Probability using combinatorics: probability doesn't sum to 1
Clash Royale CLAN TAG#URR8PPP
$begingroup$
In a certain lottery, 10,000 tickets are sold and 10 prizes are awarded. What is the probability of not getting a prize if you buy 2 tickets.
The answer to this question is simple enough: $$ frac 9990 choose 210000 choose 2 $$
I understand this answer, since there are 2 ways of picking from the 9990 tickets that don't give you a ticket. And 2 ways of picking some ticket from the 10,000 tickets. But, I wanted to verify this by calculating the probability of getting a ticket.
I used a similar method:
$$ frac10 choose 210000 choose 2 $$
However, when I use a calculator to add them up, they don't sum to 1. I repeated this exercise for buying 10 tickets, in which case the answers for not getting a ticket and getting a ticket were respectively:
$$ frac 9990 choose 1010000 choose 10 $$
$$ frac10 choose 1010000 choose 10 $$
In this case as well, the probability don't sum to 1. I understand that I'm actually decreasing the numerator in the second case. But I can't understand intuitively why this is wrong.
While calculating the probability of not getting a ticket, we divide the number of ways we can pick a ticket that doesn't give you a prize by the number of ways you can pick any ticket. So, while calculuating the probability of getting a ticket, shouldn't we divide the number of ways we can pick a ticket that gives you a prize divided the total number of ways you can pick a ticket?
probability combinatorics
edited Feb 15 at 11:49
Max
564317
asked Feb 15 at 11:43
WorldGovWorldGov
324111
$endgroup$
add a comment |
$begingroup$
In a certain lottery, 10,000 tickets are sold and 10 prizes are awarded. What is the probability of not getting a prize if you buy 2 tickets.
The answer to this question is simple enough: $$ frac 9990 choose 210000 choose 2 $$
I understand this answer, since there are 2 ways of picking from the 9990 tickets that don't give you a ticket. And 2 ways of picking some ticket from the 10,000 tickets. But, I wanted to verify this by calculating the probability of getting a ticket.
I used a similar method:
$$ frac10 choose 210000 choose 2 $$
However, when I use a calculator to add them up, they don't sum to 1. I repeated this exercise for buying 10 tickets, in which case the answers for not getting a ticket and getting a ticket were respectively:
$$ frac 9990 choose 1010000 choose 10 $$
$$ frac10 choose 1010000 choose 10 $$
In this case as well, the probability don't sum to 1. I understand that I'm actually decreasing the numerator in the second case. But I can't understand intuitively why this is wrong.
While calculating the probability of not getting a ticket, we divide the number of ways we can pick a ticket that doesn't give you a prize by the number of ways you can pick any ticket. So, while calculuating the probability of getting a ticket, shouldn't we divide the number of ways we can pick a ticket that gives you a prize divided the total number of ways you can pick a ticket?
probability combinatorics
edited Feb 15 at 11:49
Max
564317
asked Feb 15 at 11:43
WorldGovWorldGov
324111
$endgroup$
2
$begingroup$
These are good questions to ask. What if you buy $20$ tickets; what are the outcomes and their probabilities?
$endgroup$
– David K
Feb 15 at 12:08
add a comment |
$begingroup$
In a certain lottery, 10,000 tickets are sold and 10 prizes are awarded. What is the probability of not getting a prize if you buy 2 tickets.
The answer to this question is simple enough: $$ frac 9990 choose 210000 choose 2 $$
I understand this answer, since there are 2 ways of picking from the 9990 tickets that don't give you a ticket. And 2 ways of picking some ticket from the 10,000 tickets. But, I wanted to verify this by calculating the probability of getting a ticket.
I used a similar method:
$$ frac10 choose 210000 choose 2 $$
However, when I use a calculator to add them up, they don't sum to 1. I repeated this exercise for buying 10 tickets, in which case the answers for not getting a ticket and getting a ticket were respectively:
$$ frac 9990 choose 1010000 choose 10 $$
$$ frac10 choose 1010000 choose 10 $$
In this case as well, the probability don't sum to 1. I understand that I'm actually decreasing the numerator in the second case. But I can't understand intuitively why this is wrong.
While calculating the probability of not getting a ticket, we divide the number of ways we can pick a ticket that doesn't give you a prize by the number of ways you can pick any ticket. So, while calculuating the probability of getting a ticket, shouldn't we divide the number of ways we can pick a ticket that gives you a prize divided the total number of ways you can pick a ticket?
probability combinatorics
edited Feb 15 at 11:49
Max
564317
asked Feb 15 at 11:43
WorldGovWorldGov
324111
$endgroup$
In a certain lottery, 10,000 tickets are sold and 10 prizes are awarded. What is the probability of not getting a prize if you buy 2 tickets.
