Keno odds of picking 11+ numbers if 20 total are selected out of 80
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In the game of keno, there are a set number of balls in the spinner, in this case, 80. They are numbered 1-80. The goal is to guess the numbers that will be selected at random. In this particular keno variant, you are allowed to pick 20 numbers. What are the odds that you will get more than 10 correct (order does not matter)? (11-20)
probability
asked Jul 27 '17 at 22:50
JD3JD3
1014
$endgroup$
|
show 3 more comments
$begingroup$
In the game of keno, there are a set number of balls in the spinner, in this case, 80. They are numbered 1-80. The goal is to guess the numbers that will be selected at random. In this particular keno variant, you are allowed to pick 20 numbers. What are the odds that you will get more than 10 correct (order does not matter)? (11-20)
probability
asked Jul 27 '17 at 22:50
JD3JD3
1014
$endgroup$
$begingroup$
Does the order matter?
$endgroup$
– Frpzzd
Jul 27 '17 at 22:51
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The order doesn't matter.
$endgroup$
– JD3
Jul 27 '17 at 22:52
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How many balls are taken out of the spinner? I need that info to find the answer.
$endgroup$
– Frpzzd
Jul 27 '17 at 22:52
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20 total are selected out of the 80 balls in the spinner.
$endgroup$
– JD3
Jul 27 '17 at 22:53
$begingroup$
So you pick 20, and 20 are drawn at random?
$endgroup$
– Frpzzd
Jul 27 '17 at 22:54
|
show 3 more comments
$begingroup$
In the game of keno, there are a set number of balls in the spinner, in this case, 80. They are numbered 1-80. The goal is to guess the numbers that will be selected at random. In this particular keno variant, you are allowed to pick 20 numbers. What are the odds that you will get more than 10 correct (order does not matter)? (11-20)
probability
asked Jul 27 '17 at 22:50
JD3JD3
1014
$endgroup$
In the game of keno, there are a set number of balls in the spinner, in this case, 80. They are numbered 1-80. The goal is to guess the numbers that will be selected at random. In this particular keno variant, you are allowed to pick 20 numbers. What are the odds that you will get more than 10 correct (order does not matter)? (11-20)
probability
probability
asked Jul 27 '17 at 22:50
JD3JD3
1014
asked Jul 27 '17 at 22:50
JD3JD3
1014
edited Jul 27 '17 at 22:52
JD3
asked Jul 27 '17 at 22:50
JD3JD3
1014
asked Jul 27 '17 at 22:50
JD3JD3
1014
asked Jul 27 '17 at 22:50
JD3JD3
1014
1014
$begingroup$
Does the order matter?
$endgroup$
– Frpzzd
Jul 27 '17 at 22:51
$begingroup$
The order doesn't matter.
$endgroup$
– JD3
Jul 27 '17 at 22:52
$begingroup$
How many balls are taken out of the spinner? I need that info to find the answer.
$endgroup$
– Frpzzd
Jul 27 '17 at 22:52
$begingroup$
20 total are selected out of the 80 balls in the spinner.
$endgroup$
– JD3
Jul 27 '17 at 22:53
$begingroup$
So you pick 20, and 20 are drawn at random?
$endgroup$
– Frpzzd
Jul 27 '17 at 22:54
|
show 3 more comments
$begingroup$
Does the order matter?
$endgroup$
– Frpzzd
Jul 27 '17 at 22:51
$begingroup$
The order doesn't matter.
$endgroup$
– JD3
Jul 27 '17 at 22:52
$begingroup$
How many balls are taken out of the spinner? I need that info to find the answer.
$endgroup$
– Frpzzd
Jul 27 '17 at 22:52
$begingroup$
20 total are selected out of the 80 balls in the spinner.
$endgroup$
– JD3
Jul 27 '17 at 22:53
$begingroup$
So you pick 20, and 20 are drawn at random?
$endgroup$
– Frpzzd
Jul 27 '17 at 22:54
$begingroup$
Does the order matter?
$endgroup$
– Frpzzd
Jul 27 '17 at 22:51
$begingroup$
Does the order matter?
$endgroup$
– Frpzzd
Jul 27 '17 at 22:51
$begingroup$
The order doesn't matter.
$endgroup$
– JD3
Jul 27 '17 at 22:52
$begingroup$
The order doesn't matter.
