Probability of 11 members in a group of 12 choose the same outcome.
Example: Suppose there's an election and three candidates are running for the presidency. What are the odds that 11 people out of 12 will choose the same candidate? (assuming each person has a 1/3 chance to pick either candidate).
Attempt: I think that there are 3^12 possible outcomes, (3 alternatives, 12 subjects that will choose the alternatives) and 12 desirable outcomes for each candidate, so 36 overall. So my guess is 36/3^11.
probability
asked Nov 19 '18 at 2:10
Vitor Costa
32
add a comment |
Example: Suppose there's an election and three candidates are running for the presidency. What are the odds that 11 people out of 12 will choose the same candidate? (assuming each person has a 1/3 chance to pick either candidate).
Attempt: I think that there are 3^12 possible outcomes, (3 alternatives, 12 subjects that will choose the alternatives) and 12 desirable outcomes for each candidate, so 36 overall. So my guess is 36/3^11.
probability
asked Nov 19 '18 at 2:10
Vitor Costa
32
1
"Tries" is not "probability" is not "odds." Please re-write your question more carefully.
– David G. Stork
Nov 19 '18 at 2:13
I removed the last bit about tries. The probability then is approximately 0.02%, correct ?
– Vitor Costa
Nov 19 '18 at 2:18
I don't understand your $36$ at all ... '12 desirable outcomes for each candidate'? can you explain that?
– Bram28
Nov 19 '18 at 2:27
add a comment |
Example: Suppose there's an election and three candidates are running for the presidency. What are the odds that 11 people out of 12 will choose the same candidate? (assuming each person has a 1/3 chance to pick either candidate).
Attempt: I think that there are 3^12 possible outcomes, (3 alternatives, 12 subjects that will choose the alternatives) and 12 desirable outcomes for each candidate, so 36 overall. So my guess is 36/3^11.
probability
asked Nov 19 '18 at 2:10
Vitor Costa
32
Example: Suppose there's an election and three candidates are running for the presidency. What are the odds that 11 people out of 12 will choose the same candidate? (assuming each person has a 1/3 chance to pick either candidate).
Attempt: I think that there are 3^12 possible outcomes, (3 alternatives, 12 subjects that will choose the alternatives) and 12 desirable outcomes for each candidate, so 36 overall. So my guess is 36/3^11.
probability
probability
asked Nov 19 '18 at 2:10
Vitor Costa
32
asked Nov 19 '18 at 2:10
Vitor Costa
32
edited Nov 19 '18 at 2:16
asked Nov 19 '18 at 2:10
Vitor Costa
32
asked Nov 19 '18 at 2:10
Vitor Costa
32
asked Nov 19 '18 at 2:10
Vitor Costa
32
32
1
"Tries" is not "probability" is not "odds." Please re-write your question more carefully.
– David G. Stork
Nov 19 '18 at 2:13
I removed the last bit about tries. The probability then is approximately 0.02%, correct ?
– Vitor Costa
Nov 19 '18 at 2:18
I don't understand your $36$ at all ... '12 desirable outcomes for each candidate'? can you explain that?
– Bram28
Nov 19 '18 at 2:27
add a comment |
1
"Tries" is not "probability" is not "odds." Please re-write your question more carefully.
– David G. Stork
Nov 19 '18 at 2:13
I removed the last bit about tries. The probability then is approximately 0.02%, correct ?
– Vitor Costa
Nov 19 '18 at 2:18
I don't understand your $36$ at all ... '12 desirable outcomes for each candidate'? can you explain that?
– Bram28
Nov 19 '18 at 2:27
1
1
"Tries" is not "probability" is not "odds." Please re-write your question more carefully.
– David G. Stork
Nov 19 '18 at 2:13
"Tries" is not "probability" is not "odds." Please re-write your question more carefully.
– David G. Stork
Nov 19 '18 at 2:13
I removed the last bit about tries. The probability then is approximately 0.02%, correct ?
– Vitor Costa
Nov 19 '18 at 2:18
I removed the last bit about tries. The probability then is approximately 0.02%, correct ?
– Vitor Costa
Nov 19 '18 at 2:18
I don't understand your $36$ at all ... '12 desirable outcomes for each candidate'? can you explain that?
– Bram28
Nov 19 '18 at 2:27
I don't understand your $36$ at all ... '12 desirable outcomes for each candidate'? can you explain that?
