Counting Size of Event Spaces, card drawing
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Attached is a photo of the problem I'm having a hard time with.
We are trying to calculate the likelihood of drawing an ace given 10 or more draws with no aces drawn.
We are given two methods of evaluation, I'm trying to make sense of the counting method, in particular:
I'm trying to justify the 41!
It is clear to me why we choose the falling factorial 48×...×40 for the first 9 cards, but I've been trying to reason through the 41!, and I don't feel solid in my justifications.
I was thinking we have 10 cards chosen, so (52-10)! Permutations for the remaining cards but that gives us 42! And we only chose 9 cards, not 10.
What am I missing in terms of counting the event space?
probability probability-theory permutations conditional-probability elementary-probability
asked Nov 25 at 20:25
Matt1991
585
add a comment |
up vote
1
down vote
favorite
Attached is a photo of the problem I'm having a hard time with.
We are trying to calculate the likelihood of drawing an ace given 10 or more draws with no aces drawn.
We are given two methods of evaluation, I'm trying to make sense of the counting method, in particular:
I'm trying to justify the 41!
It is clear to me why we choose the falling factorial 48×...×40 for the first 9 cards, but I've been trying to reason through the 41!, and I don't feel solid in my justifications.
I was thinking we have 10 cards chosen, so (52-10)! Permutations for the remaining cards but that gives us 42! And we only chose 9 cards, not 10.
What am I missing in terms of counting the event space?
probability probability-theory permutations conditional-probability elementary-probability
asked Nov 25 at 20:25
Matt1991
585
Well, this serves as a reminder to trust my own judgement rather than accept all I read as truth. Turns out the other made a typo or a mistake. It should be 43! 39 remaining cards that are non aces and 4 that are aces. I went nearly insane trying to justify that 41!. Always check others work I suppose to see if the numbers add up.
– Matt1991
Nov 25 at 20:48
I just mentioned You already noticed by yourself. Good job!
– Peter Melech
Nov 25 at 21:25
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Attached is a photo of the problem I'm having a hard time with.
We are trying to calculate the likelihood of drawing an ace given 10 or more draws with no aces drawn.
We are given two methods of evaluation, I'm trying to make sense of the counting method, in particular:
I'm trying to justify the 41!
It is clear to me why we choose the falling factorial 48×...×40 for the first 9 cards, but I've been trying to reason through the 41!, and I don't feel solid in my justifications.
I was thinking we have 10 cards chosen, so (52-10)! Permutations for the remaining cards but that gives us 42! And we only chose 9 cards, not 10.
What am I missing in terms of counting the event space?
probability probability-theory permutations conditional-probability elementary-probability
asked Nov 25 at 20:25
Matt1991
585
Attached is a photo of the problem I'm having a hard time with.
We are trying to calculate the likelihood of drawing an ace given 10 or more draws with no aces drawn.
We are given two methods of evaluation, I'm trying to make sense of the counting method, in particular:
I'm trying to justify the 41!
It is clear to me why we choose the falling factorial 48×...×40 for the first 9 cards, but I've been trying to reason through the 41!, and I don't feel solid in my justifications.
I was thinking we have 10 cards chosen, so (52-10)! Permutations for the remaining cards but that gives us 42! And we only chose 9 cards, not 10.
What am I missing in terms of counting the event space?
probability probability-theory permutations conditional-probability elementary-probability
probability probability-theory permutations conditional-probability elementary-probability
asked Nov 25 at 20:25
Matt1991
585
asked Nov 25 at 20:25
Matt1991
585
asked Nov 25 at 20:25
Matt1991
585
asked Nov 25 at 20:25
Matt1991
585
asked Nov 25 at 20:25
Matt1991
585
585
Well, this serves as a reminder to trust my own judgement rather than accept all I read as truth. Turns out the other made a typo or a mistake. It should be 43! 39 remaining cards that are non aces and 4 that are aces. I went nearly insane trying to justify that 41!. Always check others work I suppose to see if the numbers add up.
– Matt1991
Nov 25 at 20:48
I just mentioned You already noticed by yourself. Good job!
– Peter Melech
Nov 25 at 21:25
add a comment |
Well, this serves as a reminder to trust my own judgement rather than accept all I read as truth. Turns out the other made a typo or a mistake. It should be 43! 39 remaining cards that are non aces and 4 that are aces. I went nearly insane trying to justify that 41!. Always check others work I suppose to see if the numbers add up.
– Matt1991
Nov 25 at 20:48
I just mentioned You already noticed by yourself. Good job!
– Peter Melech
Nov 25 at 21:25
Well, this serves as a reminder to trust my own judgement rather than accept all I read as truth. Turns out the other made a typo or a mistake. It should be 43! 39 remaining cards that are non aces and 4 that are aces. I went nearly insane trying to justify that 41!. Always check others work I suppose to see if the numbers add up.