$endgroup$
– JD3
Jul 27 '17 at 22:52
$begingroup$
How many balls are taken out of the spinner? I need that info to find the answer.
$endgroup$
– Frpzzd
Jul 27 '17 at 22:52
$begingroup$
How many balls are taken out of the spinner? I need that info to find the answer.
$endgroup$
– Frpzzd
Jul 27 '17 at 22:52
$begingroup$
20 total are selected out of the 80 balls in the spinner.
$endgroup$
– JD3
Jul 27 '17 at 22:53
$begingroup$
20 total are selected out of the 80 balls in the spinner.
$endgroup$
– JD3
Jul 27 '17 at 22:53
$begingroup$
So you pick 20, and 20 are drawn at random?
$endgroup$
– Frpzzd
Jul 27 '17 at 22:54
$begingroup$
So you pick 20, and 20 are drawn at random?
$endgroup$
– Frpzzd
Jul 27 '17 at 22:54
|
show 3 more comments
2 Answers
2
active
oldest
votes
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The chance of getting exactly eleven is $frac {{20 choose 11}{60 choose 9}}{80 choose 20}$ where the $20 choose 11$ is the number of ways to pick the winning balls out of your $20$, the $60 choose 9$ chooses the balls drawn that you did not pick, and the $80 choose 20$ chooses the winning balls from all balls. You should be able to do higher numbers by analogy, then add them all up to get the total probability. Alpha finds a total probability of about $0.0008$, almost all of which comes from hitting $11$ or $12$.
answered Jul 27 '17 at 23:03
Ross MillikanRoss Millikan
296k23198371
$endgroup$
add a comment |
$begingroup$
The total number of ways for you to choose $20$ of the $80$ numbers and for the operator to choose $20$ is
$$binom{80}{20}^2$$
Now suppose that you choose $20$ numbers. To find the probability that at least $11$ of these match, we must find the number of ways for the operator to choose $20$ numbers such that at least $11$ of them are the same as some $11$ of yours. There are
$$binom{20}{11}$$
sets of $11$ of your $20$ picked numbers for the operator's numbers to match, and then there will be $70$ of the $80$ numbers remaining for the operator to choose, so the number of ways for the operator to pick the other $9$ of his numbers (that may or may not match any of yours) is
$$binom{70}{9}$$
and so the number of situations in which at least $10$ of your numbers match is
$$binom{20}{11}binom{70}{9}$$
and so the probability is
$$frac{binom{20}{11}binom{70}{9}}{binom{80}{20}^2}$$
answered Jul 27 '17 at 23:02
FrpzzdFrpzzd
23k841109
$endgroup$
$begingroup$
You shouldn't be squaring the denominator. We don't care how many ways the player can pick his numbers. It is not correct to use $70 choose 10$ and claim this is the chance of at least $10$ because if the player hits $11$ numbers you will count it $11$ times, once for each group of $10$ being the first ones.
$endgroup$
– Ross Millikan
Jul 27 '17 at 23:08
$begingroup$
That's okay, because I'm counting the same number of duplicates in the numerator.
$endgroup$
– Frpzzd
Jul 27 '17 at 23:09
$begingroup$
I pointed out two distinct errors. One is squaring the denominator. The second is overcounting the winners. They don't cancel out.
$endgroup$
– Ross Millikan
Jul 27 '17 at 23:14
$begingroup$
Hmm... I disagree. Our answers are only different because I counted the probability for getting any number of balls from $11$ to $20$ correct, whereas yours counts only the ways to get $11$ of them.
$endgroup$
– Frpzzd
Jul 27 '17 at 23:17
$begingroup$
No, you are overcounting. As a simple example, choose three of ABCDEF including at least two of ABC. You would say ${3 choose 2}{4 choose 1}$ for $12$. In fact it should be ${3 choose 2}{3 choose 1}$ to get exactly two and ${3 choose 3}{3 choose 0}$ to get all three of ABC for a total of $10$. You have counted $AB+C, AC+B,$ and $BC+A$, which gives the extra two. The ABC are like the numbers the player chooses in Keno. You shouldn't square the denominator because the player has chosen and there is only one player choice active, so the sample space is $80 choose 20$
$endgroup$
– Ross Millikan
Jul 28 '17 at 0:25
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The chance of getting exactly eleven is $frac {{20 choose 11}{60 choose 9}}{80 choose 20}$ where the $20 choose 11$ is the number of ways to pick the winning balls out of your $20$, the $60 choose 9$ chooses the balls drawn that you did not pick, and the $80 choose 20$ chooses the winning balls from all balls. You should be able to do higher numbers by analogy, then add them all up to get the total probability. Alpha finds a total probability of about $0.0008$, almost all of which comes from hitting $11$ or $12$.