The answer to this question is simple enough: $$ frac 9990 choose 210000 choose 2 $$
I understand this answer, since there are 2 ways of picking from the 9990 tickets that don't give you a ticket. And 2 ways of picking some ticket from the 10,000 tickets. But, I wanted to verify this by calculating the probability of getting a ticket.
I used a similar method:
$$ frac10 choose 210000 choose 2 $$
However, when I use a calculator to add them up, they don't sum to 1. I repeated this exercise for buying 10 tickets, in which case the answers for not getting a ticket and getting a ticket were respectively:
$$ frac 9990 choose 1010000 choose 10 $$
$$ frac10 choose 1010000 choose 10 $$
In this case as well, the probability don't sum to 1. I understand that I'm actually decreasing the numerator in the second case. But I can't understand intuitively why this is wrong.
While calculating the probability of not getting a ticket, we divide the number of ways we can pick a ticket that doesn't give you a prize by the number of ways you can pick any ticket. So, while calculuating the probability of getting a ticket, shouldn't we divide the number of ways we can pick a ticket that gives you a prize divided the total number of ways you can pick a ticket?
probability combinatorics
probability combinatorics
edited Feb 15 at 11:49
Max
564317
asked Feb 15 at 11:43
WorldGovWorldGov
324111
edited Feb 15 at 11:49
Max
564317
asked Feb 15 at 11:43
WorldGovWorldGov
324111
edited Feb 15 at 11:49
Max
564317
edited Feb 15 at 11:49
Max
564317
edited Feb 15 at 11:49
Max
564317
564317
asked Feb 15 at 11:43
WorldGovWorldGov
324111
asked Feb 15 at 11:43
WorldGovWorldGov
324111
asked Feb 15 at 11:43
WorldGovWorldGov
324111
324111
2
$begingroup$
These are good questions to ask. What if you buy $20$ tickets; what are the outcomes and their probabilities?
$endgroup$
– David K
Feb 15 at 12:08
add a comment |
2
$begingroup$
These are good questions to ask. What if you buy $20$ tickets; what are the outcomes and their probabilities?
$endgroup$
– David K
Feb 15 at 12:08
2
2
$begingroup$
These are good questions to ask. What if you buy $20$ tickets; what are the outcomes and their probabilities?
$endgroup$
– David K
Feb 15 at 12:08
$begingroup$
These are good questions to ask. What if you buy $20$ tickets; what are the outcomes and their probabilities?
$endgroup$
– David K
Feb 15 at 12:08
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Note that $fracbinom102binom100002$ is the probability of getting two prizes! Hence, you are missing the case where you get just one price. If you add that, you do indeed get $1$:
$$
fracbinom99902binom100002
+fracbinom102binom100002
+fracbinom99901cdotbinom101binom100002
= 1.
$$
answered Feb 15 at 11:56
ChristophChristoph
12.5k1642
$endgroup$
$begingroup$
Shouldn't you also show the correct probability for winning at least one prize?
$endgroup$
– Wolfgang Kais
Feb 15 at 13:13
1
$begingroup$
Winning at least one prize is the same as winning exactly one or exactly two prices, so the last two summands.
$endgroup$
– Christoph
Feb 15 at 13:51
$begingroup$
Wow, that was really useful! I also used this logic to answer the question posed by @David in the comments, and they summed to 1 as well. Thank you.
$endgroup$
– WorldGov
Feb 15 at 14:00
add a comment |
$begingroup$
You can also solve by Binomial as the numbers are large. Hypergeometric distribution used by @Christoph gives the exact probabilities:
Probability that you will get a prize $= frac11000$
probability that you will not get a prize $= frac9991000$
If you buy two tickets, the probability that you will not get a prize is
$approx 2choose0(frac11000)^0(frac9991000)^2$ you will get a sum of 1 in this too.
answered Feb 15 at 12:02
Satish RamanathanSatish Ramanathan
10k31323
$endgroup$
$begingroup$
You are using the symbol "$=$" for an approximation. Please stress the fact that this answer does not give the exact probabilities!
$endgroup$
– Christoph
Feb 15 at 12:08
$begingroup$
@Christoph: Done!!
