With reference to this question: Coin Flip Probability Independent or Not?

I give you a hat which has 10 coins inside of it. 1 out of the 10 have two heads on it, and the rest of them are fair. You draw a coin at random from the jar and flip it 5 times. If you flip heads 5 times in a row, what is the probability that you get heads on your next flip?

I attempted this question using LoTP with extra conditioning, let $H$ be the event that the next flip is heads, $R$ be the event that 5 heads are flipped in a row, $F$ be the event that a fair coin was drawn and $U$ be the event that an unfair coin was drawn.

The Law of Total Probability with extra conditioning states that $$ P(B|E) = \sum_i^nP(B|A_i, E)P(A_i) $$

Following this rationale, I produced the following working: $$\begin{align*} P(H|R) &= P(H|R, F)P(F) + P(H|R, U)P(U) \\ &= \frac{1}{2} * \frac{9}{10} + \frac{1}{10} \\ &= \frac{11}{20} \end{align*}$$

While this was mentioned in the question I have referenced, I still fail to see the problem with stating that $P(H|R, F) = \frac{1}{2}$, since if it is given that it is a fair coin and there were 5 heads in a row, wouldn't the next row still be heads with probability $\frac{1}{2}$? Or is there some other part in the working that is wrong?

  • 2
    $\begingroup$ Why do you say $P(B\mid E) = \sum\limits_i^nP(B\mid A_i, E)P(A_i)$ rather than $P(B\mid E) = \sum\limits_i^nP(B\mid A_i, E)P(A_i \mid E)$? $\endgroup$
    – Henry
    Commented Aug 27, 2023 at 8:21

3 Answers 3


You're right, that P(H|R,F)=1/2. The problem lies within the LoTP formula: $$ P(B|E)=\sum\limits_i^nP(B|A_i,E)P(A_i). $$ As you're conditioning on $E$, all your events should be within $E$. So, the right formula would be: $$ P(B|E)=\sum\limits_i^nP(B|A_i,E)P(A_i|E). $$

As a rule of thumb you can think of $E$ as your new sample space $S$. Withing your sample space LoTP would be simple:

$$ P(B)=\sum\limits_i^nP(B|A_i)P(A_i), $$

but you assume that your envets $A_i$ and $B$ are within your sample space $S$. So you can also see it as:

$$ P(B|S)=\sum\limits_i^nP(B|A_i,S)P(A_i|S). $$

So when you're conditioning on $E$ it's really almost the same as you're just changing your old sample space $S$ to a new one $E$.

In the case of your problem LoTP's formula would be $$ P(H|R) = P(H|R,F)P(F|R) + P(H|R,U)P(U|R). $$

Where $P(H|R,F)=1/2$ and $P(H|R,U)=1$. The main challenge is to calculate $P(F|R)$ and $P(U|R)$. To do this you shoul use Bayes’ theorem and LoTP:

$$ P(F|R) = \frac{P(R|F)P(F)}{P(R)} = \frac{P(R|F)P(F)}{P(R|F)P(F) + P(R|U)P(U)} $$

$$ P(U|R) = \frac{P(R|U)P(U)}{P(R)} = \frac{P(R|U)P(U)}{P(R|F)P(F) + P(R|U)P(U)} $$

All that remains is to insert the values and get the answer.

  • $\begingroup$ I strongly recommend to check out "Probability and Statistics" by Morris H. DeGroot and Mark J. Schervish link to get more familiar with all this stuff. $\endgroup$
    – Claptar
    Commented Aug 27, 2023 at 9:50

Oh, indeed it is true that, $\def\P{\operatorname{\sf P}} \P(H\mid R, F)=1/2$ . That is not your issue.

When the set of events $\{A_i\cap E: i\in[[1,n]]\}$ partitions the event $E$, then the Law of Total Probability is:

$$\P(B\mid E) = \sum_{i=1}^n\P(B\mid A_i, E)\P(A_i\mid E)$$

$\P(B\mid E) = \sum_{i=1}^n\P(B\mid A_i, E)\P(A_i)$ holds only when all $A_i$ are independent from $E$.

In this case $F$ and $U$ are not independent from $R$. Rather, because $H, F$ are conditionally independent given $R$ (as are $H, U$), you require:

$$\begin{split}\P(H\mid R)&=\P(H\mid R,F)\P(F\mid R)+\P(H\mid R,U)\P(U\mid R)\\&= \P(H\mid F)\P(F\mid R)+\P(H\mid U)~\P(U\mid R)\\&= \tfrac 12\P(F\mid R)+\P(U\mid R)\end{split}$$

Use Bayes' Rule to determine the probability for using a fair coin given the result of five heads in five tosses.

$$\P(F\mid R) =\dfrac{\P(R\mid F)\P(F)}{\P(R\mid F)\P(F)+\P(R\mid U)\P(U)}$$


Why do you say $$P(B\mid E) = \sum\limits_i^nP(B\mid A_i, E)P(A_i)$$ rather than $$P(B\mid E) = \sum\limits_i^nP(B\mid A_i, E)P(A_i \mid E) \quad ?$$

The correct version would make this answer

$$\begin{align*} P(H\mid R) &= P(H\mid R, F)P(F\mid R) + P(H\mid R, U)P(U\mid R) \\ &= P(H\mid R, F)\tfrac{P(F, R)}{P(R)} + P(H\mid R, U)\tfrac{P(U, R)}{P(R)} \\ &= P(H\mid R, F)\tfrac{P(R\mid F)P(F)}{P(R\mid F)P(F)+P(R\mid U)P(U)} + P(H\mid R, U)\tfrac{P(R\mid U)P(U)}{{P(R\mid F)P(F)+P(R\mid U)P(U)}} \\ &= \frac12\tfrac{\frac1{2^5}\frac9{10}}{\frac1{2^5}\frac9{10}+1\frac1{10}} + 1\tfrac{1\frac1{10}}{{\frac1{2^5}\frac9{10}+1\frac1{10}}} \\ &= \frac{73}{82} \end{align*}$$ as in the linked answer


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .