Two weighted coins, determining which has a higher probability of landing heads

Question

A friend of mine asked me the following question, and I am not sure how to solve it:

You are given two weighted coins, $C_1$ and $C_2$. Coin $C_1$ has probability $p_1$ of landing heads and $C_2$ has probability $p_2$ of landing heads. The following experiment is preformed:

Coin $C_1$ is flipped 3 times, and lands heads 3 times.

Coin $C_2$ is flipped 10 times, and lands heads 7 times.

Based on this experiment, choose the coin which is more likely to have a higher probability of being heads. In other words, which is more likely: $p_1>p_2$ or $p_2>p_1$.

Intuition tells me coin $C_1$ is the better choice, but this could be wrong, and I am wondering how do you solve this in general. Consider the experiment, $C_1$ is flipped $n_1$ times and lands heads $m_1$ times, $C_2$ is flipped $n_2$ times and lands heads $m_2$ times.

Thanks for the help,

Edit: I think this might answer some questions: Suppose that the probabilities of the coins, $p_1$ and $p_2$ are chosen uniformly from $[0,1]$.

Answer 1

Given a specific probability $p$ of heads, the probability of getting $h$ heads and $t$ tails is just the binomial distribution: $P(H = h, T = t | n, p) = p^h (1-p)^t {n \choose h}$ (though we consider $p$ as varying, rather than $n$ and $h$.

With a uniform prior $g(p) = 1$, that probability is just the weight. The integral of this is $\frac{1}{1 + h + t}$, giving a probability density of $(1 + h + t) p^h (1-p)^t {n \choose h}$. The mean is the integral of $p$ times this, is $(1 + h + t) \int p^{(h+1)} (1-p)^t {n \choose h} dp$. Reusing the result from last time, we know that this must be $(1+h+t) {n \choose h}/{n+1 \choose h + 1}/(2+h+t) = (1+n)(h+1)/(n+1)(n+2) = (h+1)/(n+2)$. This is slightly "hedged toward the center" from the naïve estimator $h/n$ (which is the peak of the distribution).

Another common prior that a Bayesian might use is the beta distribution. It's handy because it is a conjugate prior for the binomial distribution. After collecting data generated by the binomial distribution, the probability is still in the form of a beta distribution. In fact, the uniform prior is just the beta distribution with $\alpha = \beta = 1$. Heads and tail each just add one to the parameters $\alpha$ and $\beta$ respectively. The integrals were essentially worked out above -- factorials generalize to $\Gamma(x+1)$. It's often considered that this case of $\alpha = \beta = 1$ is too conservative, and that $\alpha = \beta = 1/2$ "assumes less" and "lets the data speak more".

With the Uniform Prior (Beta(1,1)), $\overline{p} = (h+1)/(n+2)$:
$C_1$, 3 heads, 0 tails: $\overline{p} = 4/5$
$C_2$, 7 heads, 3 tails: $\overline{p} = 8/12 = 2/3$

$C_1$ is expected to do better.

With the Beta(1/2, 1/2) prior, $\overline{p} = (h+1/2)/(n+1) = (2h+1)/(2n+2)$:
$C_1$, 3 heads, 0 tails: $\overline{p} = 7/8$
$C_2$, 7 heads, 3 tails: $\overline{p} = 15/22$

$C_1$ is expected to do better.

Actually calculating the chances of $C_1$ being better than $C_2$ involve a rather nasty integral, but the calculated $p$ value is enough to tell you which is the better bet.

Answer 2

As Henry mentions, I think one needs some information about the prior distributions of the weights.

Denote by $r$ the weight of a coin. Suppose that the weight has some prior distribution $g(r)$. Let $f(r|H=h, T=t)$ be the posterior probability density function of $r$ having observed $h$ heads and $t$ tails tossed. Bayes' Theorem tells us that: $$ f(r|H=h, T=t) = \frac{Pr(H=h|r, N = h+t)g(r)}{\int_0^1 Pr(H=h|r, N = h+t)g(r)\ dr}. $$

This should allow you to answer your question. Once you have the posterior pdf for each coin, just find their respective expected weights.

Answer 3

You could do the integration which wnoise is talking about approximately using the following R code, and it is easily adapted to other cases:

> n <- 1000000                      # number of cases to simulate for integration
> prior  <- c(1,1)                  # Beta(1,1) is uniform prior
> coin_1 <- c(3,0)                  # number of heads and tails observed
> coin_2 <- c(7,3)                  # number of heads and tails observed
> p_1    <- rbeta(n, prior[1]+coin_1[1], prior[2]+coin_1[2])
> p_2    <- rbeta(n, prior[1]+coin_2[1], prior[2]+coin_2[2])
> p_diff <- p_1 - p_2
> length(p_diff[p_diff > 0]) / n    # proportion with p_1 > p_2
[1] 0.758118

Answer 4

I agree with the beta approach, but given the question, I think it makes more sense to plot out the results and compare visually:

$x$ <- seq($0,1,$length $= 1000$) #set $x$ from $0$ to $1$ since we're looking at probabilities

$y_1$ <- dbeta$(x,4,1)$ #calculate density based on a prior of $[1,1]$ and a posterior of $[3,0]$.

$y_2$ <- dbeta$(x,8,4)$ #calculate density based on a prior of $[1,1]$ and a posterior of $[7,3]$.

plot($x,y_1$,type = "l") #plot density of coin 1 in black

lines($x,y2$,col $= 2$) #plot density of coin 2 in red

This yields the following:

Think of the densities as representing the "probability of probability".

• Coin 1 (represented by the black line) has a higher probability of having a higher probability of showing heads.

• Coin 2 (represented by the red line) has a lower probability of having a higher probability of showing heads (or a higher probability of having a lower probability of showing heads...).