# 1 Biased Coin and 1 Fair Coin, probability of 2 Heads?

You have 1 fair coin and 1 coin with 2 heads. Given that the first flip was a heads what is the probability of getting another heads?

My Answer: P(2H|F=H) = P(2H|F=H, Biased Coin)*P(Biased Coin) + P(2H|F=H, Unbiased Coin)*P(Unbiased Coin) = 0.5 + 0.25 = 0.75. In my equation, F refers to the First Throw. But the answer is supposed to be 5/6 and I can't seem to understand how.

Edit: From Arthurs comment I get the following, however, I dont know if this is the correct method, despite getting the correct answer:

P(Biased|F=H) = 2/3.

P(2H|F=H) = P(2H|(Biased|F=H))*P(Biased|F=H) + P(2H|(Unbiased|F=H))*P(Unbiased|F=H) = (1*2/3) + (1/2 * 2/3) = 5/6.

Thank You

• Is it random (and unknown) which coin was flipped first? – Arthur Oct 15 '15 at 12:41
• @Arthur Yes. There is no information about that. – Jojo Oct 15 '15 at 12:42
• I think this is the biggest thing you've missed: Given that the first toss was a head, what is the proability that the first coin was the biased one? – Arthur Oct 15 '15 at 12:46
• @Arthur Could you please check the edit to my question. – Jojo Oct 15 '15 at 13:08
• @Jojo, yes, although that should be $P(2H\mid\text{ Biased} \cap F=H)$ and so forth. – Graham Kemp Oct 15 '15 at 13:11

The first flip was a head.

The $3$ heads have equal probabilities to be the head that appeared at the first flip.

$2$ of the $3$ heads have another head as other side.

$1$ of the $3$ heads has tail as other side.

So there is a chance of $\frac23.1+\frac13.\frac12=\frac56$ of throwing a second head with that coin.

• Very intuitive answer. Thanks. – Jojo Oct 15 '15 at 13:22
• Glad to help, and indeed cherish your intuition in maths. – drhab Oct 15 '15 at 13:24
• Does this assume that the same coin is flipped twice, or that one coin is flipped and then the other? I'm getting a different answer when I assume that latter, but the question is ambiguous. – Bill the Lizard Oct 15 '15 at 14:56
• It assumes that the same coin was flipped twice. – drhab Oct 15 '15 at 15:00

You missed a conditioning in your formula. Let the event of choosing biased coin be B and fair be F. Getting heads on second toss be 2H and on first toss 1H. You want to calculate the probability of getting heads on second throw given that you it landed heads on first throw, which is

P(2H/1H) = P(2H/1H,F)P(F/1H) + P(2H/1H,B)P(B/1H)

To evaluate P(F/1H) and P(B/1H) use bayes rue.

P(F/1H) = P(1H/F)*P(F)/P(1H) = 0.5*0.5/(0.5*0.5 + 1*0.5) = 1/3

P(B/1H) = 2/3

P(1H) = P(1H/F)P(F) + P(1H/B)P(B)

P(2H/1H,F) = 0.5 and P(2H/1H,B) = 1 Therefore, P(2H/1H) = 0.5*(1/3) + 1*(2/3) = 5/6