# Two biased coins. You pick a coin, flip it, and it lands heads. What is the probability it will be heads on the next flip?

You have a bag with two coins. One will come up heads $$40\%$$ of the time, and the other will come up heads $$60\%$$. You pick a coin randomly, flip it and get a head. What is the probability it will be heads on the next flip?

My approach to the problem is the following one.

We want to compute the probability $$\mathbb{P}(hh\mid h)$$. By Bayes, this is equivalent to $$\frac{\mathbb{P}(h\mid hh)\cdot\mathbb{P}(hh)}{\mathbb{P}(h)}$$ It is immediate that $$\mathbb{P}(h\mid hh)=1$$. On the other hand $$\mathbb{P}(hh)=1/2\cdot (0.6)^2 + 1/2 \cdot (0.4)^2=0.26$$ and $$\mathbb{P}(h)=1/2\cdot 0.6+1/2\cdot 0.4=0.5$$ Therefore, $$\mathbb{P}(hh\mid h)=0.52$$ Is my approach correct?

• Its $P(h|hh)=1$, and $.26/.5=.52$, but otherwise correct.
– JMP
Jun 6 at 8:46
• @JMP Thanks, I have just edited the post with the right solution. Jun 6 at 12:13