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?

  • 1
    Please format your questions so that they are read-friendly. – uniquesolution Sep 8 at 20:50
up vote 2 down vote accepted

Let's first find P(Fair | n Heads) = $P(F|nH)$, which can be written as $$P(F|nH) = \frac{P(nH|F)P(F)}{P(nH)} = \frac{(0.5)^n(0.5)}{(0.5)^n(0.5)+(0.8)^n(0.5)}$$ This should be < 0.1, and it it achieved when $$9 < (\frac{8}{5})^n$$ which is satisfied first when $ n = 5 $.

  • Okay, so you are saying that P(2H) = 0.5*0.5*0.5 + 0.8*0.8*0.5. I can't understand why this is the case. – user1436508 Sep 8 at 21:05
  • Does this mean this expression i wrote above is wrong??? P(2H|1H) = P(2H|1H, F|1H)*P(F|1H) + P(2H|1H, B|1H)*P(B|1H) Is this then supposed to be P(2H|1H) = P(2H|1H, F)*P(F) + P(2H|1H, B)*P(B) Please explain why its like this and not like what i wrote – user1436508 Sep 8 at 21:07
  • You have two ways of getting 2 heads, either you have biased or fair coin. That is $P(2H) = P(2H|F)P(F) + P(2H|B)P(B)$ – gunes Sep 8 at 21:12
  • Okay, however take a look at this question: math.stackexchange.com/questions/1481433/… They say that P(2H/1H) = P(2H/1H,F)P(F/1H) + P(2H/1H,B)P(B/1H) – user1436508 Sep 8 at 21:21
  • First of all, one should use the following notation while describing the given events: $P(A|B) = P(A,C| B) + P(A,C'|B)$, which in turn equals to $P(A|B,C)P(C|B) + P(A|B,C')P(C'|B)$, i.e. keep the occured event after | sign to impede any confusion. – gunes Sep 8 at 21:26

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.