answered Jul 27 '17 at 23:03
Ross MillikanRoss Millikan
296k23198371
$endgroup$
add a comment |
$begingroup$
The chance of getting exactly eleven is $frac {{20 choose 11}{60 choose 9}}{80 choose 20}$ where the $20 choose 11$ is the number of ways to pick the winning balls out of your $20$, the $60 choose 9$ chooses the balls drawn that you did not pick, and the $80 choose 20$ chooses the winning balls from all balls. You should be able to do higher numbers by analogy, then add them all up to get the total probability. Alpha finds a total probability of about $0.0008$, almost all of which comes from hitting $11$ or $12$.
answered Jul 27 '17 at 23:03
Ross MillikanRoss Millikan
296k23198371
$endgroup$
add a comment |
$begingroup$
The chance of getting exactly eleven is $frac {{20 choose 11}{60 choose 9}}{80 choose 20}$ where the $20 choose 11$ is the number of ways to pick the winning balls out of your $20$, the $60 choose 9$ chooses the balls drawn that you did not pick, and the $80 choose 20$ chooses the winning balls from all balls. You should be able to do higher numbers by analogy, then add them all up to get the total probability. Alpha finds a total probability of about $0.0008$, almost all of which comes from hitting $11$ or $12$.
answered Jul 27 '17 at 23:03
Ross MillikanRoss Millikan
296k23198371
$endgroup$
The chance of getting exactly eleven is $frac {{20 choose 11}{60 choose 9}}{80 choose 20}$ where the $20 choose 11$ is the number of ways to pick the winning balls out of your $20$, the $60 choose 9$ chooses the balls drawn that you did not pick, and the $80 choose 20$ chooses the winning balls from all balls. You should be able to do higher numbers by analogy, then add them all up to get the total probability. Alpha finds a total probability of about $0.0008$, almost all of which comes from hitting $11$ or $12$.
answered Jul 27 '17 at 23:03
Ross MillikanRoss Millikan
296k23198371
edited Jul 27 '17 at 23:10
answered Jul 27 '17 at 23:03
Ross MillikanRoss Millikan
296k23198371
answered Jul 27 '17 at 23:03
Ross MillikanRoss Millikan
296k23198371
answered Jul 27 '17 at 23:03
Ross MillikanRoss Millikan
296k23198371
296k23198371
add a comment |
add a comment |
$begingroup$
The total number of ways for you to choose $20$ of the $80$ numbers and for the operator to choose $20$ is
$$binom{80}{20}^2$$
Now suppose that you choose $20$ numbers. To find the probability that at least $11$ of these match, we must find the number of ways for the operator to choose $20$ numbers such that at least $11$ of them are the same as some $11$ of yours. There are
$$binom{20}{11}$$
sets of $11$ of your $20$ picked numbers for the operator's numbers to match, and then there will be $70$ of the $80$ numbers remaining for the operator to choose, so the number of ways for the operator to pick the other $9$ of his numbers (that may or may not match any of yours) is
$$binom{70}{9}$$
and so the number of situations in which at least $10$ of your numbers match is
$$binom{20}{11}binom{70}{9}$$
and so the probability is
$$frac{binom{20}{11}binom{70}{9}}{binom{80}{20}^2}$$
answered Jul 27 '17 at 23:02
FrpzzdFrpzzd
23k841109
$endgroup$
$begingroup$
You shouldn't be squaring the denominator. We don't care how many ways the player can pick his numbers. It is not correct to use $70 choose 10$ and claim this is the chance of at least $10$ because if the player hits $11$ numbers you will count it $11$ times, once for each group of $10$ being the first ones.
$endgroup$
– Ross Millikan
Jul 27 '17 at 23:08
$begingroup$
That's okay, because I'm counting the same number of duplicates in the numerator.
$endgroup$
– Frpzzd
Jul 27 '17 at 23:09
$begingroup$
I pointed out two distinct errors. One is squaring the denominator. The second is overcounting the winners. They don't cancel out.