– Bram28
Nov 19 '18 at 2:27
add a comment |
2 Answers
2
active
oldest
votes
Call the candidates $A$, $B$, and $C$. The probability you get the outcome $A = 11$, $B = 1$ and $C=0$ is given by the trinomial distribution, which here is:
$$P(A=11, B=1, C=0) = {12! over 11! 1! 0!} (1/3)^{11} (1/3)^1 (1/3)^0$$.
But you can get the criterion output if $A=1$, $B=11$, $C=0$... and likewise for other permutations. There are $3!=6$ such permutations, and you must add up their probabilities.
Thus:
$$3! {12! over 11! 1! 0!} (1/3)^{11} (1/3)^1 (1/3)^0 = frac{8}{59049} approx 0.000135481.$$
answered Nov 19 '18 at 2:26
David G. Stork
9,82021232
Thank you, this makes perfect sense. I was having trouble with the permutations of the desirable/criterion output.
– Vitor Costa
Nov 19 '18 at 2:42
Permutations: $A=11,B=1,C=0$... $A=11,B=0,C=1$... $A=1,B=11,C=0$... [can you fill in the rest of the six?]
– David G. Stork
Nov 19 '18 at 7:53
add a comment |
It looks like you calculated the odds and not the probability.
When counting successful outcomes, there are $12$ ways to get $11$ the same for each of $3$ candidates and for each of those the one different outcome can be one of two, hence:
$$P(11 text{same}) = frac{12cdot 3cdot 2}{3^{12}} = .00013548$$
answered Nov 19 '18 at 2:56
Phil H
4,0262312
add a comment |
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2 Answers
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active
oldest
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2 Answers
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active
oldest
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oldest
votes
Call the candidates $A$, $B$, and $C$. The probability you get the outcome $A = 11$, $B = 1$ and $C=0$ is given by the trinomial distribution, which here is:
$$P(A=11, B=1, C=0) = {12! over 11! 1! 0!} (1/3)^{11} (1/3)^1 (1/3)^0$$.
But you can get the criterion output if $A=1$, $B=11$, $C=0$... and likewise for other permutations. There are $3!=6$ such permutations, and you must add up their probabilities.
Thus:
$$3! {12! over 11! 1! 0!} (1/3)^{11} (1/3)^1 (1/3)^0 = frac{8}{59049} approx 0.000135481.$$
answered Nov 19 '18 at 2:26
David G. Stork
9,82021232
Thank you, this makes perfect sense. I was having trouble with the permutations of the desirable/criterion output.
– Vitor Costa
Nov 19 '18 at 2:42
Permutations: $A=11,B=1,C=0$... $A=11,B=0,C=1$... $A=1,B=11,C=0$... [can you fill in the rest of the six?]
– David G. Stork
Nov 19 '18 at 7:53
add a comment |
Call the candidates $A$, $B$, and $C$. The probability you get the outcome $A = 11$, $B = 1$ and $C=0$ is given by the trinomial distribution, which here is:
$$P(A=11, B=1, C=0) = {12! over 11! 1! 0!} (1/3)^{11} (1/3)^1 (1/3)^0$$.
But you can get the criterion output if $A=1$, $B=11$, $C=0$... and likewise for other permutations. There are $3!=6$ such permutations, and you must add up their probabilities.
Thus:
$$3! {12! over 11! 1! 0!} (1/3)^{11} (1/3)^1 (1/3)^0 = frac{8}{59049} approx 0.000135481.$$
answered Nov 19 '18 at 2:26
David G. Stork
9,82021232
Thank you, this makes perfect sense. I was having trouble with the permutations of the desirable/criterion output.
– Vitor Costa
Nov 19 '18 at 2:42
Permutations: $A=11,B=1,C=0$... $A=11,B=0,C=1$... $A=1,B=11,C=0$... [can you fill in the rest of the six?]
– David G. Stork
Nov 19 '18 at 7:53
add a comment |
Call the candidates $A$, $B$, and $C$. The probability you get the outcome $A = 11$, $B = 1$ and $C=0$ is given by the trinomial distribution, which here is:
$$P(A=11, B=1, C=0) = {12! over 11! 1! 0!} (1/3)^{11} (1/3)^1 (1/3)^0$$.
But you can get the criterion output if $A=1$, $B=11$, $C=0$... and likewise for other permutations. There are $3!=6$ such permutations, and you must add up their probabilities.
Thus:
$$3! {12! over 11! 1! 0!} (1/3)^{11} (1/3)^1 (1/3)^0 = frac{8}{59049} approx 0.000135481.$$
answered Nov 19 '18 at 2:26
David G. Stork
9,82021232
Call the candidates $A$, $B$, and $C$. The probability you get the outcome $A = 11$, $B = 1$ and $C=0$ is given by the trinomial distribution, which here is:
$$P(A=11, B=1, C=0) = {12! over 11! 1! 0!} (1/3)^{11} (1/3)^1 (1/3)^0$$.
