Three people 80% accurate at choosing a winner better than one?

skinhat

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This is a question I need answered regarding face recognition where I have three types of face recognition all with about 80% accuracy and was hoping there was a way to combine the 3 to get a better accuracy. I have rewritten the question in terms of a picking the winner of a football match instead of face recognition:

If 3 people are 80% accurate in picking the winner of a football match can you combine the 3 peoples picks to get greater a accuracy than 80%? Thought maybe if two pickers chose team A while the other person chose team B then choose team A as the winner. If you did this would you get better than 80% or would it remain at 80% accuracy?
 
This is a question I need answered regarding face recognition where I have three types of face recognition all with about 80% accuracy and was hoping there was a way to combine the 3 to get a better accuracy. I have rewritten the question in terms of a picking the winner of a football match instead of face recognition:

If 3 people are 80% accurate in picking the winner of a football match can you combine the 3 peoples picks to get greater a accuracy than 80%? Thought maybe if two pickers chose team A while the other person chose team B then choose team A as the winner. If you did this would you get better than 80% or would it remain at 80% accuracy?

I am assuming this is not a homework question. If your question is "at least one of them will recognize a certain face" then yes you can with a probability of 99.2% by using them together. But then how do you know, when two disagree with one that identifies a face, if the face is correct or not?
 
I am assuming this is not a homework question. If your question is "at least one of them will recognize a certain face" then yes you can with a probability of 99.2% by using them together. But then how do you know, when two disagree with one that identifies a face, if the face is correct or not?

No, the question is not 'at least one will recognize a certain face'. My question is better explained in the second part with the football analogy in that all three are equally good at picking a winner (picking the correct face) and all three could pick the wrong team to win.
 
No, the question is not 'at least one will recognize a certain face'. My question is better explained in the second part with the football analogy in that all three are equally good at picking a winner (picking the correct face) and all three could pick the wrong team to win.

There is a 0.8% chance all of them will be wrong, 9.6% chance exactly one of them is right, 38.4% chance two of them are right and a 51.2% chance all are right. It is your preference what to do with this information (in game theory, one would assign a utility function based on their risk adversity). This is without knowledge of what they have picked.

If you are presented the results of the choices this changes things and because p(right)>p(wrong) you should go with whatever has the most votes.

Maybe this example will help:
If at least two people pick A, what is the probability that A wins? P(exactly 2 pick A) + P(all three pick A) = (3 choose 2)*.8*.8*.2 + (3 choose 3)*.8*.8*.8 = 89.6%
 
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There is a 0.8% chance all of them will be wrong, 9.6% chance exactly one of them is right, 38.4% chance two of them are right and a 51.2% chance all are right. It is your preference what to do with this information (in game theory, one would assign a utility function based on their risk adversity). This is without knowledge of what they have picked.

If you are presented the results of the choices this changes things and because p(right)>p(wrong) you should go with whatever has the most votes.

Maybe this example will help:
If at least two people pick A, what is the probability that A wins? P(exactly 2 pick A) + P(all three pick A) = (3 choose 2)*.8*.8*.2 + (3 choose 3)*.8*.8*.8 = 89.6%

Sorry, I realise I didnt explain the question properly. A better way to explain is:

You have Alan who is 80% accurate picking a football match winner
You have Peter who is 80% accurate picking a football match winner
You have Rick who is 80% accurate picking a football match winner

Now, there are 100 football games and Alan, Peter and Rick all picks their winners. Is it possible to combine Alan, Peters and Ricks picks to get better than 80% (for example if Alan and Peter pick team A and Rick picks team B then go with team A as the winner) or on average will the accuracy remain at 80% after the 100 games despite having 3 people choosing winners at 80%? My guess it that it will remain at 80%.
 
Sorry, I realise I didnt explain the question properly. A better way to explain is:

You have Alan who is 80% accurate picking a football match winner
You have Peter who is 80% accurate picking a football match winner
You have Rick who is 80% accurate picking a football match winner

Now, there are 100 football games and Alan, Peter and Rick all picks their winners. Is it possible to combine Alan, Peters and Ricks picks to get better than 80% (for example if Alan and Peter pick team A and Rick picks team B then go with team A as the winner) or on average will the accuracy remain at 80% after the 100 games despite having 3 people choosing winners at 80%? My guess it that it will remain at 80%.

Probability is based on the law of large numbers (results on average). So your question is to find "the probability that if at least two agree, they agree on the correct answer, given each is correct 80% of the time." As stated in my above post, it certainly does improve at 89.6%. You may test it yourself, I made a (very crude) example script: Below x,y,z are chosen at random amongst the first 1000 non-negative integers, the "correct zone" arbitrarily chosen to be the first 0-799 integers (80%). The conditional statement counts the number for which at least two are correct.
Code:
s=0;
for(i=1; i<100000; ++i)
{
 x=Math.floor(Math.random()*1000);
 y=Math.floor(Math.random()*1000);
 z=Math.floor(Math.random()*1000);
if((x<800 && y<800)||(x<800 && z<800)||(z<800 && y<800))
 s+=1;
}
alert(s/100000);
 
This is a question I need answered regarding face recognition where I have three types of face recognition all with about 80% accuracy and was hoping there was a way to combine the 3 to get a better accuracy. I have rewritten the question in terms of a picking the winner of a football match instead of face recognition:

If 3 people are 80% accurate in picking the winner of a football match can you combine the 3 peoples picks to get greater a accuracy than 80%? Thought maybe if two pickers chose team A while the other person chose team B then choose team A as the winner. If you did this would you get better than 80% or would it remain at 80% accuracy?
I think Bayesian inference will give you the highest possible accuracy. Looking at this problem from that point-of-view, you need to know all the the conditional probabilities, including saying "yes" to the wrong face as well as saying "no" to the correct face.

P(says yes | correct face) = 0.80 ?
P(says no | correct face) = 0.20 ?
P(says yes | wrong face) = ??
P(says no | wrong face) = ??

Following Bayes' Theorem, you start with a "prior" probability - usually an estimate based on the total population probability of a face in the category to be recognized. (For the football match case, the prior would be 50%.) Then you would apply the first test, which modifies the prior probability. Use that as the prior when applying the second test, which modifies the prior again. Finally apply the third test to get the final probability.
 
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