Estimate number of cards needed to be drawn from deck before full house
$begingroup$
Imagine we have a well-shuffled deck of cards and we keep drawing cards until there is at least one full house in the drawn cards. How many cards will we draw on average? I would be interested in both the exact solution (which is something around 12) but, more importantly, in a good quick way of estimating this value.
probability puzzle card-games
asked Dec 14 '18 at 23:46
user132290user132290
536
$endgroup$
|
show 2 more comments
$begingroup$
Imagine we have a well-shuffled deck of cards and we keep drawing cards until there is at least one full house in the drawn cards. How many cards will we draw on average? I would be interested in both the exact solution (which is something around 12) but, more importantly, in a good quick way of estimating this value.
probability puzzle card-games
asked Dec 14 '18 at 23:46
user132290user132290
536
$endgroup$
$begingroup$
If this is too hard, I would be interested in the same question but for other simple poker hands like a single pair or a three-of-a-kind
$endgroup$
– user132290
Dec 14 '18 at 23:48
$begingroup$
And the value of 12 quoted above comes from my computer simulations.
$endgroup$
– user132290
Dec 15 '18 at 12:13
$begingroup$
My computer simulation gives an average of $13.9$ with a standard deviation of $3.7$.
$endgroup$
– Jens
Dec 15 '18 at 13:13
$begingroup$
@Jens Indeed. The exact expectation I got is $13.893157dots$, which is amazingly close to your simulation result. The interesting question is how to get it with a back of the envelope calculation (I just honestly wrote the combinatorial sum and told my PC to compute it)..
$endgroup$
– fedja
Dec 15 '18 at 17:39
$begingroup$
How do you guys simulate this? My simulation gives 13.55 after 1 million runs. Do you also actually simulate a deck of cards, or do you calculate the probability somehow?
$endgroup$
– user132290
Dec 17 '18 at 21:51
|
show 2 more comments
$begingroup$
Imagine we have a well-shuffled deck of cards and we keep drawing cards until there is at least one full house in the drawn cards. How many cards will we draw on average? I would be interested in both the exact solution (which is something around 12) but, more importantly, in a good quick way of estimating this value.
probability puzzle card-games
asked Dec 14 '18 at 23:46
user132290user132290
536
$endgroup$
Imagine we have a well-shuffled deck of cards and we keep drawing cards until there is at least one full house in the drawn cards. How many cards will we draw on average? I would be interested in both the exact solution (which is something around 12) but, more importantly, in a good quick way of estimating this value.
probability puzzle card-games
probability puzzle card-games
asked Dec 14 '18 at 23:46
user132290user132290
536
asked Dec 14 '18 at 23:46
user132290user132290
536
asked Dec 14 '18 at 23:46
user132290user132290
536
asked Dec 14 '18 at 23:46
user132290user132290
536
asked Dec 14 '18 at 23:46
user132290user132290
536
536
$begingroup$
If this is too hard, I would be interested in the same question but for other simple poker hands like a single pair or a three-of-a-kind
$endgroup$
– user132290
Dec 14 '18 at 23:48
$begingroup$
And the value of 12 quoted above comes from my computer simulations.
$endgroup$
– user132290
Dec 15 '18 at 12:13
$begingroup$
My computer simulation gives an average of $13.9$ with a standard deviation of $3.7$.
$endgroup$
– Jens
Dec 15 '18 at 13:13
$begingroup$
@Jens Indeed. The exact expectation I got is $13.893157dots$, which is amazingly close to your simulation result. The interesting question is how to get it with a back of the envelope calculation (I just honestly wrote the combinatorial sum and told my PC to compute it)..
$endgroup$
– fedja
Dec 15 '18 at 17:39
$begingroup$
How do you guys simulate this? My simulation gives 13.55 after 1 million runs. Do you also actually simulate a deck of cards, or do you calculate the probability somehow?
$endgroup$
– user132290
Dec 17 '18 at 21:51
|
show 2 more comments
$begingroup$
If this is too hard, I would be interested in the same question but for other simple poker hands like a single pair or a three-of-a-kind
$endgroup$
– user132290
Dec 14 '18 at 23:48
$begingroup$
And the value of 12 quoted above comes from my computer simulations.
$endgroup$
– user132290
Dec 15 '18 at 12:13
$begingroup$
My computer simulation gives an average of $13.9$ with a standard deviation of $3.7$.
$endgroup$
– Jens
Dec 15 '18 at 13:13
$begingroup$
@Jens Indeed. The exact expectation I got is $13.893157dots$, which is amazingly close to your simulation result. The interesting question is how to get it with a back of the envelope calculation (I just honestly wrote the combinatorial sum and told my PC to compute it)..
