Please be patient with my English, as I am a native Korean and English is not my first language.

I want to write a story about a political organization built on fortune-telling.

Let us assume there are fortune-tellers who accurately predict the future more than 50% of the time, without any prior or inside information.

And, at this time, 'the probability' is not the probability of the fortuneteller's 'right foretell' but the probability that fortune-tellers present 'a right foretell'. [Edit: this sentence left as written.]

Here is my question.

If there is two fortune-tellers who have 60% chance of accurately answering a question about the future, is there a way to employ such fortune-tellers to increase the odds above 60%.

Would increasing the number of fortune-tellers to ten or more raise the odds of always making "right decisions?"

If possible, to optimize probability of accuracy, how should fortune-tellers foretell? [Edit: open-endedness of original preserved.]

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please... Now I'm in the envoronment that cannot use my language... english is terribly difficult. though this, i had to write this question to solve my terrible this question.. please give me a answer... T.T – user36187 Jul 20 '12 at 11:44
So, the question is: given that any fortune-teller's probability in predicting the outcome of an event (of binary type) is about 60%, how many fortune-tellers should a political organization hire to make the right choices? Is that right? – Raskolnikov Jul 20 '12 at 11:48
Yup! your think is right. – user36187 Jul 20 '12 at 11:53
daniell// I'm sorry. I really didn't know where to go to solve my question... :( – user36187 Jul 20 '12 at 11:55
My bad. I think Raskolnikov has interpreted your question correctly. – daniel Jul 20 '12 at 12:08

Here is how I understand your question: suppose the politician have to take a decision, having to choose between 2 possibilities. They ask the fortune tellers: "which decision should we take, A or B ?"

Now, the hypothesis are that the fortune tellers are correct with probability 60%, and that their answer are independant from what the other fortune tellers say.

If 1 fortune teller gives an answer: the probability to be correct is 60%

If 2 fortune tellers give an answer:

• it is possible neither is correct (probability of this happening is 16%),
• it is possible one of them is correct (probability is 48%),
• it is possible both are correct (p = 36%).

Is it what you want to know?

In general, if there are $n$ fortune tellers and you want to know the probability that $k$ of them are correct provided each of them has probability $p$ of being correct, the formula is:

$${n \choose k}p^k(1-p)^{n-k}$$

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This does not seem to answer the question, though. The original question was to design a methodology to increase your chances above 60% using multiple fortune tellers, or to prove it is not possible. – gt6989b Jul 20 '12 at 13:01

I think that you want to make a decision based on the fortune tellers responses. A logical choice might be majority rule. So for any n calculate the sum of the binomial probabilities given by Djaian where you sum from n/2 to n. If you solve for this as a function of n you can determine if there is an n where this probability is maximized or if it increases to a limit as n approaches infinity. We can get a good approximation for large n using the normal distribution approximation to the binomial distribution knowing the mean is np and the variance is np(1-p) where in your example p=0.60.

Now for any p>.5 the probability approaches 1 and for p<.5 the probability approaches 0. To see this using majority rule the normal approximation would integrate the normal density from .5n to n. This amounts to integrating the standard normal from (0.5-p)√n/√{p(1-p)} to (1-p)√n/√{p(1-p)} for p>0.5 this probability goes to 1 because the lower limit approaches minus infinity and the upper limit approaches infinity. But for p<0.5 the lower limit approaches plus infinity and hence the probability goes to 0. For p=0.5 the probability is always 1/2 and hence for the binomial the probability approaches 0.5. It is equal to .5 for n odd and greater than .5 for n even assuming if exactly half yes the decision is yes or you flip a fair coin to decide when there is a tie.

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+1, but what do you do when you have a tie? – gt6989b Jul 20 '12 at 13:10
It doesn't matter what you do in a tie. You can pick at random as I wrote in my revised answer. – Michael Chernick Jul 20 '12 at 13:13