$endgroup$
– Ross Millikan
Jul 27 '17 at 23:14
$begingroup$
Hmm... I disagree. Our answers are only different because I counted the probability for getting any number of balls from $11$ to $20$ correct, whereas yours counts only the ways to get $11$ of them.
$endgroup$
– Frpzzd
Jul 27 '17 at 23:17
$begingroup$
No, you are overcounting. As a simple example, choose three of ABCDEF including at least two of ABC. You would say ${3 choose 2}{4 choose 1}$ for $12$. In fact it should be ${3 choose 2}{3 choose 1}$ to get exactly two and ${3 choose 3}{3 choose 0}$ to get all three of ABC for a total of $10$. You have counted $AB+C, AC+B,$ and $BC+A$, which gives the extra two. The ABC are like the numbers the player chooses in Keno. You shouldn't square the denominator because the player has chosen and there is only one player choice active, so the sample space is $80 choose 20$
$endgroup$
– Ross Millikan
Jul 28 '17 at 0:25
add a comment |
$begingroup$
The total number of ways for you to choose $20$ of the $80$ numbers and for the operator to choose $20$ is
$$binom{80}{20}^2$$
Now suppose that you choose $20$ numbers. To find the probability that at least $11$ of these match, we must find the number of ways for the operator to choose $20$ numbers such that at least $11$ of them are the same as some $11$ of yours. There are
$$binom{20}{11}$$
sets of $11$ of your $20$ picked numbers for the operator's numbers to match, and then there will be $70$ of the $80$ numbers remaining for the operator to choose, so the number of ways for the operator to pick the other $9$ of his numbers (that may or may not match any of yours) is
$$binom{70}{9}$$
and so the number of situations in which at least $10$ of your numbers match is
$$binom{20}{11}binom{70}{9}$$
and so the probability is
$$frac{binom{20}{11}binom{70}{9}}{binom{80}{20}^2}$$
answered Jul 27 '17 at 23:02
FrpzzdFrpzzd
23k841109
$endgroup$
$begingroup$
You shouldn't be squaring the denominator. We don't care how many ways the player can pick his numbers. It is not correct to use $70 choose 10$ and claim this is the chance of at least $10$ because if the player hits $11$ numbers you will count it $11$ times, once for each group of $10$ being the first ones.
$endgroup$
– Ross Millikan
Jul 27 '17 at 23:08
$begingroup$
That's okay, because I'm counting the same number of duplicates in the numerator.
$endgroup$
– Frpzzd
Jul 27 '17 at 23:09
$begingroup$
I pointed out two distinct errors. One is squaring the denominator. The second is overcounting the winners. They don't cancel out.
$endgroup$
– Ross Millikan
Jul 27 '17 at 23:14
$begingroup$
Hmm... I disagree. Our answers are only different because I counted the probability for getting any number of balls from $11$ to $20$ correct, whereas yours counts only the ways to get $11$ of them.
$endgroup$
– Frpzzd
Jul 27 '17 at 23:17
$begingroup$
No, you are overcounting. As a simple example, choose three of ABCDEF including at least two of ABC. You would say ${3 choose 2}{4 choose 1}$ for $12$. In fact it should be ${3 choose 2}{3 choose 1}$ to get exactly two and ${3 choose 3}{3 choose 0}$ to get all three of ABC for a total of $10$. You have counted $AB+C, AC+B,$ and $BC+A$, which gives the extra two. The ABC are like the numbers the player chooses in Keno. You shouldn't square the denominator because the player has chosen and there is only one player choice active, so the sample space is $80 choose 20$
$endgroup$
– Ross Millikan
Jul 28 '17 at 0:25
add a comment |
$begingroup$
The total number of ways for you to choose $20$ of the $80$ numbers and for the operator to choose $20$ is
$$binom{80}{20}^2$$
Now suppose that you choose $20$ numbers. To find the probability that at least $11$ of these match, we must find the number of ways for the operator to choose $20$ numbers such that at least $11$ of them are the same as some $11$ of yours. There are
$$binom{20}{11}$$
