choose which card you have $3$ of
- then, choose $3$ of the $4$ available cards
- then, the card you have $2$ of
- then, $2$ of $4$ cards
and they get to
$$binom{13}{1} binom{4}{3} binom{12}{1} binom{4}{2}$$
for the numerator.
My question is: doesn't this imply some kind of ordering?
By following the same logic, I would think that for the denominator
- pick any of $52$ cards
- then, pick any of $51$ remaining cards
and so on, getting $52 cdot 51 cdot 50 cdot 49 cdot 48$. But instead, the correct value of the denominator is $binom{52}{5}$.
Why are those cases different?
probability combinatorics combinations
probability combinatorics combinations
asked Nov 25 '18 at 13:17
blue_noteblue_note
45948
asked Nov 25 '18 at 13:17
blue_noteblue_note
45948
asked Nov 25 '18 at 13:17
blue_noteblue_note
45948
asked Nov 25 '18 at 13:17
blue_noteblue_note
45948
asked Nov 25 '18 at 13:17
blue_noteblue_note
45948
45948
$begingroup$
There is no order imposed by the construction. A full house is determined by the rank of the triple, the suits within that rank, the rank of the pair, the suits within that rank. No order involved.
$endgroup$
– lulu
Nov 25 '18 at 13:18
$begingroup$
By contrast, your second calculation does impose an order...since you choose the first card, then the second, and so on.
$endgroup$
– lulu
Nov 25 '18 at 13:19
$begingroup$
@lulu: i realise that for the second calculation, but I don't see why the first one does not. how could I visualize it?
$endgroup$
– blue_note
Nov 25 '18 at 13:22
1
$begingroup$
Work with a smaller hand. Say you want to count the two card hands consisting of an ace and a queen. There are four ways to choose the ace and four to choose the queen, so $4times 4=16$. If you order them, there are $32$.
$endgroup$
– lulu
Nov 25 '18 at 13:25
2
$begingroup$
Build up from two cards. Say I want to count the unordered hands with two spades and a heart. Convince yourself that this is $binom {13}2times binom {13}1$.
$endgroup$
– lulu
Nov 25 '18 at 13:27
|
show 1 more comment
$begingroup$
There is no order imposed by the construction. A full house is determined by the rank of the triple, the suits within that rank, the rank of the pair, the suits within that rank. No order involved.
$endgroup$
– lulu
Nov 25 '18 at 13:18
$begingroup$
By contrast, your second calculation does impose an order...since you choose the first card, then the second, and so on.
$endgroup$
– lulu
Nov 25 '18 at 13:19
$begingroup$
@lulu: i realise that for the second calculation, but I don't see why the first one does not. how could I visualize it?
$endgroup$
– blue_note
Nov 25 '18 at 13:22
1
$begingroup$
Work with a smaller hand. Say you want to count the two card hands consisting of an ace and a queen. There are four ways to choose the ace and four to choose the queen, so $4times 4=16$. If you order them, there are $32$.
$endgroup$
– lulu
Nov 25 '18 at 13:25
2
$begingroup$
Build up from two cards. Say I want to count the unordered hands with two spades and a heart. Convince yourself that this is $binom {13}2times binom {13}1$.
$endgroup$
– lulu
Nov 25 '18 at 13:27
$begingroup$
There is no order imposed by the construction. A full house is determined by the rank of the triple, the suits within that rank, the rank of the pair, the suits within that rank. No order involved.
$endgroup$
– lulu
Nov 25 '18 at 13:18
$begingroup$
There is no order imposed by the construction. A full house is determined by the rank of the triple, the suits within that rank, the rank of the pair, the suits within that rank. No order involved.
$endgroup$
– lulu
Nov 25 '18 at 13:18
$begingroup$
By contrast, your second calculation does impose an order...since you choose the first card, then the second, and so on.
$endgroup$
– lulu
Nov 25 '18 at 13:19
$begingroup$
By contrast, your second calculation does impose an order...since you choose the first card, then the second, and so on.
$endgroup$
– lulu
Nov 25 '18 at 13:19
$begingroup$
@lulu: i realise that for the second calculation, but I don't see why the first one does not. how could I visualize it?
$endgroup$
– blue_note
Nov 25 '18 at 13:22
$begingroup$
@lulu: i realise that for the second calculation, but I don't see why the first one does not. how could I visualize it?
$endgroup$
– blue_note
Nov 25 '18 at 13:22
1
1
$begingroup$
Work with a smaller hand. Say you want to count the two card hands consisting of an ace and a queen. There are four ways to choose the ace and four to choose the queen, so $4times 4=16$. If you order them, there are $32$.
$endgroup$
– lulu
Nov 25 '18 at 13:25
$begingroup$
Work with a smaller hand. Say you want to count the two card hands consisting of an ace and a queen. There are four ways to choose the ace and four to choose the queen, so $4times 4=16$. If you order them, there are $32$.
$endgroup$
– lulu
Nov 25 '18 at 13:25
2
2
$begingroup$
Build up from two cards. Say I want to count the unordered hands with two spades and a heart. Convince yourself that this is $binom {13}2times binom {13}1$.
$endgroup$
– lulu
Nov 25 '18 at 13:27
$begingroup$
Build up from two cards. Say I want to count the unordered hands with two spades and a heart. Convince yourself that this is $binom {13}2times binom {13}1$.
$endgroup$
– lulu
Nov 25 '18 at 13:27
|
show 1 more comment
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$begingroup$
There is no order imposed by the construction. A full house is determined by the rank of the triple, the suits within that rank, the rank of the pair, the suits within that rank. No order involved.
$endgroup$
– lulu
Nov 25 '18 at 13:18
$begingroup$
By contrast, your second calculation does impose an order...since you choose the first card, then the second, and so on.
$endgroup$
– lulu
Nov 25 '18 at 13:19
$begingroup$
@lulu: i realise that for the second calculation, but I don't see why the first one does not. how could I visualize it?
$endgroup$
– blue_note
Nov 25 '18 at 13:22
1
$begingroup$
Work with a smaller hand. Say you want to count the two card hands consisting of an ace and a queen. There are four ways to choose the ace and four to choose the queen, so $4times 4=16$. If you order them, there are $32$.
$endgroup$
– lulu
Nov 25 '18 at 13:25
2
$begingroup$
Build up from two cards. Say I want to count the unordered hands with two spades and a heart. Convince yourself that this is $binom {13}2times binom {13}1$.
$endgroup$
– lulu
Nov 25 '18 at 13:27