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Xtaki

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There is a game between two teams - team A and team B. An expert with an 80% success rate of guessing the right team to win is saying that team A will win. Another expert with an 80% success rate is saying that the same team (team A) will win. What are the odds that they will be correct? is it still 80% or is it getting higher?
 
They are either both correct or both incorrect.

[MATH]P[\text{both correct}] = 1 - P[\text{both incorrect}] = 1 - (1-0.8)^2 = 1-0.04 = 0.96[/MATH]
 
Romsek said:
They are either both correct or both incorrect.

[MATH]P[\text{both correct}] = 1 - P[\text{both incorrect}] = 1 - (1-0.8)^2 = 1-0.04 = 0.96[/MATH]
Click to expand...

But P[\text{both correct}] = 0.8 * 0.8 = 0.64

so how is it possible?
 
Xtaki said:
But P[\text{both correct}] = 0.8 * 0.8 = 0.64

so how is it possible?
Click to expand...

My post above wasn't correct. The problem is more complicated and is missing information.

There are two ways they both can be correct. Predicting A wins and A does win. Or predicting A loses and A does lose.
We aren't given the probability that the experts correctly predict A losing.

If we say the probability of A winning is p, and the probability the experts predict A loses when A does lose is q, we have

P[both correct] = 0.8^2 p + q^2 (1-p) = p(0.8^2 - q^2) + q^2

Let's go ahead and assume the experts are equivalently good at choosing wins and losses so that q=0.8

P[both correct] = 0.8^2 p + 0.8^2 (1-p) = 0.8^2 = 0.64

But now as a counterexample let's assume the experts aren't so good at picking losses. Let q = 0.2

P[both correct] = 0.8^2 p + 0.2^2 (1-p) = p(0.8^2 - 0.2^2) + 0.2^2 = 0.6p + 0.04

and this depends on how likely a priori it is for A to win.
 
Romsek said:
My post above wasn't correct. The problem is more complicated and is missing information.

There are two ways they both can be correct. Predicting A wins and A does win. Or predicting A loses and A does lose.
We aren't given the probability that the experts correctly predict A losing.

If we say the probability of A winning is p, and the probability the experts predict A loses when A does lose is q, we have

P[both correct] = 0.8^2 p + q^2 (1-p) = p(0.8^2 - q^2) + q^2

Let's go ahead and assume the experts are equivalently good at choosing wins and losses so that q=0.8

P[both correct] = 0.8^2 p + 0.8^2 (1-p) = 0.8^2 = 0.64

But now as a counterexample let's assume the experts aren't so good at picking losses. Let q = 0.2

P[both correct] = 0.8^2 p + 0.2^2 (1-p) = p(0.8^2 - 0.2^2) + 0.2^2 = 0.6p + 0.04

and this depends on how likely a priori it is for A to win.
Click to expand...

So what is the final answer?
Each expert is right 80% of the time and wrong 20% of the time. They both are betting on the same team. What are the odds that they are correct? The minimum is 80%. My question is if the reason that two experts saying the same thing is it making the odds higher? And if yes how much?
 
Xtaki said:
There is a game between two teams - team A and team B. An expert with an 80% success rate of guessing the right team to win is saying that team A will win. Another expert with an 80% success rate is saying that the same team (team A) will win. What are the odds that they will be correct? is it still 80% or is it getting higher?
Click to expand...
One complicating factor is that each expert is not just guessing randomly; they are using some data to make their predictions. Moreover, their predictions probably are not independent. So I'm not sure probability even applies here. But a Bayesian approach may be appropriate.

I disagree with Romsek's statement that "We aren't given the probability that the experts correctly predict A losing." We are given that each has "an 80% success rate of guessing the right team to win", and if they guess that a given team will win, they are at the same time predicting that the other team loses.

I'm going to change the question to one about an area I'm more familiar with. Suppose that Romsek and I both solve a math problem. Each of us has an 80% success rate in getting the math right; that means 20% of the time we make at least one mistake. Now we both solve the same problem, and we get the same answer, say 17. How likely is it that we are right? [This is different from the question asked, because there are many possible wrong answers, rather than just a choice between A and B.]

Suppose that 17 is the wrong answer. Then both of us made mistakes. That happens 0.20*0.20 = 0.04 of the time.

Suppose that 17 is the right answer. Then both of us were right. That happens 0.80*0.80 = 0.64 of the time.

Those are the only two possibilities (ignoring the fact that we might also get the right answer by incorrect work ...), so the probability that what happened would happen is 0.04 + 0.64 = 0.68. Therefore the probability that 17 is the right answer is 0.64/0.68 = 0.94.

Does that sound better? I'm not entirely convinced, but I think it's the best answer I've found.
 
Dr.Peterson said:
One complicating factor is that each expert is not just guessing randomly; they are using some data to make their predictions. Moreover, their predictions probably are not independent. So I'm not sure probability even applies here. But a Bayesian approach may be appropriate.

I disagree with Romsek's statement that "We aren't given the probability that the experts correctly predict A losing." We are given that each has "an 80% success rate of guessing the right team to win", and if they guess that a given team will win, they are at the same time predicting that the other team loses.

I'm going to change the question to one about an area I'm more familiar with. Suppose that Romsek and I both solve a math problem. Each of us has an 80% success rate in getting the math right; that means 20% of the time we make at least one mistake. Now we both solve the same problem, and we get the same answer, say 17. How likely is it that we are right? [This is different from the question asked, because there are many possible wrong answers, rather than just a choice between A and B.]

Suppose that 17 is the wrong answer. Then both of us made mistakes. That happens 0.20*0.20 = 0.04 of the time.

Suppose that 17 is the right answer. Then both of us were right. That happens 0.80*0.80 = 0.64 of the time.

Those are the only two possibilities (ignoring the fact that we might also get the right answer by incorrect work ...), so the probability that what happened would happen is 0.04 + 0.64 = 0.68. Therefore the probability that 17 is the right answer is 0.64/0.68 = 0.94.

Does that sound better? I'm not entirely convinced, but I think it's the best answer I've found.
Click to expand...
Interesting! Thank you
 
I agree with Dr Peterson that the two experts choices are not independent. For example if they both make their decision based on the same method then the chance that they are both correct in choosing team A is 80%.

Consider this. Suppose there are 100 students in a probability class with a great instructor. The instructor has a method on picking the correct team 80% of the time and teaches this method to her students. On an exam she asks her students to predict if team A or B will win giving them all the information to use her method. Given that she is a great teacher all her students arrive at the same answer. The question is if all 100 (or is it 101 including the teacher?) pick the same team is the probability that team A wins still 80%. The answer is yes, it is still 80%.

Now on the other hand if two people use two completely different methods, each with an 80% success rate, then certainly the chance that team A will win will be greater than 80%.
 
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