sets of $11$ of your $20$ picked numbers for the operator's numbers to match, and then there will be $70$ of the $80$ numbers remaining for the operator to choose, so the number of ways for the operator to pick the other $9$ of his numbers (that may or may not match any of yours) is
$$binom{70}{9}$$
and so the number of situations in which at least $10$ of your numbers match is
$$binom{20}{11}binom{70}{9}$$
and so the probability is
$$frac{binom{20}{11}binom{70}{9}}{binom{80}{20}^2}$$
answered Jul 27 '17 at 23:02
FrpzzdFrpzzd
23k841109
$endgroup$
The total number of ways for you to choose $20$ of the $80$ numbers and for the operator to choose $20$ is
$$binom{80}{20}^2$$
Now suppose that you choose $20$ numbers. To find the probability that at least $11$ of these match, we must find the number of ways for the operator to choose $20$ numbers such that at least $11$ of them are the same as some $11$ of yours. There are
$$binom{20}{11}$$
sets of $11$ of your $20$ picked numbers for the operator's numbers to match, and then there will be $70$ of the $80$ numbers remaining for the operator to choose, so the number of ways for the operator to pick the other $9$ of his numbers (that may or may not match any of yours) is
$$binom{70}{9}$$
and so the number of situations in which at least $10$ of your numbers match is
$$binom{20}{11}binom{70}{9}$$
and so the probability is
$$frac{binom{20}{11}binom{70}{9}}{binom{80}{20}^2}$$
answered Jul 27 '17 at 23:02
FrpzzdFrpzzd
23k841109
edited Jul 27 '17 at 23:17
answered Jul 27 '17 at 23:02
FrpzzdFrpzzd
23k841109
answered Jul 27 '17 at 23:02
FrpzzdFrpzzd
23k841109
answered Jul 27 '17 at 23:02
FrpzzdFrpzzd
23k841109
23k841109
$begingroup$
You shouldn't be squaring the denominator. We don't care how many ways the player can pick his numbers. It is not correct to use $70 choose 10$ and claim this is the chance of at least $10$ because if the player hits $11$ numbers you will count it $11$ times, once for each group of $10$ being the first ones.
$endgroup$
– Ross Millikan
Jul 27 '17 at 23:08
$begingroup$
That's okay, because I'm counting the same number of duplicates in the numerator.
$endgroup$
– Frpzzd
Jul 27 '17 at 23:09
$begingroup$
I pointed out two distinct errors. One is squaring the denominator. The second is overcounting the winners. They don't cancel out.
$endgroup$
– Ross Millikan
Jul 27 '17 at 23:14
$begingroup$
Hmm... I disagree. Our answers are only different because I counted the probability for getting any number of balls from $11$ to $20$ correct, whereas yours counts only the ways to get $11$ of them.
$endgroup$
– Frpzzd
Jul 27 '17 at 23:17
$begingroup$
No, you are overcounting. As a simple example, choose three of ABCDEF including at least two of ABC. You would say ${3 choose 2}{4 choose 1}$ for $12$. In fact it should be ${3 choose 2}{3 choose 1}$ to get exactly two and ${3 choose 3}{3 choose 0}$ to get all three of ABC for a total of $10$. You have counted $AB+C, AC+B,$ and $BC+A$, which gives the extra two. The ABC are like the numbers the player chooses in Keno. You shouldn't square the denominator because the player has chosen and there is only one player choice active, so the sample space is $80 choose 20$
$endgroup$
– Ross Millikan
Jul 28 '17 at 0:25
add a comment |
$begingroup$
You shouldn't be squaring the denominator. We don't care how many ways the player can pick his numbers. It is not correct to use $70 choose 10$ and claim this is the chance of at least $10$ because if the player hits $11$ numbers you will count it $11$ times, once for each group of $10$ being the first ones.
$endgroup$
– Ross Millikan
Jul 27 '17 at 23:08
$begingroup$
That's okay, because I'm counting the same number of duplicates in the numerator.
$endgroup$
– Frpzzd
Jul 27 '17 at 23:09
$begingroup$
I pointed out two distinct errors. One is squaring the denominator. The second is overcounting the winners. They don't cancel out.
$endgroup$
– Ross Millikan
Jul 27 '17 at 23:14
$begingroup$
Hmm... I disagree. Our answers are only different because I counted the probability for getting any number of balls from $11$ to $20$ correct, whereas yours counts only the ways to get $11$ of them.
