Probability that first card is spade given that second and third card are spades
$begingroup$
I am trying to solve the following probability problem:
Given a regular deck of playing cards containing 52 cards, we draw 3 cards without replacement. Given that the second and third cards are spades, what is the probability that the first card is also a spade?
I am aware that the same question has been asked here
I understand the solution that is being presented in the referenced link.
However I attempted to solve the problem using combination rather than permutation:
Let $ S_i $ denote the event that the $ i^{th} $ card is a spade and $ S_{i}^{c} $ otherwise.
By Bayes' theorem:
$$
begin{aligned}
P(S_{1} | S_{2}S_{3}) &= frac{P(S_{1}S_{2}S_{3})}{P(S_{2}P_{3})} \ \
&= frac{P(S_{1}S_{2}S_{3})}{P(S_{1}S_{2}P_{3}) + P(S_{1}^{c}S_{2}P_{3})} \ \
&= largefrac{frac{C(13,3)}{C(52,3)}}{frac{C(13,3)}{C(52,3)} + frac{C(39,1) cdot C(13,2)}{C(52,3)}} \ \
&= frac{11}{128}
end{aligned}
$$
Correct answer is $ frac{11}{50} $
After some checking, I realized that the term $ frac{C(39,1) cdot C(13,2)}{C(52,3)} $ is causing my method to produce the wrong answer. But I can't seem to understand why this term is incorrect. Could someone please explain to me?
probability combinations bayes-theorem
edited Apr 13 '17 at 12:21
Community♦
1
asked Feb 19 '17 at 14:55
LanceHAOHLanceHAOH
3971413
$endgroup$
|
show 4 more comments
$begingroup$
I am trying to solve the following probability problem:
Given a regular deck of playing cards containing 52 cards, we draw 3 cards without replacement. Given that the second and third cards are spades, what is the probability that the first card is also a spade?
I am aware that the same question has been asked here
I understand the solution that is being presented in the referenced link.
However I attempted to solve the problem using combination rather than permutation:
Let $ S_i $ denote the event that the $ i^{th} $ card is a spade and $ S_{i}^{c} $ otherwise.
By Bayes' theorem:
$$
begin{aligned}
P(S_{1} | S_{2}S_{3}) &= frac{P(S_{1}S_{2}S_{3})}{P(S_{2}P_{3})} \ \
&= frac{P(S_{1}S_{2}S_{3})}{P(S_{1}S_{2}P_{3}) + P(S_{1}^{c}S_{2}P_{3})} \ \
&= largefrac{frac{C(13,3)}{C(52,3)}}{frac{C(13,3)}{C(52,3)} + frac{C(39,1) cdot C(13,2)}{C(52,3)}} \ \
&= frac{11}{128}
end{aligned}
$$
Correct answer is $ frac{11}{50} $
After some checking, I realized that the term $ frac{C(39,1) cdot C(13,2)}{C(52,3)} $ is causing my method to produce the wrong answer. But I can't seem to understand why this term is incorrect. Could someone please explain to me?
probability combinations bayes-theorem
edited Apr 13 '17 at 12:21
Community♦
1
asked Feb 19 '17 at 14:55
LanceHAOHLanceHAOH
3971413
$endgroup$
3
$begingroup$
Since the cards are drawn without replacement. Knowing the 2nd and 3rd cards are spade, there is only 11 spade cards in the other total 50 cards. So the probability is $frac{11}{50}$. You don't need Bayes theorem for this.
$endgroup$
– Guangliang
Feb 19 '17 at 15:20
$begingroup$
@Guangliang I think LanceHAOH wants to solve this problem with Bayes' Rule.
$endgroup$
– gobucksmath
Feb 19 '17 at 15:29
2
$begingroup$
That term does not compute the probability you intend. Rather you want $frac {C(39,1)times C(13,2)}{C(52,1)times C(51,2)}$
$endgroup$
– lulu
Feb 19 '17 at 15:29
1
$begingroup$
No...that's not the problem. The problem is that by using combinations, and disregarding order, you are losing the special role of the first card. To be clear, the probability that the first card is not a spade, but the next two are, is $frac {39}{52}times frac {13}{51}times frac {12}{50}$.
$endgroup$
– lulu
Feb 19 '17 at 15:41
1
$begingroup$
The point is that introducing unordered combinations is a poor idea in this case (even though it can be salvaged). That's because the problem explicitly concerns the order of the first three choices.
$endgroup$
– lulu
Feb 19 '17 at 15:58
|
show 4 more comments
$begingroup$
I am trying to solve the following probability problem:
Given a regular deck of playing cards containing 52 cards, we draw 3 cards without replacement. Given that the second and third cards are spades, what is the probability that the first card is also a spade?
I am aware that the same question has been asked here
I understand the solution that is being presented in the referenced link.
However I attempted to solve the problem using combination rather than permutation:
Let $ S_i $ denote the event that the $ i^{th} $ card is a spade and $ S_{i}^{c} $ otherwise.
By Bayes' theorem:
$$
begin{aligned}
P(S_{1} | S_{2}S_{3}) &= frac{P(S_{1}S_{2}S_{3})}{P(S_{2}P_{3})} \ \
&= frac{P(S_{1}S_{2}S_{3})}{P(S_{1}S_{2}P_{3}) + P(S_{1}^{c}S_{2}P_{3})} \ \
&= largefrac{frac{C(13,3)}{C(52,3)}}{frac{C(13,3)}{C(52,3)} + frac{C(39,1) cdot C(13,2)}{C(52,3)}} \ \
&= frac{11}{128}
end{aligned}
$$
Correct answer is $ frac{11}{50} $
After some checking, I realized that the term $ frac{C(39,1) cdot C(13,2)}{C(52,3)} $ is causing my method to produce the wrong answer. But I can't seem to understand why this term is incorrect. Could someone please explain to me?
probability combinations bayes-theorem
edited Apr 13 '17 at 12:21
Community♦
1
asked Feb 19 '17 at 14:55
LanceHAOHLanceHAOH
3971413
$endgroup$
I am trying to solve the following probability problem:
Given a regular deck of playing cards containing 52 cards, we draw 3 cards without replacement. Given that the second and third cards are spades, what is the probability that the first card is also a spade?
