Say I have a coin with a chance $p$ of landing heads where $p$ is drawn from the uniform distribution from 0 to 1. Now let's say I have some fixed number $x \in (0,1)$. My question is what is the probability that the coin lands on heads, given that $p \leq x$? In other words, what is $\Pr(H | p \leq x)$?

I first attempted to use Baye's rule $$\Pr(H | p \leq x) = \frac{\Pr(p \leq x | H)\Pr(H)}{\Pr(p \leq x)}$$ but I couldn't deduce the value of $\Pr(p \leq x | H)$ or the value of $\Pr(H)$.

I then tried using an integral to solve this. My idea was that the possible values of $p$ range from 0 to $x$ and for each, they have a probability $p$ of landing on heads. Hence: $$\int_0^x p \text{ }dp = \frac{x^2}{2}$$ However, my friend tells me the answer is $\frac{x}{2}$ because I need to divide the integral by $x$ since I need to "average" it. I'm afraid I don't understand what he means.

An explanation of who is right and why would be much appreciated. Also, if one could explain how to find the values of $\Pr(p \leq x | H)$ and $\Pr(H)$ that would be much appreciated as well.

  • 1
    $\begingroup$ Your friend is right :-(. Hope you did not bet on something unaffordable... Do you want step by step guidance, or just the answer? $\endgroup$
    – Ute
    Commented Jul 19, 2023 at 0:24
  • $\begingroup$ @Ute Thank you for the response my friend. Step by step guidance would be very appreciated because I have thought a great deal about his comments but have not been able to figure out why it is true. $\endgroup$
    – timeinbaku
    Commented Jul 19, 2023 at 0:26
  • $\begingroup$ You could set the self-study tag then :-). But let's see if we can do this in a few minutes: There are two random mechanisms in the whole thing, right? Can you tell me what these are? $\endgroup$
    – Ute
    Commented Jul 19, 2023 at 0:27
  • $\begingroup$ First the selection of $p$ and then the event the coin lands on heads given the selected $p$ $\endgroup$
    – timeinbaku
    Commented Jul 19, 2023 at 0:31
  • $\begingroup$ Yeah! (I am taking a bit long, because I did not know where is best to start) $$ $$ I think your integral approach is easier to start with. How do you usually calculate the mean of a random variable $X$ when you know the density function $f(x)$? $\endgroup$
    – Ute
    Commented Jul 19, 2023 at 0:46

1 Answer 1


You can continue here if you like.

You said that you need to random variables to generate the outcome, the probability, and then the head-or-tail outcome.

To model this mathematically, we introduce a bit notation.

You have a random variable that gives you the head probability for the coin. Let's call it $P$.

Then you have a random variable that gives head or tail. Let's encode this by a r.v. $Z$ with values in $0, 1$ (tail head). $Z$ has a Binomial distribution with size $1$ and probability $P$. The mean of $Z$ is the probability to get head, because $Z$ counts successes $$ Pr(H) = \mathbb{E} Z. $$

Bayes formula

This approach would become quite complicated. You would need to look at the joint distribution of $P$ and $Z$.

but I couldn't deduce the value of $Pr(p\leq x|H)$ or the value of $Pr(H)$.

To be able to calculate $Pr(p \leq x \mid H)$, which is in terms of the new variables $Pr(P \leq x\mid Z=1)$, you would actually need to calculate $\mathbb{E} Z$ - you don't win anything with the Bayes formula here. But it looks correct :-)

integral approach

As I can see, you have applied the correct formula $$ \mathbb{E} Z = \mathbb{E}(\mathbb{E}(Z \mid P)) $$ and used used that $$ \mathbb{E}(Z \mid P) = P \implies \mathbb{E} Z = \mathbb{E}(\mathbb{E}(Z \mid P)) = \mathbb{E}P. $$ Here, in calculating $\mathbb{E}P$, seems to lie the problem,

  1. How do you usually calculate the mean of a random variable X when you know the density function f(x)?

$E(X) = \int xf(x) d x$

  1. You know that has a uniform distribution on $[0,x]$. What is its density function $f$?

the density is a zero outside the interval $[0,x]$ and a constant inside $[0,x]$. The constant must be $1/x$, so that the density integrates to 1. Thus we have $f(p) = \frac{1}{x}$ for $p\in[0,x]$ and $f(p) = 0$ if $p$ is not in the interval.

  1. can you now calculate the mean of $P$ with the integral form?

$\mathbb{E}P = \int p\cdot f(p) d p =\int_0^x p\cdot \frac{1}{x} d p = \frac{1}{x}\int_0^x p d p = \frac{1}{x}\cdot \frac{1}{2}x^2 = \frac{x}{2} $


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