I randomly chose between 2 coins. One of the coins has a 0.8 chance of heads and a 0.2 chance of tails. The other is a fair coin that has a 0.5 chance of either heads or tails.

I flip this coin twice and get 2 heads. What's the probability that my next flip is heads?

I tried using Bayes' Theorem:

$$ P(H|HH) = \frac{P(HH|H)P(H)}{P(HH)} $$

But then $P(HH|H)$ is not easy to solve...


2 Answers 2


Let $HH$ be the event that we get $2$ heads in a row. Let $H_3$ be the event the third toss is a head. We want $\Pr(H_3|HH)$. Maybe we start from something a little simpler than the Bayes' Theorem that you used, essentially the definition of conditional probability: $$\Pr(H_3|HH)=\frac{\Pr(H_3 \cap HH)}{\Pr(HH)}.$$

We calculate the two probabilities on the right. For $\Pr(HH)$, note that two heads in a row happens with probability $(4/5)^2$ if we use the funny coin, and with probability $(1/2)^2$ if the coin is the ordinary coin. It follows that $$\Pr(HH)=\frac{1}{2}\left(\frac{4}{5}\right)^2+\frac{1}{2}\left(\frac{1}{2}\right)^2.$$

A similar calculation gives us the probability of $HH$, followed by $H_3$: $$\Pr(HH\cap H_3)=\frac{1}{2}\left(\frac{4}{5}\right)^3+\frac{1}{2}\left(\frac{1}{2}\right)^3.$$ Divide.

  • $\begingroup$ Thank you for fixing typo. $\endgroup$ Oct 14, 2012 at 4:52
  • $\begingroup$ because isn't this the case ? $$Pr(H_1, H_2) = Pr(H_1) * Pr(H_2) = (Pr(H_1|C_1) * Pr(C_1) + Pr(H_1|C_2) * Pr(C_2)) * (Pr(H_2|C_1) * Pr(C_1) + Pr(H_2|C_2) * Pr(C_2))$$ where $C_i$ represents selecting coin $i$ $\endgroup$ Jan 10, 2015 at 13:26
  • $\begingroup$ I guess you did this, $$Pr(H1,H2) = \sum_i Pr(H1, H2 | C_i) P(C_i) = \sum_i Pr(H1 | C_i) Pr( H2 | C_i) P(C_i)$$ which makes more sense, but may I know what makes the equation in my previous comment wrong ? I mean, aren't the heads independent ? $\endgroup$ Jan 10, 2015 at 13:54
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    $\begingroup$ The answer you had proposed would be correct if we picked a coin, tossed it, replaced it, picked and tossed again. But in the problem we pick a coin and flip this coin twice. $\endgroup$ Jan 10, 2015 at 15:13

If you had coin $A$ (the funny one), you would have a 0.64 chance of two heads; if you had coin $B$ (the fair one), you would have a 0.25 chance of two heads. Assuming you selected between $A$ and $B$ equiprobably, your probability of having coin $A$ is $${0.64\over 0.25+0.64} = {64\over 89}$$

and your probability of having coin $B$ is $1-{64\over 89} = {25\over 89}$.

Your chance of getting a head on the next flip is therefore $${64\over 89}\cdot \frac45 + {25\over 89}\cdot \frac12={637\over 890}\approx 71.6\%.$$


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