# You have a fair coin and a biased coin. What is the probability the second flip will be heads if the first flip was?

Derived from example 2.5.10 from Blitzstein and Hwang (page 65).

Assume $$P( heads | fair )=1/2$$ and $$P( heads | biased )=1/4$$.

If we let F be the event we draw the fair coin, $$A_1$$ be the event the first flip is heads, and $$A_2$$ be the event the second flip is heads, how can I find $$P(A_2|A_1)$$?

I find myself running into a lot of recursion when trying to parse these conditional probability problems, i.e., $$P(A_2|A_1)={P(A_1|A_2)P(A_2)\over P(A_1)}$$

Which doesn't really get me anywhere.

By LOTP, $$P(A_i)=P(A_i|F)P(F) + P(A_i|F^c)P(F^c)$$, of course, but the logical jump that I need to do to incorporate the information about F from $$A_1$$ is escaping me.

Any help?

• Plug the equation for the total probability into the denominator of your first equation. Commented Sep 21, 2022 at 19:42
• Does this end up being $P(A_2|A_1,F)P(F|A_1) + P(A_2|A_1,F^c)P(F|A_2)$, where $P(A_2|A_1,F) = P(A_2|F)$?
– jmc
Commented Sep 21, 2022 at 19:58
• Do you see the $P(A_1)$ on the left of the total probability equation? You are given all of the values on the right. Multiply and add to see what the number on the left is. Plug that number into the denominator of the first equation. You know the numbers for the terms in the numerator. Combining all of that will give you the final answer. Commented Sep 21, 2022 at 20:10
• "You know the numbers for the terms in the numerator" This is what I was talking about when I said I was running into recursive trouble. I couldn't say confidently what $P(A_1|A_2)$ is, based on the information provided. My formula in the above comment was attempt to get $P(A_1|A_2)$ in terms of probabilities I do know - is it correct?
– jmc
Commented Sep 21, 2022 at 20:47
• I see what you mean. I thought that was given. Commented Sep 21, 2022 at 21:15

Let $$B$$ be the event that the coin is biased and $$F$$ that the coin is fair. We also assume that the probability of choosing a biased coin or a fair coin are equal.

Then $$P(A_2|A_1)=\frac{P(A_2,A_1)}{P(A_1)}=\frac{P(A_2,A_1|F)P(F)+P(A_2,A_1|B)P(B)}{P(A_1)}=\frac{\frac{1}{4}\frac{1}{2}+\frac{1}{16}\frac{1}{2}}{P(A_1)}=\frac{5}{32P(A_1)}$$

Plugging in the values for $$P(A_1)=P(A_1|F)P(F)+P(A_1|B)P(B)$$ gives $$P(A_1)=\frac{3}{8}$$

Combining everything gives $$P(A_2|A_1)=\frac{5}{12}$$

• Thanks so much!
– jmc
Commented Sep 22, 2022 at 15:47
• @YaYaYaYaYaYa I do have one question for you. Are we allowed to assume there is an even probability of getting either a fair coin or a biased coin? You didn't say in your post. Commented Sep 22, 2022 at 15:50
• Yes, $P(F)=P(F^c)=0.5$.
– jmc
Commented Sep 22, 2022 at 20:07