I am aware that the same question has been asked here
I understand the solution that is being presented in the referenced link.
However I attempted to solve the problem using combination rather than permutation:
Let $ S_i $ denote the event that the $ i^{th} $ card is a spade and $ S_{i}^{c} $ otherwise.
By Bayes' theorem:
$$
begin{aligned}
P(S_{1} | S_{2}S_{3}) &= frac{P(S_{1}S_{2}S_{3})}{P(S_{2}P_{3})} \ \
&= frac{P(S_{1}S_{2}S_{3})}{P(S_{1}S_{2}P_{3}) + P(S_{1}^{c}S_{2}P_{3})} \ \
&= largefrac{frac{C(13,3)}{C(52,3)}}{frac{C(13,3)}{C(52,3)} + frac{C(39,1) cdot C(13,2)}{C(52,3)}} \ \
&= frac{11}{128}
end{aligned}
$$
Correct answer is $ frac{11}{50} $
After some checking, I realized that the term $ frac{C(39,1) cdot C(13,2)}{C(52,3)} $ is causing my method to produce the wrong answer. But I can't seem to understand why this term is incorrect. Could someone please explain to me?
probability combinations bayes-theorem
probability combinations bayes-theorem
edited Apr 13 '17 at 12:21
Community♦
1
asked Feb 19 '17 at 14:55
LanceHAOHLanceHAOH
3971413
edited Apr 13 '17 at 12:21
Community♦
1
asked Feb 19 '17 at 14:55
LanceHAOHLanceHAOH
3971413
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
1
asked Feb 19 '17 at 14:55
LanceHAOHLanceHAOH
3971413
asked Feb 19 '17 at 14:55
LanceHAOHLanceHAOH
3971413
asked Feb 19 '17 at 14:55
LanceHAOHLanceHAOH
3971413
3971413
3
$begingroup$
Since the cards are drawn without replacement. Knowing the 2nd and 3rd cards are spade, there is only 11 spade cards in the other total 50 cards. So the probability is $frac{11}{50}$. You don't need Bayes theorem for this.
$endgroup$
– Guangliang
Feb 19 '17 at 15:20
$begingroup$
@Guangliang I think LanceHAOH wants to solve this problem with Bayes' Rule.
$endgroup$
– gobucksmath
Feb 19 '17 at 15:29
2
$begingroup$
That term does not compute the probability you intend. Rather you want $frac {C(39,1)times C(13,2)}{C(52,1)times C(51,2)}$
$endgroup$
– lulu
Feb 19 '17 at 15:29
1
$begingroup$
No...that's not the problem. The problem is that by using combinations, and disregarding order, you are losing the special role of the first card. To be clear, the probability that the first card is not a spade, but the next two are, is $frac {39}{52}times frac {13}{51}times frac {12}{50}$.
$endgroup$
– lulu
Feb 19 '17 at 15:41
1
$begingroup$
The point is that introducing unordered combinations is a poor idea in this case (even though it can be salvaged). That's because the problem explicitly concerns the order of the first three choices.
$endgroup$
– lulu
Feb 19 '17 at 15:58
|
show 4 more comments
3
$begingroup$
Since the cards are drawn without replacement. Knowing the 2nd and 3rd cards are spade, there is only 11 spade cards in the other total 50 cards. So the probability is $frac{11}{50}$. You don't need Bayes theorem for this.
$endgroup$
– Guangliang
Feb 19 '17 at 15:20
$begingroup$
@Guangliang I think LanceHAOH wants to solve this problem with Bayes' Rule.
$endgroup$
– gobucksmath
Feb 19 '17 at 15:29
2
$begingroup$
That term does not compute the probability you intend. Rather you want $frac {C(39,1)times C(13,2)}{C(52,1)times C(51,2)}$
$endgroup$
– lulu
Feb 19 '17 at 15:29
1
$begingroup$
No...that's not the problem. The problem is that by using combinations, and disregarding order, you are losing the special role of the first card. To be clear, the probability that the first card is not a spade, but the next two are, is $frac {39}{52}times frac {13}{51}times frac {12}{50}$.
$endgroup$
– lulu
Feb 19 '17 at 15:41
1
$begingroup$
The point is that introducing unordered combinations is a poor idea in this case (even though it can be salvaged). That's because the problem explicitly concerns the order of the first three choices.
$endgroup$
– lulu
Feb 19 '17 at 15:58
3
3
$begingroup$
Since the cards are drawn without replacement. Knowing the 2nd and 3rd cards are spade, there is only 11 spade cards in the other total 50 cards. So the probability is $frac{11}{50}$. You don't need Bayes theorem for this.
$endgroup$
– Guangliang
Feb 19 '17 at 15:20
$begingroup$
Since the cards are drawn without replacement. Knowing the 2nd and 3rd cards are spade, there is only 11 spade cards in the other total 50 cards. So the probability is $frac{11}{50}$. You don't need Bayes theorem for this.
$endgroup$
– Guangliang
Feb 19 '17 at 15:20
$begingroup$
@Guangliang I think LanceHAOH wants to solve this problem with Bayes' Rule.
$endgroup$
– gobucksmath
Feb 19 '17 at 15:29
$begingroup$
@Guangliang I think LanceHAOH wants to solve this problem with Bayes' Rule.
$endgroup$
– gobucksmath
Feb 19 '17 at 15:29
2
2
$begingroup$
That term does not compute the probability you intend. Rather you want $frac {C(39,1)times C(13,2)}{C(52,1)times C(51,2)}$
$endgroup$
– lulu
Feb 19 '17 at 15:29
$begingroup$
That term does not compute the probability you intend. Rather you want $frac {C(39,1)times C(13,2)}{C(52,1)times C(51,2)}$
$endgroup$
– lulu
Feb 19 '17 at 15:29
1
1
$begingroup$
No...that's not the problem. The problem is that by using combinations, and disregarding order, you are losing the special role of the first card. To be clear, the probability that the first card is not a spade, but the next two are, is $frac {39}{52}times frac {13}{51}times frac {12}{50}$.