$endgroup$
– Satish Ramanathan
Feb 15 at 12:10
$begingroup$
Why a downvote, I am introducing another way to look at the problem!!
$endgroup$
– Satish Ramanathan
Feb 15 at 12:27
$begingroup$
I downvoted because this is not an answer to the question "Why do my probabilities do not sum up to $1$?"
$endgroup$
– Christoph
Feb 15 at 12:35
$begingroup$
Although it does not answer the question that the OP asked, it is still a valid answer to the problem posed by OP. I am going to leave it even if it receives further downvotes.
$endgroup$
– Satish Ramanathan
Feb 15 at 12:38
|
show 1 more comment
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that $fracbinom102binom100002$ is the probability of getting two prizes! Hence, you are missing the case where you get just one price. If you add that, you do indeed get $1$:
$$
fracbinom99902binom100002
+fracbinom102binom100002
+fracbinom99901cdotbinom101binom100002
= 1.
$$
answered Feb 15 at 11:56
ChristophChristoph
12.5k1642
$endgroup$
$begingroup$
Shouldn't you also show the correct probability for winning at least one prize?
$endgroup$
– Wolfgang Kais
Feb 15 at 13:13
1
$begingroup$
Winning at least one prize is the same as winning exactly one or exactly two prices, so the last two summands.
$endgroup$
– Christoph
Feb 15 at 13:51
$begingroup$
Wow, that was really useful! I also used this logic to answer the question posed by @David in the comments, and they summed to 1 as well. Thank you.
$endgroup$
– WorldGov
Feb 15 at 14:00
add a comment |
$begingroup$
Note that $fracbinom102binom100002$ is the probability of getting two prizes! Hence, you are missing the case where you get just one price. If you add that, you do indeed get $1$:
$$
fracbinom99902binom100002
+fracbinom102binom100002
+fracbinom99901cdotbinom101binom100002
= 1.
$$
answered Feb 15 at 11:56
ChristophChristoph
12.5k1642
$endgroup$
$begingroup$
Shouldn't you also show the correct probability for winning at least one prize?
$endgroup$
– Wolfgang Kais
Feb 15 at 13:13
1
$begingroup$
Winning at least one prize is the same as winning exactly one or exactly two prices, so the last two summands.
$endgroup$
– Christoph
Feb 15 at 13:51
$begingroup$
Wow, that was really useful! I also used this logic to answer the question posed by @David in the comments, and they summed to 1 as well. Thank you.
$endgroup$
– WorldGov
Feb 15 at 14:00
add a comment |
$begingroup$
Note that $fracbinom102binom100002$ is the probability of getting two prizes! Hence, you are missing the case where you get just one price. If you add that, you do indeed get $1$:
$$
fracbinom99902binom100002
+fracbinom102binom100002
+fracbinom99901cdotbinom101binom100002
= 1.
$$
answered Feb 15 at 11:56
ChristophChristoph
12.5k1642
$endgroup$
Note that $fracbinom102binom100002$ is the probability of getting two prizes! Hence, you are missing the case where you get just one price. If you add that, you do indeed get $1$:
$$
fracbinom99902binom100002
+fracbinom102binom100002
+fracbinom99901cdotbinom101binom100002
= 1.
$$
answered Feb 15 at 11:56
ChristophChristoph
12.5k1642
answered Feb 15 at 11:56
ChristophChristoph
12.5k1642
answered Feb 15 at 11:56
ChristophChristoph
12.5k1642
answered Feb 15 at 11:56
ChristophChristoph
12.5k1642
12.5k1642
$begingroup$
Shouldn't you also show the correct probability for winning at least one prize?
$endgroup$
– Wolfgang Kais
Feb 15 at 13:13
1
$begingroup$
Winning at least one prize is the same as winning exactly one or exactly two prices, so the last two summands.
$endgroup$
– Christoph
Feb 15 at 13:51
$begingroup$
Wow, that was really useful! I also used this logic to answer the question posed by @David in the comments, and they summed to 1 as well. Thank you.
$endgroup$
– WorldGov
Feb 15 at 14:00
add a comment |
$begingroup$
Shouldn't you also show the correct probability for winning at least one prize?
$endgroup$
– Wolfgang Kais
Feb 15 at 13:13
1
$begingroup$
Winning at least one prize is the same as winning exactly one or exactly two prices, so the last two summands.
$endgroup$
– Christoph
Feb 15 at 13:51
$begingroup$
Wow, that was really useful! I also used this logic to answer the question posed by @David in the comments, and they summed to 1 as well. Thank you.
