This is a very challenging problem of which I'm stuck.
I have a fair coin with a probability of getting head as 0.5, and a bias coin with probability of getting head as 0.8
I grab a coin and I keep flipping it, and I keep getting heads. How many flips will it take for the probability of the coin being fair to drop < 0.1
Let our notation be:
- P(1H) = Probability of getting heads in first flip
- P(2H|1H) = Probability of getting heads on first flip, and heads on second flip
- P(F) = Probability of it being a fair coin
- P(B) = Probability of it being biased coin
I know that P(1H) = 0.5*0.5 + 0.5*0.8 = 0.65
P(F|1H) = Probability of first coin being fair if i got heads on first flip = P(1H|F)*P(F) / P(H) = (0.5*0.5)/0.65 = 0.3846
P(B|1H) = (0.8*0.5)/0.65 0.6154 given same calculations
P(2H|1H) = P(2H|1H, F|1H)*P(F|1H) + P(2H|1H, B|1H)*P(B|1H) = 0.5*0.3846 + 0.8*0.6154 = 0.6846
P(F|(2H|1H)) = P(2H|1H, F)*P(F) / P(2H|1H) = 0.5*0.5 / 0.6846 = 0.3652
So I know the probability of getting a fair coin with two flips is as follows
But i could keep going, and this expression would get more and more complicated. How on earth do i solve this?