$endgroup$
– lulu
Feb 19 '17 at 15:41
$begingroup$
No...that's not the problem. The problem is that by using combinations, and disregarding order, you are losing the special role of the first card. To be clear, the probability that the first card is not a spade, but the next two are, is $frac {39}{52}times frac {13}{51}times frac {12}{50}$.
$endgroup$
– lulu
Feb 19 '17 at 15:41
1
1
$begingroup$
The point is that introducing unordered combinations is a poor idea in this case (even though it can be salvaged). That's because the problem explicitly concerns the order of the first three choices.
$endgroup$
– lulu
Feb 19 '17 at 15:58
$begingroup$
The point is that introducing unordered combinations is a poor idea in this case (even though it can be salvaged). That's because the problem explicitly concerns the order of the first three choices.
$endgroup$
– lulu
Feb 19 '17 at 15:58
|
show 4 more comments
3 Answers
3
active
oldest
votes
$begingroup$
I'm not sure that that term is the one causing you trouble! It is the other term. The key is that order of cards does matter. In a typical poker hand order doesn't matter.
This gives us that you should be tripling your values for each collection of three spades; the collection Ace of Spades, 2 of Spades, and 4 of Spades has three distinct choices for first card.
The correct calculation is
begin{equation*}
mathbb{P}(S1|S2S3)=frac{3cdot C(13,3)}{3cdot C(13,3)+C(39,1)cdot C(13,2)}
end{equation*}
To simplify, imagine that you have a twelve card deck with four suits of three cards. The desired probability should be $tfrac{1}{10}$. Choose the first card, and then choose the remaining two together:
Four hands begin with a spade as a choice, and there is only one way to end up with all spades. Using your calculation, you would only have one way to arrive at this sequence.
Twelve hands begin with no spade, and there are $C(3,2)$ ways to draw the next two spades.
begin{equation*}
mathbb{P}(S1|S2S3)=frac{4}{4+36}=frac{1}{10}
end{equation*}
answered Feb 19 '17 at 15:27
gobucksmathgobucksmath
74537
$endgroup$
add a comment |
$begingroup$
It's true that
$$ P(S_1 S_2 S_3) = frac{C(13,3)}{C(52,3)}$$
and also that
$$ P(S_1^c S_2 S_3) neq frac{C(39,1) cdot C(13,2)}{C(52,3)}.$$
In each case, the denominator on the right-hand side counts the number of possible three-card hands, ignoring the order in which the cards are drawn.
The numerator $C(13,3)$ counts the number of hands with three spades;
although it again disregards the order in which the cards were drawn,
the event that you draw a spade followed by two others is exactly the event that you draw three spades, so indeed the order of drawing does not matter for this particular probability.
But suppose we wanted the probability that you drew a hand with exactly two spades in it. Examples of such hands are
$(2spadesuit, 9spadesuit, 3diamondsuit)$,
$(5spadesuit, 9heartsuit, mathrm Jspadesuit)$, and
$(9heartsuit, 4spadesuit, mathrm 8spadesuit)$ -- shown here in the order the cards were drawn, but for this particular probability all these hands are considered "two spades and one non-spade."
To compute the probability of any hand with exactly two spades,
combinations are fine: to get a random three card hand we can simply line up the $52$ cards and pick three at random, with $C(52,3)$ equally-likely possible outcomes.
The hands with exactly two spades will occur when two of the three choices
occurred among the spades (there are $C(13,2)$ possible ways to do this)
and one choice occurred among the other three suits
(there are $C(39,1)$ possible ways to do that).
The total number of hands with exactly two spades, disregarding the order of drawing them, is $C(39,1) cdot C(13,2)$, and the probability of drawing such a hand is
$$
frac{C(39,1) cdot C(13,2)}{C(52,3)}.
$$
The probability that all three cards are spades, given that at least two of the cards are spades, is in fact
$$frac{C(39,1) cdot C(13,2)}{C(52,3)}.$$
The only thing wrong with that answer is that the question specifically said the second and third cards must be spades.
In the usual language of card probabilities, that means the second and third cards in the particular sequence in which the cards were drawn.
Cards drawn in sequences such as
$(2spadesuit, 9spadesuit, 3diamondsuit)$ and
$(5spadesuit, 9heartsuit, mathrm Jspadesuit)$
are not cases in which the second and third cards are spades but the first is not; they are not part of the event $S_1^c S_2 S_3.$
There are several ways to fix this. One way is to forget about using combinations and just use permutations for everything, since there are cases where the order of drawing matters.
Another way to fix this is to use the probability
$$frac{C(39,1) cdot C(13,2)}{C(52,3)},$$
but also account for the fact that independent of which set of three cards was chosen for the hand, there is only a $frac 13$ probability that the single non-spade in that hand happened to have been drawn first.
Therefore
$$P(S_1^c S_2 S_3) = frac13 cdot frac{C(39,1) cdot C(13,2)}{C(52,3)}.$$
Another fix is to partially consider the order of drawing cards when drawing a non-spade followed by two spades.
That is, in the denominator we consider all the choices that could have been made for the first card ($C(52,1)$), and all the choices for the other two cards from the remaining cards after the first is chosen ($C(51,2)$).
That is, the number of ways to choose one "first" card and then choose two others is $C(52,1)cdot C(51,2),$ so
$$P(S_1^c S_2 S_3) = frac{C(39,1) cdot C(13,2)}{C(52,1)cdot C(51,2)}.$$
Personally, I find all of the above (including the idea of using permutations) to be more complicated than necessary.
We have a simple sequence of independent events:
draw a non-spade, then draw a spade from the remaining cards, then draw one of the remaining spades from the remaining cards.
Taking these events one at a time, the probability comes to
$$P(S_1^c S_2 S_3) = frac{39}{52} cdot frac{13}{51} cdot frac{12}{50}.$$
One can easily confirm that all these ways of computing $P(S_1^c S_2 S_3)$
give the same answer.
answered May 7 '18 at 13:05
David KDavid K
55.9k345121
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However I attempted to solve the problem using combination rather than permutation:
The problem is, the card of interest is in a particular position.
