A biased coin

Probability Level 4

You have an unfair coin, with a probability, $p$ , of landing on heads.

This probability is a linear distribution in the interval $[0,1]$ defined by the following function.

$P(p) = 2p$

You flip it and get heads.

What is the probability that $p>\frac{1}{2}$ ?

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\frac{1}{2}

\frac{3}{4}

\frac{7}{8}

\frac{2}{3}

\frac{1}{4}

None of these answers

1

0

1 solution

Geoff Pilling
Nov 5, 2016

Let $H =$ The event that heads is flipped.

Using Baye's Theorem :

$P(p>\frac{1}{2}|H) = \frac{P(H|p>\frac{1}{2})\times P(p>\frac{1}{2})}{P(H)} = \frac{\int_\frac{1}{2}^1 r^2 \, dr}{\int_0^1 r^2 \, dr} = \boxed{\frac{7}{8}}$

But if the probability of the probability of getting heads being p is 2p, then isn't it impossible for p to be greater than 1/2, considering that the probability of p being any value greater than 0.5 is greater than 1?

Jacob Swenberg - 4 years, 7 months ago

Good question... However, this is a probability distribution. i.e. It is a way of determining the probability of being between any two numbers. It is not a graph showing what the probability of any given value is. i.e. The probability of p being any particular number, say 2/3, is precisely zero. In order to use it, you need to integrate between the two numbers of interest.

@Calvin Lin can you think of, perhaps, a more clearer way that the question could be stated to avoid this confusion?

Geoff Pilling - 4 years, 7 months ago

It isn't about having a "clearer way to state the confusion". This is a very common misconception, confusing discrete probability with uniform probability. Under discrete, we want the sum of non-negative numbers to be 1, which is why there is an upper bound of 1. Under uniform, we want the integral of non-negative numbers to be 1, but there is no apriori upper bound on the values.

For example, if we consider the uniform distribution on the interval $[0, \frac{1}{2} ]$ , then the uniform probability means $P ( X = x ) = p$ for $x \in [ 0, \frac{1}{2} ]$ , and $\int_{\mathbb{R}} P(X=x) \, dx = 1$ would then imply that $p = 2$ . It is not true that "The probability that $P ( X = \frac{1}{4} )$ is 2." In fact, the probability that $P ( X = \frac{1}{4} ) = 0$ , because we're supposed to take the integral of the domain.

Calvin Lin Staff - 4 years, 7 months ago

A biased coin

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