$endgroup$
– fedja
Dec 15 '18 at 17:39
$begingroup$
How do you guys simulate this? My simulation gives 13.55 after 1 million runs. Do you also actually simulate a deck of cards, or do you calculate the probability somehow?
$endgroup$
– user132290
Dec 17 '18 at 21:51
$begingroup$
If this is too hard, I would be interested in the same question but for other simple poker hands like a single pair or a three-of-a-kind
$endgroup$
– user132290
Dec 14 '18 at 23:48
$begingroup$
If this is too hard, I would be interested in the same question but for other simple poker hands like a single pair or a three-of-a-kind
$endgroup$
– user132290
Dec 14 '18 at 23:48
$begingroup$
And the value of 12 quoted above comes from my computer simulations.
$endgroup$
– user132290
Dec 15 '18 at 12:13
$begingroup$
And the value of 12 quoted above comes from my computer simulations.
$endgroup$
– user132290
Dec 15 '18 at 12:13
$begingroup$
My computer simulation gives an average of $13.9$ with a standard deviation of $3.7$.
$endgroup$
– Jens
Dec 15 '18 at 13:13
$begingroup$
My computer simulation gives an average of $13.9$ with a standard deviation of $3.7$.
$endgroup$
– Jens
Dec 15 '18 at 13:13
$begingroup$
@Jens Indeed. The exact expectation I got is $13.893157dots$, which is amazingly close to your simulation result. The interesting question is how to get it with a back of the envelope calculation (I just honestly wrote the combinatorial sum and told my PC to compute it)..
$endgroup$
– fedja
Dec 15 '18 at 17:39
$begingroup$
@Jens Indeed. The exact expectation I got is $13.893157dots$, which is amazingly close to your simulation result. The interesting question is how to get it with a back of the envelope calculation (I just honestly wrote the combinatorial sum and told my PC to compute it)..
$endgroup$
– fedja
Dec 15 '18 at 17:39
$begingroup$
How do you guys simulate this? My simulation gives 13.55 after 1 million runs. Do you also actually simulate a deck of cards, or do you calculate the probability somehow?
$endgroup$
– user132290
Dec 17 '18 at 21:51
$begingroup$
How do you guys simulate this? My simulation gives 13.55 after 1 million runs. Do you also actually simulate a deck of cards, or do you calculate the probability somehow?
$endgroup$
– user132290
Dec 17 '18 at 21:51
|
show 2 more comments
1 Answer
1
active
oldest
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$begingroup$
A simpler question to address is, having drawn n cards, what is the probability of having a full house. Calculating this is equivalent to counting the number of possible hands with a full house, and then dividing my total possible hands.
To get this, we can count the invalid hands through casework, and then subtract this from the total possible hands:
Case 1 - no doubles/ triples:
$$ 4^n cdot binom{13}{n} $$
Case 2 - a triple/quadruple:
$$ 13 cdot left(4 cdot 4^{n-3} binom{12}{n-3} + 4^{n-4} binom{12}{n-4}right) $$
Case 3 - having k doubles:
$$ binom{13}{k} binom{13-k}{n-2k} binom{4}{2}^k 4^{n-2k} $$
Thus, the chance that we get a straight flush drawing n cards is:
$$ frac{binom{52}{n}- left( 4^n binom{13}{n} + 13 cdot left(4 cdot 4^{n-3} binom{12}{n-3} + 4^{n-4} binom{12}{n-4}right) + sum_{k=1}^n binom{13}{k} binom{13-k}{n-2k} binom{4}{2}^k 4^{n-2k} right)}{binom{52}{n}}$$
Putting this into WolframAlpha, we get these values:
Now, this doesn't account from when it happens, only how likely it is to happen after a certain number of draws. To get a closer number on that, you'd have to do some stricter case work, involving ordering.
Essentially, just calculate the number of ordered decks which will has a royal flush at exactly the $n$-th card.
Let $f(n)$ be the number of orderings of $n$ cards containing a royal flush. Using the methods of before, but not diving by $binom{52}{n}$ we get $f(n)$. From here, we can calculate exactly how many decks get a royal flush in exactly the $n$-th draw. We shall denote this number as g(n).
$$ g(n) = f(n)(52-n)! - g(n-1) $$
Then, simply take the sum:
$$ sum_{n=1}^{52} frac{n cdot g(n)}{52!} $$
Which, unfortunately, I am too lazy to evaluate... consider it an exercise :)
answered Jan 4 at 4:29
Zachary HunterZachary Hunter
57611
$endgroup$
add a comment |
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1 Answer
1
active
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1 Answer
1
active
oldest
votes
active
oldest
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oldest
votes
$begingroup$
A simpler question to address is, having drawn n cards, what is the probability of having a full house. Calculating this is equivalent to counting the number of possible hands with a full house, and then dividing my total possible hands.