– Matt1991
Nov 25 at 20:48
Well, this serves as a reminder to trust my own judgement rather than accept all I read as truth. Turns out the other made a typo or a mistake. It should be 43! 39 remaining cards that are non aces and 4 that are aces. I went nearly insane trying to justify that 41!. Always check others work I suppose to see if the numbers add up.
– Matt1991
Nov 25 at 20:48
I just mentioned You already noticed by yourself. Good job!
– Peter Melech
Nov 25 at 21:25
I just mentioned You already noticed by yourself. Good job!
– Peter Melech
Nov 25 at 21:25
add a comment |
1 Answer
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1
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You found a mistake in the book and Your reasoning is correct . Here however "10 or more cards before the first ace appears" is obviously interpreted as "the first $9$ cards are no aces", quite incorrectly if You take "before" literally. In this case after drawing the first $9$ cards, that are no aces, there are $43$ cards left and ideed
$$P(A)=frac{48cdot74cdot.46cdotcdotcdotcdot40cdot 43!}{52!}approx 0.4559$$
as You can easily check.( Actually even the equation $P(A)=48cdots40cdot41!/52!=0.4559$ is wrong computationally, this expression gives a much smaller value.)
answered Nov 25 at 21:15
Peter Melech
2,519813
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
You found a mistake in the book and Your reasoning is correct . Here however "10 or more cards before the first ace appears" is obviously interpreted as "the first $9$ cards are no aces", quite incorrectly if You take "before" literally. In this case after drawing the first $9$ cards, that are no aces, there are $43$ cards left and ideed
$$P(A)=frac{48cdot74cdot.46cdotcdotcdotcdot40cdot 43!}{52!}approx 0.4559$$
as You can easily check.( Actually even the equation $P(A)=48cdots40cdot41!/52!=0.4559$ is wrong computationally, this expression gives a much smaller value.)
answered Nov 25 at 21:15
Peter Melech
2,519813
add a comment |
up vote
1
down vote
You found a mistake in the book and Your reasoning is correct . Here however "10 or more cards before the first ace appears" is obviously interpreted as "the first $9$ cards are no aces", quite incorrectly if You take "before" literally. In this case after drawing the first $9$ cards, that are no aces, there are $43$ cards left and ideed
$$P(A)=frac{48cdot74cdot.46cdotcdotcdotcdot40cdot 43!}{52!}approx 0.4559$$
as You can easily check.( Actually even the equation $P(A)=48cdots40cdot41!/52!=0.4559$ is wrong computationally, this expression gives a much smaller value.)
answered Nov 25 at 21:15
Peter Melech
2,519813
add a comment |
up vote
1
down vote
up vote
1
down vote
You found a mistake in the book and Your reasoning is correct . Here however "10 or more cards before the first ace appears" is obviously interpreted as "the first $9$ cards are no aces", quite incorrectly if You take "before" literally. In this case after drawing the first $9$ cards, that are no aces, there are $43$ cards left and ideed
$$P(A)=frac{48cdot74cdot.46cdotcdotcdotcdot40cdot 43!}{52!}approx 0.4559$$
as You can easily check.( Actually even the equation $P(A)=48cdots40cdot41!/52!=0.4559$ is wrong computationally, this expression gives a much smaller value.)
answered Nov 25 at 21:15
Peter Melech
2,519813
You found a mistake in the book and Your reasoning is correct . Here however "10 or more cards before the first ace appears" is obviously interpreted as "the first $9$ cards are no aces", quite incorrectly if You take "before" literally. In this case after drawing the first $9$ cards, that are no aces, there are $43$ cards left and ideed
$$P(A)=frac{48cdot74cdot.46cdotcdotcdotcdot40cdot 43!}{52!}approx 0.4559$$
as You can easily check.( Actually even the equation $P(A)=48cdots40cdot41!/52!=0.4559$ is wrong computationally, this expression gives a much smaller value.)
answered Nov 25 at 21:15
Peter Melech
2,519813
edited Nov 25 at 21:21
answered Nov 25 at 21:15
Peter Melech
2,519813
answered Nov 25 at 21:15
Peter Melech
2,519813
answered Nov 25 at 21:15
Peter Melech
2,519813
2,519813
add a comment |
add a comment |
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Well, this serves as a reminder to trust my own judgement rather than accept all I read as truth. Turns out the other made a typo or a mistake. It should be 43! 39 remaining cards that are non aces and 4 that are aces. I went nearly insane trying to justify that 41!. Always check others work I suppose to see if the numbers add up.
– Matt1991
Nov 25 at 20:48
I just mentioned You already noticed by yourself. Good job!
– Peter Melech
Nov 25 at 21:25