Probability after $n$ execution with two different probability distributions
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Lets say there's a statement and you want to know of, whether it is correct or not. Given an algorithm which outputs "probably true" or "probably false". In case the statement is correct the probability of "probably true" is $frac{9}{10}$ and in case the statement is incorrect the probability of "probably false" is $frac{6}{10}$.
I see that executing the algorithm many times will in-/decrease the probability of the statement being correct depending on the outputs. But what is the exact probability after $n$ executions?
probability
asked Nov 26 at 21:34
upe
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up vote
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Lets say there's a statement and you want to know of, whether it is correct or not. Given an algorithm which outputs "probably true" or "probably false". In case the statement is correct the probability of "probably true" is $frac{9}{10}$ and in case the statement is incorrect the probability of "probably false" is $frac{6}{10}$.
I see that executing the algorithm many times will in-/decrease the probability of the statement being correct depending on the outputs. But what is the exact probability after $n$ executions?
probability
asked Nov 26 at 21:34
upe
123
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Lets say there's a statement and you want to know of, whether it is correct or not. Given an algorithm which outputs "probably true" or "probably false". In case the statement is correct the probability of "probably true" is $frac{9}{10}$ and in case the statement is incorrect the probability of "probably false" is $frac{6}{10}$.
I see that executing the algorithm many times will in-/decrease the probability of the statement being correct depending on the outputs. But what is the exact probability after $n$ executions?
probability
asked Nov 26 at 21:34
upe
123
Lets say there's a statement and you want to know of, whether it is correct or not. Given an algorithm which outputs "probably true" or "probably false". In case the statement is correct the probability of "probably true" is $frac{9}{10}$ and in case the statement is incorrect the probability of "probably false" is $frac{6}{10}$.
I see that executing the algorithm many times will in-/decrease the probability of the statement being correct depending on the outputs. But what is the exact probability after $n$ executions?
probability
probability
asked Nov 26 at 21:34
upe
123
asked Nov 26 at 21:34
upe
123
asked Nov 26 at 21:34
upe
123
asked Nov 26 at 21:34
upe
123
asked Nov 26 at 21:34
upe
123
123
add a comment |
add a comment |
1 Answer
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This is more easily done as an odds calculation. The odds of the statement being correct based solely on a single output of "probably true" are $9:4$ (since the algorithm outputs "probably true" with probability $frac{9}{10}$ for a correct statement and $frac{4}{10}$ for an incorrect statement). The odds based on a single output of "probably false" are $1:6$ for the same reason.
If different runs of the algorithm are independent, then we can just multiply these together. For example, if the algorithm says "probably true" $3$ times and "probably false" twice, then the combined odds are $$(9:4)^3 cdot (1:6)^2 = 9^3 : 4^3cdot6^2 = 729: 2304 = 81:256$$ so the probability of the statement being correct is $frac{81}{81+256}$.
This calculation assumes that initially the odds are $1:1$ and the statement is equally likely to be true or false without further information. If you have other information about the odds, you can just use that instead by multiplying that in, as well.
answered Nov 26 at 21:40
Misha Lavrov
42.8k555101
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
This is more easily done as an odds calculation. The odds of the statement being correct based solely on a single output of "probably true" are $9:4$ (since the algorithm outputs "probably true" with probability $frac{9}{10}$ for a correct statement and $frac{4}{10}$ for an incorrect statement). The odds based on a single output of "probably false" are $1:6$ for the same reason.
If different runs of the algorithm are independent, then we can just multiply these together. For example, if the algorithm says "probably true" $3$ times and "probably false" twice, then the combined odds are $$(9:4)^3 cdot (1:6)^2 = 9^3 : 4^3cdot6^2 = 729: 2304 = 81:256$$ so the probability of the statement being correct is $frac{81}{81+256}$.
This calculation assumes that initially the odds are $1:1$ and the statement is equally likely to be true or false without further information. If you have other information about the odds, you can just use that instead by multiplying that in, as well.
answered Nov 26 at 21:40
Misha Lavrov
42.8k555101
add a comment |
up vote
0
down vote
accepted
This is more easily done as an odds calculation. The odds of the statement being correct based solely on a single output of "probably true" are $9:4$ (since the algorithm outputs "probably true" with probability $frac{9}{10}$ for a correct statement and $frac{4}{10}$ for an incorrect statement). The odds based on a single output of "probably false" are $1:6$ for the same reason.
If different runs of the algorithm are independent, then we can just multiply these together. For example, if the algorithm says "probably true" $3$ times and "probably false" twice, then the combined odds are $$(9:4)^3 cdot (1:6)^2 = 9^3 : 4^3cdot6^2 = 729: 2304 = 81:256$$ so the probability of the statement being correct is $frac{81}{81+256}$.
This calculation assumes that initially the odds are $1:1$ and the statement is equally likely to be true or false without further information. If you have other information about the odds, you can just use that instead by multiplying that in, as well.
answered Nov 26 at 21:40
Misha Lavrov
42.8k555101
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
This is more easily done as an odds calculation. The odds of the statement being correct based solely on a single output of "probably true" are $9:4$ (since the algorithm outputs "probably true" with probability $frac{9}{10}$ for a correct statement and $frac{4}{10}$ for an incorrect statement). The odds based on a single output of "probably false" are $1:6$ for the same reason.
If different runs of the algorithm are independent, then we can just multiply these together. For example, if the algorithm says "probably true" $3$ times and "probably false" twice, then the combined odds are $$(9:4)^3 cdot (1:6)^2 = 9^3 : 4^3cdot6^2 = 729: 2304 = 81:256$$ so the probability of the statement being correct is $frac{81}{81+256}$.
This calculation assumes that initially the odds are $1:1$ and the statement is equally likely to be true or false without further information. If you have other information about the odds, you can just use that instead by multiplying that in, as well.
answered Nov 26 at 21:40
Misha Lavrov
42.8k555101
This is more easily done as an odds calculation. The odds of the statement being correct based solely on a single output of "probably true" are $9:4$ (since the algorithm outputs "probably true" with probability $frac{9}{10}$ for a correct statement and $frac{4}{10}$ for an incorrect statement). The odds based on a single output of "probably false" are $1:6$ for the same reason.
If different runs of the algorithm are independent, then we can just multiply these together. For example, if the algorithm says "probably true" $3$ times and "probably false" twice, then the combined odds are $$(9:4)^3 cdot (1:6)^2 = 9^3 : 4^3cdot6^2 = 729: 2304 = 81:256$$ so the probability of the statement being correct is $frac{81}{81+256}$.
This calculation assumes that initially the odds are $1:1$ and the statement is equally likely to be true or false without further information. If you have other information about the odds, you can just use that instead by multiplying that in, as well.
answered Nov 26 at 21:40
Misha Lavrov
42.8k555101
answered Nov 26 at 21:40
Misha Lavrov
42.8k555101
answered Nov 26 at 21:40
Misha Lavrov
42.8k555101
answered Nov 26 at 21:40
Misha Lavrov
42.8k555101
42.8k555101
add a comment |
add a comment |
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