$endgroup$
– Frpzzd
Jul 27 '17 at 23:17
$begingroup$
No, you are overcounting. As a simple example, choose three of ABCDEF including at least two of ABC. You would say ${3 choose 2}{4 choose 1}$ for $12$. In fact it should be ${3 choose 2}{3 choose 1}$ to get exactly two and ${3 choose 3}{3 choose 0}$ to get all three of ABC for a total of $10$. You have counted $AB+C, AC+B,$ and $BC+A$, which gives the extra two. The ABC are like the numbers the player chooses in Keno. You shouldn't square the denominator because the player has chosen and there is only one player choice active, so the sample space is $80 choose 20$
$endgroup$
– Ross Millikan
Jul 28 '17 at 0:25
$begingroup$
You shouldn't be squaring the denominator. We don't care how many ways the player can pick his numbers. It is not correct to use $70 choose 10$ and claim this is the chance of at least $10$ because if the player hits $11$ numbers you will count it $11$ times, once for each group of $10$ being the first ones.
$endgroup$
– Ross Millikan
Jul 27 '17 at 23:08
$begingroup$
You shouldn't be squaring the denominator. We don't care how many ways the player can pick his numbers. It is not correct to use $70 choose 10$ and claim this is the chance of at least $10$ because if the player hits $11$ numbers you will count it $11$ times, once for each group of $10$ being the first ones.
$endgroup$
– Ross Millikan
Jul 27 '17 at 23:08
$begingroup$
That's okay, because I'm counting the same number of duplicates in the numerator.
$endgroup$
– Frpzzd
Jul 27 '17 at 23:09
$begingroup$
That's okay, because I'm counting the same number of duplicates in the numerator.
$endgroup$
– Frpzzd
Jul 27 '17 at 23:09
$begingroup$
I pointed out two distinct errors. One is squaring the denominator. The second is overcounting the winners. They don't cancel out.
$endgroup$
– Ross Millikan
Jul 27 '17 at 23:14
$begingroup$
I pointed out two distinct errors. One is squaring the denominator. The second is overcounting the winners. They don't cancel out.
$endgroup$
– Ross Millikan
Jul 27 '17 at 23:14
$begingroup$
Hmm... I disagree. Our answers are only different because I counted the probability for getting any number of balls from $11$ to $20$ correct, whereas yours counts only the ways to get $11$ of them.
$endgroup$
– Frpzzd
Jul 27 '17 at 23:17
$begingroup$
Hmm... I disagree. Our answers are only different because I counted the probability for getting any number of balls from $11$ to $20$ correct, whereas yours counts only the ways to get $11$ of them.
$endgroup$
– Frpzzd
Jul 27 '17 at 23:17
$begingroup$
No, you are overcounting. As a simple example, choose three of ABCDEF including at least two of ABC. You would say ${3 choose 2}{4 choose 1}$ for $12$. In fact it should be ${3 choose 2}{3 choose 1}$ to get exactly two and ${3 choose 3}{3 choose 0}$ to get all three of ABC for a total of $10$. You have counted $AB+C, AC+B,$ and $BC+A$, which gives the extra two. The ABC are like the numbers the player chooses in Keno. You shouldn't square the denominator because the player has chosen and there is only one player choice active, so the sample space is $80 choose 20$
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– Ross Millikan
Jul 28 '17 at 0:25
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No, you are overcounting. As a simple example, choose three of ABCDEF including at least two of ABC. You would say ${3 choose 2}{4 choose 1}$ for $12$. In fact it should be ${3 choose 2}{3 choose 1}$ to get exactly two and ${3 choose 3}{3 choose 0}$ to get all three of ABC for a total of $10$. You have counted $AB+C, AC+B,$ and $BC+A$, which gives the extra two. The ABC are like the numbers the player chooses in Keno. You shouldn't square the denominator because the player has chosen and there is only one player choice active, so the sample space is $80 choose 20$
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– Ross Millikan
Jul 28 '17 at 0:25
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Does the order matter?
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– Frpzzd
Jul 27 '17 at 22:51
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The order doesn't matter.
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– JD3
Jul 27 '17 at 22:52
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How many balls are taken out of the spinner? I need that info to find the answer.
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– Frpzzd
Jul 27 '17 at 22:52
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20 total are selected out of the 80 balls in the spinner.
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– JD3
Jul 27 '17 at 22:53
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So you pick 20, and 20 are drawn at random?
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– Frpzzd
Jul 27 '17 at 22:54