A man claims to have extrasensory perception. As a test, a fair coin is flipped 23 times, and the man is asked

Question

A man claims to have extrasensory perception. As a test, a fair coin is flipped 23 times, and the man is asked to predict the outcome in advance. He gets 19 out of 23 correct. What is the probability that he would have done at least this well if he had no ESP?

Answer 1

WRONG!!!!! You people are forgetting the binomial theorem. The probability is NOT 50% because there are many ways to get 19 out of 23 flips, NOT just one.

Take 2 coin flips. What is the probability of getting 2 heads? It is obviously 1/4 (HH, TH, HT, and TT) but that is NOT 50%. In the same way, the probability of getting 19 out of 23 flips is NOT 50%.

It is 23 choose 19 divided by 2^23 = 8855/2^23 = .00105555983

If you don't believe me, then look up the Binomial Theorem.

Answer 2

For readibility i will denote as
n
C =( n C r )
r

probability of getting exactly 19 right
= (23 C 19) 0.5^19 0.5^4
= (23 C 19) 0.5^23

probability of getting at least 19 right =
= Summation (i from 19->23) (23 C i) *0.5^i *0.5^23-i
= Summation (i from 19->23) (23 C i) *0.5^23

my trusty calculator gives me
= (8855+ 1771 + 253 + 23 + 1) * 0.5^23
= 10903 *0.5^23
= 0.00129973

~ 1 in 1000 He is mildly ESP sensitive... i would say ::P

Answer 3

4 out of 23 :)

Answer 4

50%

Answer 5

50%