$endgroup$
– WorldGov
Feb 15 at 14:00
$begingroup$
Shouldn't you also show the correct probability for winning at least one prize?
$endgroup$
– Wolfgang Kais
Feb 15 at 13:13
$begingroup$
Shouldn't you also show the correct probability for winning at least one prize?
$endgroup$
– Wolfgang Kais
Feb 15 at 13:13
1
1
$begingroup$
Winning at least one prize is the same as winning exactly one or exactly two prices, so the last two summands.
$endgroup$
– Christoph
Feb 15 at 13:51
$begingroup$
Winning at least one prize is the same as winning exactly one or exactly two prices, so the last two summands.
$endgroup$
– Christoph
Feb 15 at 13:51
$begingroup$
Wow, that was really useful! I also used this logic to answer the question posed by @David in the comments, and they summed to 1 as well. Thank you.
$endgroup$
– WorldGov
Feb 15 at 14:00
$begingroup$
Wow, that was really useful! I also used this logic to answer the question posed by @David in the comments, and they summed to 1 as well. Thank you.
$endgroup$
– WorldGov
Feb 15 at 14:00
add a comment |
$begingroup$
You can also solve by Binomial as the numbers are large. Hypergeometric distribution used by @Christoph gives the exact probabilities:
Probability that you will get a prize $= frac11000$
probability that you will not get a prize $= frac9991000$
If you buy two tickets, the probability that you will not get a prize is
$approx 2choose0(frac11000)^0(frac9991000)^2$ you will get a sum of 1 in this too.
answered Feb 15 at 12:02
Satish RamanathanSatish Ramanathan
10k31323
$endgroup$
$begingroup$
You are using the symbol "$=$" for an approximation. Please stress the fact that this answer does not give the exact probabilities!
$endgroup$
– Christoph
Feb 15 at 12:08
$begingroup$
@Christoph: Done!!
$endgroup$
– Satish Ramanathan
Feb 15 at 12:10
$begingroup$
Why a downvote, I am introducing another way to look at the problem!!
$endgroup$
– Satish Ramanathan
Feb 15 at 12:27
$begingroup$
I downvoted because this is not an answer to the question "Why do my probabilities do not sum up to $1$?"
$endgroup$
– Christoph
Feb 15 at 12:35
$begingroup$
Although it does not answer the question that the OP asked, it is still a valid answer to the problem posed by OP. I am going to leave it even if it receives further downvotes.
$endgroup$
– Satish Ramanathan
Feb 15 at 12:38
|
show 1 more comment
$begingroup$
You can also solve by Binomial as the numbers are large. Hypergeometric distribution used by @Christoph gives the exact probabilities:
Probability that you will get a prize $= frac11000$
probability that you will not get a prize $= frac9991000$
If you buy two tickets, the probability that you will not get a prize is
$approx 2choose0(frac11000)^0(frac9991000)^2$ you will get a sum of 1 in this too.
answered Feb 15 at 12:02
Satish RamanathanSatish Ramanathan
10k31323
$endgroup$
$begingroup$
You are using the symbol "$=$" for an approximation. Please stress the fact that this answer does not give the exact probabilities!
$endgroup$
– Christoph
Feb 15 at 12:08
$begingroup$
@Christoph: Done!!
$endgroup$
– Satish Ramanathan
Feb 15 at 12:10
$begingroup$
Why a downvote, I am introducing another way to look at the problem!!
$endgroup$
– Satish Ramanathan
Feb 15 at 12:27
$begingroup$
I downvoted because this is not an answer to the question "Why do my probabilities do not sum up to $1$?"
$endgroup$
– Christoph
Feb 15 at 12:35
$begingroup$
Although it does not answer the question that the OP asked, it is still a valid answer to the problem posed by OP. I am going to leave it even if it receives further downvotes.