Well, if we let $A$ be the event that the second card is a spade, and $B$ the event that the first and third are spades, then $Acap B$ is the event that the first three cards are all spades. $$mathsf P(Amid B)=dfrac{mathsf P(Acap B)}{mathsf P(B)}=dfrac{left.dbinom{13}3middle/dbinom {52}{3}right.}{left.dbinom{13}{2}middle/dbinom {52}2right.}=dfrac{dfrac{13!}{3!10!}dfrac{52!}{50!2!}}{dfrac{13!}{2!11!}dfrac{52!}{49!3!}}=dfrac{11}{50}$$
Now look at the denominator: $dfrac{binom{13}2}{binom{52}2}=dfrac{binom{13}3}{binom{52}3}+dfrac{binom{13}2binom{39}1}{binom{52}2binom{50}1}$. It is the probability for selecting three spades into all three positions, or two spades and one non-spade into a particular position (second). If you like, we can also express that as: $dfrac{binom{13}2}{binom{52}2}=dfrac{binom{13}3}{binom{52}3}+dfrac{binom{13}2binom{39}1/binom 31}{binom{52}3}$.
$$mathsf P(Amid B)=dfrac{dbinom{13}3}{dbinom{13}3+left.dbinom{13}2dbinom{39}1middle/dbinom 31right.}=dfrac{11}{50}$$
answered Nov 21 '18 at 0:26
Graham KempGraham Kemp
88.1k43579
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3 Answers
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$begingroup$
I'm not sure that that term is the one causing you trouble! It is the other term. The key is that order of cards does matter. In a typical poker hand order doesn't matter.
This gives us that you should be tripling your values for each collection of three spades; the collection Ace of Spades, 2 of Spades, and 4 of Spades has three distinct choices for first card.
The correct calculation is
begin{equation*}
mathbb{P}(S1|S2S3)=frac{3cdot C(13,3)}{3cdot C(13,3)+C(39,1)cdot C(13,2)}
end{equation*}
To simplify, imagine that you have a twelve card deck with four suits of three cards. The desired probability should be $tfrac{1}{10}$. Choose the first card, and then choose the remaining two together:
Four hands begin with a spade as a choice, and there is only one way to end up with all spades. Using your calculation, you would only have one way to arrive at this sequence.
Twelve hands begin with no spade, and there are $C(3,2)$ ways to draw the next two spades.
begin{equation*}
mathbb{P}(S1|S2S3)=frac{4}{4+36}=frac{1}{10}
end{equation*}
answered Feb 19 '17 at 15:27
gobucksmathgobucksmath
74537
$endgroup$
add a comment |
$begingroup$
I'm not sure that that term is the one causing you trouble! It is the other term. The key is that order of cards does matter. In a typical poker hand order doesn't matter.
This gives us that you should be tripling your values for each collection of three spades; the collection Ace of Spades, 2 of Spades, and 4 of Spades has three distinct choices for first card.
The correct calculation is
begin{equation*}
mathbb{P}(S1|S2S3)=frac{3cdot C(13,3)}{3cdot C(13,3)+C(39,1)cdot C(13,2)}
end{equation*}
To simplify, imagine that you have a twelve card deck with four suits of three cards. The desired probability should be $tfrac{1}{10}$. Choose the first card, and then choose the remaining two together:
Four hands begin with a spade as a choice, and there is only one way to end up with all spades. Using your calculation, you would only have one way to arrive at this sequence.
Twelve hands begin with no spade, and there are $C(3,2)$ ways to draw the next two spades.
begin{equation*}
mathbb{P}(S1|S2S3)=frac{4}{4+36}=frac{1}{10}
end{equation*}
answered Feb 19 '17 at 15:27
gobucksmathgobucksmath
74537
$endgroup$
add a comment |
$begingroup$
I'm not sure that that term is the one causing you trouble! It is the other term. The key is that order of cards does matter. In a typical poker hand order doesn't matter.
This gives us that you should be tripling your values for each collection of three spades; the collection Ace of Spades, 2 of Spades, and 4 of Spades has three distinct choices for first card.
The correct calculation is
begin{equation*}
mathbb{P}(S1|S2S3)=frac{3cdot C(13,3)}{3cdot C(13,3)+C(39,1)cdot C(13,2)}
end{equation*}
To simplify, imagine that you have a twelve card deck with four suits of three cards. The desired probability should be $tfrac{1}{10}$. Choose the first card, and then choose the remaining two together:
Four hands begin with a spade as a choice, and there is only one way to end up with all spades. Using your calculation, you would only have one way to arrive at this sequence.
Twelve hands begin with no spade, and there are $C(3,2)$ ways to draw the next two spades.
begin{equation*}
mathbb{P}(S1|S2S3)=frac{4}{4+36}=frac{1}{10}
end{equation*}
answered Feb 19 '17 at 15:27
gobucksmathgobucksmath
74537
$endgroup$
I'm not sure that that term is the one causing you trouble! It is the other term. The key is that order of cards does matter. In a typical poker hand order doesn't matter.
This gives us that you should be tripling your values for each collection of three spades; the collection Ace of Spades, 2 of Spades, and 4 of Spades has three distinct choices for first card.
The correct calculation is
begin{equation*}
mathbb{P}(S1|S2S3)=frac{3cdot C(13,3)}{3cdot C(13,3)+C(39,1)cdot C(13,2)}
end{equation*}
To simplify, imagine that you have a twelve card deck with four suits of three cards. The desired probability should be $tfrac{1}{10}$. Choose the first card, and then choose the remaining two together:
Four hands begin with a spade as a choice, and there is only one way to end up with all spades. Using your calculation, you would only have one way to arrive at this sequence.
