# 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? Commented Oct 15, 2015 at 12:41
• @Arthur Yes. There is no information about that.
– Jojo
Commented Oct 15, 2015 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? Commented Oct 15, 2015 at 12:46
• @Arthur Could you please check the edit to my question.
– Jojo
Commented Oct 15, 2015 at 13:08
• @Jojo, yes, although that should be $P(2H\mid\text{ Biased} \cap F=H)$ and so forth. Commented Oct 15, 2015 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.

– Jojo
Commented Oct 15, 2015 at 13:22
• Glad to help, and indeed cherish your intuition in maths. Commented Oct 15, 2015 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. Commented Oct 15, 2015 at 14:56
• It assumes that the same coin was flipped twice. Commented Oct 15, 2015 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