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Combining confidence levels Combining confidence levels
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03-20-2018 , 07:40 AM
#1
Hackdeath
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adept
Join Date: Sep 2009 Posts: 848
Was playing badminton with some friends and this situation came up. The shuttle cock was hit close to the line and we weren't sure if it was in or out.

Person A said he was 60% sure it was in. Person B also said they were 60% sure it was in. On the assumption that the confidence levels are independent of each other, how sure does person C have to be that the shuttle was hit out for there to be an overall 50% chance of the shuttle being in or out?

Spoiler:
I thought the answer would be 84% because there is a 16% chance that the person is wrong, which is the same chance that both of the others are wrong (0.4^2).

But then it was pointed out later that if the 2 people were 50% sure of their call, then using the same method that would need the other person to be 75% sure, when logic states that the answer should be 50%.

So what is the right way to solve this question?
Combining confidence levels Quote
03-20-2018 , 03:18 PM
#2
NewOldGuy
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Pooh-Bah
Join Date: Mar 2009 Posts: 5,935
Quote:
Originally Posted by Hackdeath
Was playing badminton with some friends and this situation came up. The shuttle cock was hit close to the line and we weren't sure if it was in or out.

Person A said he was 60% sure it was in. Person B also said they were 60% sure it was in. On the assumption that the confidence levels are independent of each other, how sure does person C have to be that the shuttle was hit out for there to be an overall 50% chance of the shuttle being in or out?

Spoiler:
I thought the answer would be 84% because there is a 16% chance that the person is wrong, which is the same chance that both of the others are wrong (0.4^2).

But then it was pointed out later that if the 2 people were 50% sure of their call, then using the same method that would need the other person to be 75% sure, when logic states that the answer should be 50%.

So what is the right way to solve this question?
This isn't really a combination of probabilities, It's a combination of opinions about one probability. So the method of taking one minus the sum of the complements of the probabilities would not be correct.

I think the appropriate method here is called combining forecasts. Normally you would use some kind of weighted average, but with no other information about the participants a simple average is all you can do.
Last edited by NewOldGuy; 03-20-2018 at 03:24 PM.
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03-21-2018 , 08:02 AM
#3
Hackdeath
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Quote:
Originally Posted by NewOldGuy

I think the appropriate method here is called combining forecasts. Normally you would use some kind of weighted average, but with no other information about the participants a simple average is all you can do.
So the answer is 60%? That seems counter intuitive to me; surely as the number of people who think the call was in rather than out increases, so does the likelihood of it being in?
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03-21-2018 , 09:38 AM
#4
NewOldGuy
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Quote:
Originally Posted by Hackdeath
So the answer is 60%? That seems counter intuitive to me; surely as the number of people who think the call was in rather than out increases, so does the likelihood of it being in?
Nobody said they think it was in. They said they think there is a 60% chance it was in. That literally means that over many identical subective observations, they think 40% of the time it will actually be out.

If on the other hand each observer had to simply vote in or out, your calculation would work.
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