Twelve hands begin with no spade, and there are $C(3,2)$ ways to draw the next two spades.
begin{equation*}
mathbb{P}(S1|S2S3)=frac{4}{4+36}=frac{1}{10}
end{equation*}
answered Feb 19 '17 at 15:27
gobucksmathgobucksmath
74537
answered Feb 19 '17 at 15:27
gobucksmathgobucksmath
74537
answered Feb 19 '17 at 15:27
gobucksmathgobucksmath
74537
answered Feb 19 '17 at 15:27
gobucksmathgobucksmath
74537
74537
add a comment |
add a comment |
$begingroup$
It's true that
$$ P(S_1 S_2 S_3) = frac{C(13,3)}{C(52,3)}$$
and also that
$$ P(S_1^c S_2 S_3) neq frac{C(39,1) cdot C(13,2)}{C(52,3)}.$$
In each case, the denominator on the right-hand side counts the number of possible three-card hands, ignoring the order in which the cards are drawn.
The numerator $C(13,3)$ counts the number of hands with three spades;
although it again disregards the order in which the cards were drawn,
the event that you draw a spade followed by two others is exactly the event that you draw three spades, so indeed the order of drawing does not matter for this particular probability.
But suppose we wanted the probability that you drew a hand with exactly two spades in it. Examples of such hands are
$(2spadesuit, 9spadesuit, 3diamondsuit)$,
$(5spadesuit, 9heartsuit, mathrm Jspadesuit)$, and
$(9heartsuit, 4spadesuit, mathrm 8spadesuit)$ -- shown here in the order the cards were drawn, but for this particular probability all these hands are considered "two spades and one non-spade."
To compute the probability of any hand with exactly two spades,
combinations are fine: to get a random three card hand we can simply line up the $52$ cards and pick three at random, with $C(52,3)$ equally-likely possible outcomes.
The hands with exactly two spades will occur when two of the three choices
occurred among the spades (there are $C(13,2)$ possible ways to do this)
and one choice occurred among the other three suits
(there are $C(39,1)$ possible ways to do that).
The total number of hands with exactly two spades, disregarding the order of drawing them, is $C(39,1) cdot C(13,2)$, and the probability of drawing such a hand is
$$
frac{C(39,1) cdot C(13,2)}{C(52,3)}.
$$
The probability that all three cards are spades, given that at least two of the cards are spades, is in fact
$$frac{C(39,1) cdot C(13,2)}{C(52,3)}.$$
The only thing wrong with that answer is that the question specifically said the second and third cards must be spades.
In the usual language of card probabilities, that means the second and third cards in the particular sequence in which the cards were drawn.
Cards drawn in sequences such as
$(2spadesuit, 9spadesuit, 3diamondsuit)$ and
$(5spadesuit, 9heartsuit, mathrm Jspadesuit)$
are not cases in which the second and third cards are spades but the first is not; they are not part of the event $S_1^c S_2 S_3.$
There are several ways to fix this. One way is to forget about using combinations and just use permutations for everything, since there are cases where the order of drawing matters.
Another way to fix this is to use the probability
$$frac{C(39,1) cdot C(13,2)}{C(52,3)},$$
but also account for the fact that independent of which set of three cards was chosen for the hand, there is only a $frac 13$ probability that the single non-spade in that hand happened to have been drawn first.
Therefore
$$P(S_1^c S_2 S_3) = frac13 cdot frac{C(39,1) cdot C(13,2)}{C(52,3)}.$$
Another fix is to partially consider the order of drawing cards when drawing a non-spade followed by two spades.
That is, in the denominator we consider all the choices that could have been made for the first card ($C(52,1)$), and all the choices for the other two cards from the remaining cards after the first is chosen ($C(51,2)$).
That is, the number of ways to choose one "first" card and then choose two others is $C(52,1)cdot C(51,2),$ so
$$P(S_1^c S_2 S_3) = frac{C(39,1) cdot C(13,2)}{C(52,1)cdot C(51,2)}.$$
Personally, I find all of the above (including the idea of using permutations) to be more complicated than necessary.
We have a simple sequence of independent events:
draw a non-spade, then draw a spade from the remaining cards, then draw one of the remaining spades from the remaining cards.
Taking these events one at a time, the probability comes to
$$P(S_1^c S_2 S_3) = frac{39}{52} cdot frac{13}{51} cdot frac{12}{50}.$$
One can easily confirm that all these ways of computing $P(S_1^c S_2 S_3)$
give the same answer.
answered May 7 '18 at 13:05
David KDavid K
55.9k345121
$endgroup$
add a comment |
$begingroup$
It's true that
$$ P(S_1 S_2 S_3) = frac{C(13,3)}{C(52,3)}$$
and also that
$$ P(S_1^c S_2 S_3) neq frac{C(39,1) cdot C(13,2)}{C(52,3)}.$$
In each case, the denominator on the right-hand side counts the number of possible three-card hands, ignoring the order in which the cards are drawn.
The numerator $C(13,3)$ counts the number of hands with three spades;
although it again disregards the order in which the cards were drawn,
the event that you draw a spade followed by two others is exactly the event that you draw three spades, so indeed the order of drawing does not matter for this particular probability.
But suppose we wanted the probability that you drew a hand with exactly two spades in it. Examples of such hands are
$(2spadesuit, 9spadesuit, 3diamondsuit)$,
$(5spadesuit, 9heartsuit, mathrm Jspadesuit)$, and
$(9heartsuit, 4spadesuit, mathrm 8spadesuit)$ -- shown here in the order the cards were drawn, but for this particular probability all these hands are considered "two spades and one non-spade."
To compute the probability of any hand with exactly two spades,
combinations are fine: to get a random three card hand we can simply line up the $52$ cards and pick three at random, with $C(52,3)$ equally-likely possible outcomes.
The hands with exactly two spades will occur when two of the three choices
occurred among the spades (there are $C(13,2)$ possible ways to do this)
and one choice occurred among the other three suits
(there are $C(39,1)$ possible ways to do that).
The total number of hands with exactly two spades, disregarding the order of drawing them, is $C(39,1) cdot C(13,2)$, and the probability of drawing such a hand is
$$
frac{C(39,1) cdot C(13,2)}{C(52,3)}.
$$
The probability that all three cards are spades, given that at least two of the cards are spades, is in fact
$$frac{C(39,1) cdot C(13,2)}{C(52,3)}.$$
The only thing wrong with that answer is that the question specifically said the second and third cards must be spades.