But you can get the criterion output if $A=1$, $B=11$, $C=0$... and likewise for other permutations. There are $3!=6$ such permutations, and you must add up their probabilities.
Thus:
$$3! {12! over 11! 1! 0!} (1/3)^{11} (1/3)^1 (1/3)^0 = frac{8}{59049} approx 0.000135481.$$
answered Nov 19 '18 at 2:26
David G. Stork
9,82021232
answered Nov 19 '18 at 2:26
David G. Stork
9,82021232
answered Nov 19 '18 at 2:26
David G. Stork
9,82021232
answered Nov 19 '18 at 2:26
David G. Stork
9,82021232
9,82021232
Thank you, this makes perfect sense. I was having trouble with the permutations of the desirable/criterion output.
– Vitor Costa
Nov 19 '18 at 2:42
Permutations: $A=11,B=1,C=0$... $A=11,B=0,C=1$... $A=1,B=11,C=0$... [can you fill in the rest of the six?]
– David G. Stork
Nov 19 '18 at 7:53
add a comment |
Thank you, this makes perfect sense. I was having trouble with the permutations of the desirable/criterion output.
– Vitor Costa
Nov 19 '18 at 2:42
Permutations: $A=11,B=1,C=0$... $A=11,B=0,C=1$... $A=1,B=11,C=0$... [can you fill in the rest of the six?]
– David G. Stork
Nov 19 '18 at 7:53
Thank you, this makes perfect sense. I was having trouble with the permutations of the desirable/criterion output.
– Vitor Costa
Nov 19 '18 at 2:42
Thank you, this makes perfect sense. I was having trouble with the permutations of the desirable/criterion output.
– Vitor Costa
Nov 19 '18 at 2:42
Permutations: $A=11,B=1,C=0$... $A=11,B=0,C=1$... $A=1,B=11,C=0$... [can you fill in the rest of the six?]
– David G. Stork
Nov 19 '18 at 7:53
Permutations: $A=11,B=1,C=0$... $A=11,B=0,C=1$... $A=1,B=11,C=0$... [can you fill in the rest of the six?]
– David G. Stork
Nov 19 '18 at 7:53
add a comment |
It looks like you calculated the odds and not the probability.
When counting successful outcomes, there are $12$ ways to get $11$ the same for each of $3$ candidates and for each of those the one different outcome can be one of two, hence:
$$P(11 text{same}) = frac{12cdot 3cdot 2}{3^{12}} = .00013548$$
answered Nov 19 '18 at 2:56
Phil H
4,0262312
add a comment |
It looks like you calculated the odds and not the probability.
When counting successful outcomes, there are $12$ ways to get $11$ the same for each of $3$ candidates and for each of those the one different outcome can be one of two, hence:
$$P(11 text{same}) = frac{12cdot 3cdot 2}{3^{12}} = .00013548$$
answered Nov 19 '18 at 2:56
Phil H
4,0262312
add a comment |
It looks like you calculated the odds and not the probability.
When counting successful outcomes, there are $12$ ways to get $11$ the same for each of $3$ candidates and for each of those the one different outcome can be one of two, hence:
$$P(11 text{same}) = frac{12cdot 3cdot 2}{3^{12}} = .00013548$$
answered Nov 19 '18 at 2:56
Phil H
4,0262312
It looks like you calculated the odds and not the probability.
When counting successful outcomes, there are $12$ ways to get $11$ the same for each of $3$ candidates and for each of those the one different outcome can be one of two, hence:
$$P(11 text{same}) = frac{12cdot 3cdot 2}{3^{12}} = .00013548$$
answered Nov 19 '18 at 2:56
Phil H
4,0262312
answered Nov 19 '18 at 2:56
Phil H
4,0262312
answered Nov 19 '18 at 2:56
Phil H
4,0262312
answered Nov 19 '18 at 2:56
Phil H
4,0262312
4,0262312
add a comment |
add a comment |
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1
"Tries" is not "probability" is not "odds." Please re-write your question more carefully.
– David G. Stork
Nov 19 '18 at 2:13
I removed the last bit about tries. The probability then is approximately 0.02%, correct ?
– Vitor Costa
Nov 19 '18 at 2:18
I don't understand your $36$ at all ... '12 desirable outcomes for each candidate'? can you explain that?
– Bram28
Nov 19 '18 at 2:27