$endgroup$
– Satish Ramanathan
Feb 15 at 12:38
|
show 1 more comment
$begingroup$
You can also solve by Binomial as the numbers are large. Hypergeometric distribution used by @Christoph gives the exact probabilities:
Probability that you will get a prize $= frac11000$
probability that you will not get a prize $= frac9991000$
If you buy two tickets, the probability that you will not get a prize is
$approx 2choose0(frac11000)^0(frac9991000)^2$ you will get a sum of 1 in this too.
answered Feb 15 at 12:02
Satish RamanathanSatish Ramanathan
10k31323
$endgroup$
You can also solve by Binomial as the numbers are large. Hypergeometric distribution used by @Christoph gives the exact probabilities:
Probability that you will get a prize $= frac11000$
probability that you will not get a prize $= frac9991000$
If you buy two tickets, the probability that you will not get a prize is
$approx 2choose0(frac11000)^0(frac9991000)^2$ you will get a sum of 1 in this too.
answered Feb 15 at 12:02
Satish RamanathanSatish Ramanathan
10k31323
edited Feb 15 at 12:09
answered Feb 15 at 12:02
Satish RamanathanSatish Ramanathan
10k31323
answered Feb 15 at 12:02
Satish RamanathanSatish Ramanathan
10k31323
answered Feb 15 at 12:02
Satish RamanathanSatish Ramanathan
10k31323
10k31323
$begingroup$
You are using the symbol "$=$" for an approximation. Please stress the fact that this answer does not give the exact probabilities!
$endgroup$
– Christoph
Feb 15 at 12:08
$begingroup$
@Christoph: Done!!
$endgroup$
– Satish Ramanathan
Feb 15 at 12:10
$begingroup$
Why a downvote, I am introducing another way to look at the problem!!
$endgroup$
– Satish Ramanathan
Feb 15 at 12:27
$begingroup$
I downvoted because this is not an answer to the question "Why do my probabilities do not sum up to $1$?"
$endgroup$
– Christoph
Feb 15 at 12:35
$begingroup$
Although it does not answer the question that the OP asked, it is still a valid answer to the problem posed by OP. I am going to leave it even if it receives further downvotes.
$endgroup$
– Satish Ramanathan
Feb 15 at 12:38
|
show 1 more comment
$begingroup$
You are using the symbol "$=$" for an approximation. Please stress the fact that this answer does not give the exact probabilities!
$endgroup$
– Christoph
Feb 15 at 12:08
$begingroup$
@Christoph: Done!!
$endgroup$
– Satish Ramanathan
Feb 15 at 12:10
$begingroup$
Why a downvote, I am introducing another way to look at the problem!!
$endgroup$
– Satish Ramanathan
Feb 15 at 12:27
$begingroup$
I downvoted because this is not an answer to the question "Why do my probabilities do not sum up to $1$?"
$endgroup$
– Christoph
Feb 15 at 12:35
$begingroup$
Although it does not answer the question that the OP asked, it is still a valid answer to the problem posed by OP. I am going to leave it even if it receives further downvotes.
$endgroup$
– Satish Ramanathan
Feb 15 at 12:38
$begingroup$
You are using the symbol "$=$" for an approximation. Please stress the fact that this answer does not give the exact probabilities!
$endgroup$
– Christoph
Feb 15 at 12:08
$begingroup$
You are using the symbol "$=$" for an approximation. Please stress the fact that this answer does not give the exact probabilities!
$endgroup$
– Christoph
Feb 15 at 12:08
$begingroup$
@Christoph: Done!!
$endgroup$
– Satish Ramanathan
Feb 15 at 12:10
$begingroup$
@Christoph: Done!!
$endgroup$
– Satish Ramanathan
Feb 15 at 12:10
$begingroup$
Why a downvote, I am introducing another way to look at the problem!!
$endgroup$
– Satish Ramanathan
Feb 15 at 12:27
$begingroup$
Why a downvote, I am introducing another way to look at the problem!!
$endgroup$
– Satish Ramanathan
Feb 15 at 12:27
$begingroup$
I downvoted because this is not an answer to the question "Why do my probabilities do not sum up to $1$?"
$endgroup$
– Christoph
Feb 15 at 12:35
$begingroup$
I downvoted because this is not an answer to the question "Why do my probabilities do not sum up to $1$?"
$endgroup$
– Christoph
Feb 15 at 12:35
$begingroup$
Although it does not answer the question that the OP asked, it is still a valid answer to the problem posed by OP. I am going to leave it even if it receives further downvotes.
$endgroup$
– Satish Ramanathan
Feb 15 at 12:38
$begingroup$
Although it does not answer the question that the OP asked, it is still a valid answer to the problem posed by OP. I am going to leave it even if it receives further downvotes.
$endgroup$
– Satish Ramanathan
Feb 15 at 12:38
|
show 1 more comment
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2
$begingroup$
These are good questions to ask. What if you buy $20$ tickets; what are the outcomes and their probabilities?
$endgroup$
– David K
Feb 15 at 12:08