In the usual language of card probabilities, that means the second and third cards in the particular sequence in which the cards were drawn.
Cards drawn in sequences such as
$(2spadesuit, 9spadesuit, 3diamondsuit)$ and
$(5spadesuit, 9heartsuit, mathrm Jspadesuit)$
are not cases in which the second and third cards are spades but the first is not; they are not part of the event $S_1^c S_2 S_3.$
There are several ways to fix this. One way is to forget about using combinations and just use permutations for everything, since there are cases where the order of drawing matters.
Another way to fix this is to use the probability
$$frac{C(39,1) cdot C(13,2)}{C(52,3)},$$
but also account for the fact that independent of which set of three cards was chosen for the hand, there is only a $frac 13$ probability that the single non-spade in that hand happened to have been drawn first.
Therefore
$$P(S_1^c S_2 S_3) = frac13 cdot frac{C(39,1) cdot C(13,2)}{C(52,3)}.$$
Another fix is to partially consider the order of drawing cards when drawing a non-spade followed by two spades.
That is, in the denominator we consider all the choices that could have been made for the first card ($C(52,1)$), and all the choices for the other two cards from the remaining cards after the first is chosen ($C(51,2)$).
That is, the number of ways to choose one "first" card and then choose two others is $C(52,1)cdot C(51,2),$ so
$$P(S_1^c S_2 S_3) = frac{C(39,1) cdot C(13,2)}{C(52,1)cdot C(51,2)}.$$
Personally, I find all of the above (including the idea of using permutations) to be more complicated than necessary.
We have a simple sequence of independent events:
draw a non-spade, then draw a spade from the remaining cards, then draw one of the remaining spades from the remaining cards.
Taking these events one at a time, the probability comes to
$$P(S_1^c S_2 S_3) = frac{39}{52} cdot frac{13}{51} cdot frac{12}{50}.$$
One can easily confirm that all these ways of computing $P(S_1^c S_2 S_3)$
give the same answer.
answered May 7 '18 at 13:05
David KDavid K
55.9k345121
$endgroup$
add a comment |
$begingroup$
It's true that
$$ P(S_1 S_2 S_3) = frac{C(13,3)}{C(52,3)}$$
and also that
$$ P(S_1^c S_2 S_3) neq frac{C(39,1) cdot C(13,2)}{C(52,3)}.$$
In each case, the denominator on the right-hand side counts the number of possible three-card hands, ignoring the order in which the cards are drawn.
The numerator $C(13,3)$ counts the number of hands with three spades;
although it again disregards the order in which the cards were drawn,
the event that you draw a spade followed by two others is exactly the event that you draw three spades, so indeed the order of drawing does not matter for this particular probability.
But suppose we wanted the probability that you drew a hand with exactly two spades in it. Examples of such hands are
$(2spadesuit, 9spadesuit, 3diamondsuit)$,
$(5spadesuit, 9heartsuit, mathrm Jspadesuit)$, and
$(9heartsuit, 4spadesuit, mathrm 8spadesuit)$ -- shown here in the order the cards were drawn, but for this particular probability all these hands are considered "two spades and one non-spade."
To compute the probability of any hand with exactly two spades,
combinations are fine: to get a random three card hand we can simply line up the $52$ cards and pick three at random, with $C(52,3)$ equally-likely possible outcomes.
The hands with exactly two spades will occur when two of the three choices
occurred among the spades (there are $C(13,2)$ possible ways to do this)
and one choice occurred among the other three suits
(there are $C(39,1)$ possible ways to do that).
The total number of hands with exactly two spades, disregarding the order of drawing them, is $C(39,1) cdot C(13,2)$, and the probability of drawing such a hand is
$$
frac{C(39,1) cdot C(13,2)}{C(52,3)}.
$$
The probability that all three cards are spades, given that at least two of the cards are spades, is in fact
$$frac{C(39,1) cdot C(13,2)}{C(52,3)}.$$
The only thing wrong with that answer is that the question specifically said the second and third cards must be spades.
In the usual language of card probabilities, that means the second and third cards in the particular sequence in which the cards were drawn.
Cards drawn in sequences such as
$(2spadesuit, 9spadesuit, 3diamondsuit)$ and
$(5spadesuit, 9heartsuit, mathrm Jspadesuit)$
are not cases in which the second and third cards are spades but the first is not; they are not part of the event $S_1^c S_2 S_3.$
There are several ways to fix this. One way is to forget about using combinations and just use permutations for everything, since there are cases where the order of drawing matters.
Another way to fix this is to use the probability
$$frac{C(39,1) cdot C(13,2)}{C(52,3)},$$
but also account for the fact that independent of which set of three cards was chosen for the hand, there is only a $frac 13$ probability that the single non-spade in that hand happened to have been drawn first.
Therefore
$$P(S_1^c S_2 S_3) = frac13 cdot frac{C(39,1) cdot C(13,2)}{C(52,3)}.$$
Another fix is to partially consider the order of drawing cards when drawing a non-spade followed by two spades.
That is, in the denominator we consider all the choices that could have been made for the first card ($C(52,1)$), and all the choices for the other two cards from the remaining cards after the first is chosen ($C(51,2)$).
That is, the number of ways to choose one "first" card and then choose two others is $C(52,1)cdot C(51,2),$ so
$$P(S_1^c S_2 S_3) = frac{C(39,1) cdot C(13,2)}{C(52,1)cdot C(51,2)}.$$
Personally, I find all of the above (including the idea of using permutations) to be more complicated than necessary.
We have a simple sequence of independent events:
draw a non-spade, then draw a spade from the remaining cards, then draw one of the remaining spades from the remaining cards.