To get this, we can count the invalid hands through casework, and then subtract this from the total possible hands:
Case 1 - no doubles/ triples:
$$ 4^n cdot binom{13}{n} $$
Case 2 - a triple/quadruple:
$$ 13 cdot left(4 cdot 4^{n-3} binom{12}{n-3} + 4^{n-4} binom{12}{n-4}right) $$
Case 3 - having k doubles:
$$ binom{13}{k} binom{13-k}{n-2k} binom{4}{2}^k 4^{n-2k} $$
Thus, the chance that we get a straight flush drawing n cards is:
$$ frac{binom{52}{n}- left( 4^n binom{13}{n} + 13 cdot left(4 cdot 4^{n-3} binom{12}{n-3} + 4^{n-4} binom{12}{n-4}right) + sum_{k=1}^n binom{13}{k} binom{13-k}{n-2k} binom{4}{2}^k 4^{n-2k} right)}{binom{52}{n}}$$
Putting this into WolframAlpha, we get these values:
Now, this doesn't account from when it happens, only how likely it is to happen after a certain number of draws. To get a closer number on that, you'd have to do some stricter case work, involving ordering.
Essentially, just calculate the number of ordered decks which will has a royal flush at exactly the $n$-th card.
Let $f(n)$ be the number of orderings of $n$ cards containing a royal flush. Using the methods of before, but not diving by $binom{52}{n}$ we get $f(n)$. From here, we can calculate exactly how many decks get a royal flush in exactly the $n$-th draw. We shall denote this number as g(n).
$$ g(n) = f(n)(52-n)! - g(n-1) $$
Then, simply take the sum:
$$ sum_{n=1}^{52} frac{n cdot g(n)}{52!} $$
Which, unfortunately, I am too lazy to evaluate... consider it an exercise :)
answered Jan 4 at 4:29
Zachary HunterZachary Hunter
57611
$endgroup$
add a comment |
$begingroup$
A simpler question to address is, having drawn n cards, what is the probability of having a full house. Calculating this is equivalent to counting the number of possible hands with a full house, and then dividing my total possible hands.
To get this, we can count the invalid hands through casework, and then subtract this from the total possible hands:
Case 1 - no doubles/ triples:
$$ 4^n cdot binom{13}{n} $$
Case 2 - a triple/quadruple:
$$ 13 cdot left(4 cdot 4^{n-3} binom{12}{n-3} + 4^{n-4} binom{12}{n-4}right) $$
Case 3 - having k doubles:
$$ binom{13}{k} binom{13-k}{n-2k} binom{4}{2}^k 4^{n-2k} $$
Thus, the chance that we get a straight flush drawing n cards is:
$$ frac{binom{52}{n}- left( 4^n binom{13}{n} + 13 cdot left(4 cdot 4^{n-3} binom{12}{n-3} + 4^{n-4} binom{12}{n-4}right) + sum_{k=1}^n binom{13}{k} binom{13-k}{n-2k} binom{4}{2}^k 4^{n-2k} right)}{binom{52}{n}}$$
Putting this into WolframAlpha, we get these values:
Now, this doesn't account from when it happens, only how likely it is to happen after a certain number of draws. To get a closer number on that, you'd have to do some stricter case work, involving ordering.
Essentially, just calculate the number of ordered decks which will has a royal flush at exactly the $n$-th card.
Let $f(n)$ be the number of orderings of $n$ cards containing a royal flush. Using the methods of before, but not diving by $binom{52}{n}$ we get $f(n)$. From here, we can calculate exactly how many decks get a royal flush in exactly the $n$-th draw. We shall denote this number as g(n).
$$ g(n) = f(n)(52-n)! - g(n-1) $$
Then, simply take the sum:
$$ sum_{n=1}^{52} frac{n cdot g(n)}{52!} $$
Which, unfortunately, I am too lazy to evaluate... consider it an exercise :)
answered Jan 4 at 4:29
Zachary HunterZachary Hunter
57611
$endgroup$
add a comment |
$begingroup$
A simpler question to address is, having drawn n cards, what is the probability of having a full house. Calculating this is equivalent to counting the number of possible hands with a full house, and then dividing my total possible hands.