Taking these events one at a time, the probability comes to
$$P(S_1^c S_2 S_3) = frac{39}{52} cdot frac{13}{51} cdot frac{12}{50}.$$
One can easily confirm that all these ways of computing $P(S_1^c S_2 S_3)$
give the same answer.
answered May 7 '18 at 13:05
David KDavid K
55.9k345121
$endgroup$
It's true that
$$ P(S_1 S_2 S_3) = frac{C(13,3)}{C(52,3)}$$
and also that
$$ P(S_1^c S_2 S_3) neq frac{C(39,1) cdot C(13,2)}{C(52,3)}.$$
In each case, the denominator on the right-hand side counts the number of possible three-card hands, ignoring the order in which the cards are drawn.
The numerator $C(13,3)$ counts the number of hands with three spades;
although it again disregards the order in which the cards were drawn,
the event that you draw a spade followed by two others is exactly the event that you draw three spades, so indeed the order of drawing does not matter for this particular probability.
But suppose we wanted the probability that you drew a hand with exactly two spades in it. Examples of such hands are
$(2spadesuit, 9spadesuit, 3diamondsuit)$,
$(5spadesuit, 9heartsuit, mathrm Jspadesuit)$, and
$(9heartsuit, 4spadesuit, mathrm 8spadesuit)$ -- shown here in the order the cards were drawn, but for this particular probability all these hands are considered "two spades and one non-spade."
To compute the probability of any hand with exactly two spades,
combinations are fine: to get a random three card hand we can simply line up the $52$ cards and pick three at random, with $C(52,3)$ equally-likely possible outcomes.
The hands with exactly two spades will occur when two of the three choices
occurred among the spades (there are $C(13,2)$ possible ways to do this)
and one choice occurred among the other three suits
(there are $C(39,1)$ possible ways to do that).
The total number of hands with exactly two spades, disregarding the order of drawing them, is $C(39,1) cdot C(13,2)$, and the probability of drawing such a hand is
$$
frac{C(39,1) cdot C(13,2)}{C(52,3)}.
$$
The probability that all three cards are spades, given that at least two of the cards are spades, is in fact
$$frac{C(39,1) cdot C(13,2)}{C(52,3)}.$$
The only thing wrong with that answer is that the question specifically said the second and third cards must be spades.
In the usual language of card probabilities, that means the second and third cards in the particular sequence in which the cards were drawn.
Cards drawn in sequences such as
$(2spadesuit, 9spadesuit, 3diamondsuit)$ and
$(5spadesuit, 9heartsuit, mathrm Jspadesuit)$
are not cases in which the second and third cards are spades but the first is not; they are not part of the event $S_1^c S_2 S_3.$
There are several ways to fix this. One way is to forget about using combinations and just use permutations for everything, since there are cases where the order of drawing matters.
Another way to fix this is to use the probability
$$frac{C(39,1) cdot C(13,2)}{C(52,3)},$$
but also account for the fact that independent of which set of three cards was chosen for the hand, there is only a $frac 13$ probability that the single non-spade in that hand happened to have been drawn first.
Therefore
$$P(S_1^c S_2 S_3) = frac13 cdot frac{C(39,1) cdot C(13,2)}{C(52,3)}.$$
Another fix is to partially consider the order of drawing cards when drawing a non-spade followed by two spades.
That is, in the denominator we consider all the choices that could have been made for the first card ($C(52,1)$), and all the choices for the other two cards from the remaining cards after the first is chosen ($C(51,2)$).
That is, the number of ways to choose one "first" card and then choose two others is $C(52,1)cdot C(51,2),$ so
$$P(S_1^c S_2 S_3) = frac{C(39,1) cdot C(13,2)}{C(52,1)cdot C(51,2)}.$$
Personally, I find all of the above (including the idea of using permutations) to be more complicated than necessary.
We have a simple sequence of independent events:
draw a non-spade, then draw a spade from the remaining cards, then draw one of the remaining spades from the remaining cards.
Taking these events one at a time, the probability comes to
$$P(S_1^c S_2 S_3) = frac{39}{52} cdot frac{13}{51} cdot frac{12}{50}.$$
One can easily confirm that all these ways of computing $P(S_1^c S_2 S_3)$
give the same answer.
answered May 7 '18 at 13:05
David KDavid K
55.9k345121
answered May 7 '18 at 13:05
David KDavid K
55.9k345121
answered May 7 '18 at 13:05
David KDavid K
55.9k345121
answered May 7 '18 at 13:05
David KDavid K
55.9k345121
55.9k345121
add a comment |
add a comment |
$begingroup$
However I attempted to solve the problem using combination rather than permutation:
The problem is, the card of interest is in a particular position.
Well, if we let $A$ be the event that the second card is a spade, and $B$ the event that the first and third are spades, then $Acap B$ is the event that the first three cards are all spades. $$mathsf P(Amid B)=dfrac{mathsf P(Acap B)}{mathsf P(B)}=dfrac{left.dbinom{13}3middle/dbinom {52}{3}right.}{left.dbinom{13}{2}middle/dbinom {52}2right.}=dfrac{dfrac{13!}{3!10!}dfrac{52!}{50!2!}}{dfrac{13!}{2!11!}dfrac{52!}{49!3!}}=dfrac{11}{50}$$
Now look at the denominator: $dfrac{binom{13}2}{binom{52}2}=dfrac{binom{13}3}{binom{52}3}+dfrac{binom{13}2binom{39}1}{binom{52}2binom{50}1}$. It is the probability for selecting three spades into all three positions, or two spades and one non-spade into a particular position (second). If you like, we can also express that as: $dfrac{binom{13}2}{binom{52}2}=dfrac{binom{13}3}{binom{52}3}+dfrac{binom{13}2binom{39}1/binom 31}{binom{52}3}$.
$$mathsf P(Amid B)=dfrac{dbinom{13}3}{dbinom{13}3+left.dbinom{13}2dbinom{39}1middle/dbinom 31right.}=dfrac{11}{50}$$
answered Nov 21 '18 at 0:26
Graham KempGraham Kemp
88.1k43579
$endgroup$
add a comment |
$begingroup$
However I attempted to solve the problem using combination rather than permutation:
The problem is, the card of interest is in a particular position.