To get this, we can count the invalid hands through casework, and then subtract this from the total possible hands:
Case 1 - no doubles/ triples:
$$ 4^n cdot binom{13}{n} $$
Case 2 - a triple/quadruple:
$$ 13 cdot left(4 cdot 4^{n-3} binom{12}{n-3} + 4^{n-4} binom{12}{n-4}right) $$
Case 3 - having k doubles:
$$ binom{13}{k} binom{13-k}{n-2k} binom{4}{2}^k 4^{n-2k} $$
Thus, the chance that we get a straight flush drawing n cards is:
$$ frac{binom{52}{n}- left( 4^n binom{13}{n} + 13 cdot left(4 cdot 4^{n-3} binom{12}{n-3} + 4^{n-4} binom{12}{n-4}right) + sum_{k=1}^n binom{13}{k} binom{13-k}{n-2k} binom{4}{2}^k 4^{n-2k} right)}{binom{52}{n}}$$
Putting this into WolframAlpha, we get these values:
Now, this doesn't account from when it happens, only how likely it is to happen after a certain number of draws. To get a closer number on that, you'd have to do some stricter case work, involving ordering.
Essentially, just calculate the number of ordered decks which will has a royal flush at exactly the $n$-th card.
Let $f(n)$ be the number of orderings of $n$ cards containing a royal flush. Using the methods of before, but not diving by $binom{52}{n}$ we get $f(n)$. From here, we can calculate exactly how many decks get a royal flush in exactly the $n$-th draw. We shall denote this number as g(n).
$$ g(n) = f(n)(52-n)! - g(n-1) $$
Then, simply take the sum:
$$ sum_{n=1}^{52} frac{n cdot g(n)}{52!} $$
Which, unfortunately, I am too lazy to evaluate... consider it an exercise :)
answered Jan 4 at 4:29
Zachary HunterZachary Hunter
57611
$endgroup$
A simpler question to address is, having drawn n cards, what is the probability of having a full house. Calculating this is equivalent to counting the number of possible hands with a full house, and then dividing my total possible hands.
To get this, we can count the invalid hands through casework, and then subtract this from the total possible hands:
Case 1 - no doubles/ triples:
$$ 4^n cdot binom{13}{n} $$
Case 2 - a triple/quadruple:
$$ 13 cdot left(4 cdot 4^{n-3} binom{12}{n-3} + 4^{n-4} binom{12}{n-4}right) $$
Case 3 - having k doubles:
$$ binom{13}{k} binom{13-k}{n-2k} binom{4}{2}^k 4^{n-2k} $$
Thus, the chance that we get a straight flush drawing n cards is:
$$ frac{binom{52}{n}- left( 4^n binom{13}{n} + 13 cdot left(4 cdot 4^{n-3} binom{12}{n-3} + 4^{n-4} binom{12}{n-4}right) + sum_{k=1}^n binom{13}{k} binom{13-k}{n-2k} binom{4}{2}^k 4^{n-2k} right)}{binom{52}{n}}$$
Putting this into WolframAlpha, we get these values:
Now, this doesn't account from when it happens, only how likely it is to happen after a certain number of draws. To get a closer number on that, you'd have to do some stricter case work, involving ordering.
Essentially, just calculate the number of ordered decks which will has a royal flush at exactly the $n$-th card.
Let $f(n)$ be the number of orderings of $n$ cards containing a royal flush. Using the methods of before, but not diving by $binom{52}{n}$ we get $f(n)$. From here, we can calculate exactly how many decks get a royal flush in exactly the $n$-th draw. We shall denote this number as g(n).
$$ g(n) = f(n)(52-n)! - g(n-1) $$
Then, simply take the sum:
$$ sum_{n=1}^{52} frac{n cdot g(n)}{52!} $$
Which, unfortunately, I am too lazy to evaluate... consider it an exercise :)
answered Jan 4 at 4:29
Zachary HunterZachary Hunter
57611
answered Jan 4 at 4:29
Zachary HunterZachary Hunter
57611
answered Jan 4 at 4:29
Zachary HunterZachary Hunter
57611
answered Jan 4 at 4:29
Zachary HunterZachary Hunter
57611
57611
add a comment |
add a comment |
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$begingroup$
If this is too hard, I would be interested in the same question but for other simple poker hands like a single pair or a three-of-a-kind
$endgroup$
– user132290
Dec 14 '18 at 23:48
$begingroup$
And the value of 12 quoted above comes from my computer simulations.
$endgroup$
– user132290
Dec 15 '18 at 12:13
$begingroup$
My computer simulation gives an average of $13.9$ with a standard deviation of $3.7$.
$endgroup$
– Jens
Dec 15 '18 at 13:13
$begingroup$
@Jens Indeed. The exact expectation I got is $13.893157dots$, which is amazingly close to your simulation result. The interesting question is how to get it with a back of the envelope calculation (I just honestly wrote the combinatorial sum and told my PC to compute it)..
$endgroup$
– fedja
Dec 15 '18 at 17:39
$begingroup$
How do you guys simulate this? My simulation gives 13.55 after 1 million runs. Do you also actually simulate a deck of cards, or do you calculate the probability somehow?
$endgroup$
– user132290
Dec 17 '18 at 21:51