Well, if we let $A$ be the event that the second card is a spade, and $B$ the event that the first and third are spades, then $Acap B$ is the event that the first three cards are all spades. $$mathsf P(Amid B)=dfrac{mathsf P(Acap B)}{mathsf P(B)}=dfrac{left.dbinom{13}3middle/dbinom {52}{3}right.}{left.dbinom{13}{2}middle/dbinom {52}2right.}=dfrac{dfrac{13!}{3!10!}dfrac{52!}{50!2!}}{dfrac{13!}{2!11!}dfrac{52!}{49!3!}}=dfrac{11}{50}$$
Now look at the denominator: $dfrac{binom{13}2}{binom{52}2}=dfrac{binom{13}3}{binom{52}3}+dfrac{binom{13}2binom{39}1}{binom{52}2binom{50}1}$. It is the probability for selecting three spades into all three positions, or two spades and one non-spade into a particular position (second). If you like, we can also express that as: $dfrac{binom{13}2}{binom{52}2}=dfrac{binom{13}3}{binom{52}3}+dfrac{binom{13}2binom{39}1/binom 31}{binom{52}3}$.
$$mathsf P(Amid B)=dfrac{dbinom{13}3}{dbinom{13}3+left.dbinom{13}2dbinom{39}1middle/dbinom 31right.}=dfrac{11}{50}$$
answered Nov 21 '18 at 0:26
Graham KempGraham Kemp
88.1k43579
$endgroup$
add a comment |
$begingroup$
However I attempted to solve the problem using combination rather than permutation:
The problem is, the card of interest is in a particular position.
Well, if we let $A$ be the event that the second card is a spade, and $B$ the event that the first and third are spades, then $Acap B$ is the event that the first three cards are all spades. $$mathsf P(Amid B)=dfrac{mathsf P(Acap B)}{mathsf P(B)}=dfrac{left.dbinom{13}3middle/dbinom {52}{3}right.}{left.dbinom{13}{2}middle/dbinom {52}2right.}=dfrac{dfrac{13!}{3!10!}dfrac{52!}{50!2!}}{dfrac{13!}{2!11!}dfrac{52!}{49!3!}}=dfrac{11}{50}$$
Now look at the denominator: $dfrac{binom{13}2}{binom{52}2}=dfrac{binom{13}3}{binom{52}3}+dfrac{binom{13}2binom{39}1}{binom{52}2binom{50}1}$. It is the probability for selecting three spades into all three positions, or two spades and one non-spade into a particular position (second). If you like, we can also express that as: $dfrac{binom{13}2}{binom{52}2}=dfrac{binom{13}3}{binom{52}3}+dfrac{binom{13}2binom{39}1/binom 31}{binom{52}3}$.
$$mathsf P(Amid B)=dfrac{dbinom{13}3}{dbinom{13}3+left.dbinom{13}2dbinom{39}1middle/dbinom 31right.}=dfrac{11}{50}$$
answered Nov 21 '18 at 0:26
Graham KempGraham Kemp
88.1k43579
$endgroup$
However I attempted to solve the problem using combination rather than permutation:
The problem is, the card of interest is in a particular position.
Well, if we let $A$ be the event that the second card is a spade, and $B$ the event that the first and third are spades, then $Acap B$ is the event that the first three cards are all spades. $$mathsf P(Amid B)=dfrac{mathsf P(Acap B)}{mathsf P(B)}=dfrac{left.dbinom{13}3middle/dbinom {52}{3}right.}{left.dbinom{13}{2}middle/dbinom {52}2right.}=dfrac{dfrac{13!}{3!10!}dfrac{52!}{50!2!}}{dfrac{13!}{2!11!}dfrac{52!}{49!3!}}=dfrac{11}{50}$$
Now look at the denominator: $dfrac{binom{13}2}{binom{52}2}=dfrac{binom{13}3}{binom{52}3}+dfrac{binom{13}2binom{39}1}{binom{52}2binom{50}1}$. It is the probability for selecting three spades into all three positions, or two spades and one non-spade into a particular position (second). If you like, we can also express that as: $dfrac{binom{13}2}{binom{52}2}=dfrac{binom{13}3}{binom{52}3}+dfrac{binom{13}2binom{39}1/binom 31}{binom{52}3}$.
$$mathsf P(Amid B)=dfrac{dbinom{13}3}{dbinom{13}3+left.dbinom{13}2dbinom{39}1middle/dbinom 31right.}=dfrac{11}{50}$$
answered Nov 21 '18 at 0:26
Graham KempGraham Kemp
88.1k43579
answered Nov 21 '18 at 0:26
Graham KempGraham Kemp
88.1k43579
answered Nov 21 '18 at 0:26
Graham KempGraham Kemp
88.1k43579
answered Nov 21 '18 at 0:26
Graham KempGraham Kemp
88.1k43579
88.1k43579
add a comment |
add a comment |
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3
$begingroup$
Since the cards are drawn without replacement. Knowing the 2nd and 3rd cards are spade, there is only 11 spade cards in the other total 50 cards. So the probability is $frac{11}{50}$. You don't need Bayes theorem for this.
$endgroup$
– Guangliang
Feb 19 '17 at 15:20
$begingroup$
@Guangliang I think LanceHAOH wants to solve this problem with Bayes' Rule.
$endgroup$
– gobucksmath
Feb 19 '17 at 15:29
2
$begingroup$
That term does not compute the probability you intend. Rather you want $frac {C(39,1)times C(13,2)}{C(52,1)times C(51,2)}$
$endgroup$
– lulu
Feb 19 '17 at 15:29
1
$begingroup$
No...that's not the problem. The problem is that by using combinations, and disregarding order, you are losing the special role of the first card. To be clear, the probability that the first card is not a spade, but the next two are, is $frac {39}{52}times frac {13}{51}times frac {12}{50}$.
$endgroup$
– lulu
Feb 19 '17 at 15:41
1
$begingroup$
The point is that introducing unordered combinations is a poor idea in this case (even though it can be salvaged). That's because the problem explicitly concerns the order of the first three choices.
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– lulu
Feb 19 '17 at 15:58