A coin is enough

You have a coin that comes out heads with probability $p \in (0, 1)$ . But you don't need it; you need to generate an event with probability $q \in [0, 1]$ . (You know $p, q$ .) But you don't have any other source of randomness, so you have to use your coin.

You want to devise a strategy (an algorithm) that uses the coin to decide whether your event happens or not. You toss the coin; depending on that result and the previous results, you can either toss again, declare success, or declare failure.

Your strategy should declare success with probability exactly $q$ . And since you don't want to wait until the heat death of the universe, you should at least have some guarantee that this will finish.

Prove that there exists a strategy with finite expected number of tosses.
(open) Find the minimum expected number of tosses (by using the best strategy possible) in terms of $p, q$ .

This is inspired from another Brilliant problem (I forgot which one, though), where $p = \frac{1}{2}, q = \frac{1}{5}$ . I have a solution for part 2 for $p = \frac{1}{2}$ , but other values of $p$ , I don't know.

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Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Are p and q both rational?

Julian Poon - 3 years, 10 months ago

Not necessarily. In fact, my solution for $p = \frac{1}{2}$ works no matter whether $q$ is rational or not.

Ivan Koswara - 3 years, 10 months ago

I haven't had time to think about the second part of the problem, but here's my solution to the first part:

Algorithm: Consider a number line representing the reals from 0 to 1. On that line, place a marking on point $q$ .

Divide the line into 2 by placing a marking at $p$ . Now throw the coin. If it lands on heads, take the left section of the line. If it lands on tails, take the right section of the line:

If the section you have taken does not contain the marking on $q$ , terminate the algorithm. Otherwise, repeat the above steps by dividing the section you have taken into 2 smaller sections of length ratio $\frac{p}{p-1}$ , with the section corresponding to ratio $p$ to the left. Thereafter, toss the coin again and follow the same rules to indicate when to terminate:

Continue this algorithm until you terminate. There is a probability $q$ that you terminated on the left of marking $q$ and probability $1-q$ that you terminated on the right of marking $q$ .

@Julian Poon – Proving that it terminates is easy; I don't see you proving that it has a finite expected number of tosses, though. (But yes, that algorithm does terminate in finite expected number of tosses.)

@Ivan Koswara – Opps, I forgot that part of the solution :P

The expected number of throws to terminate for probability $q$ is $\displaystyle \mathbf{E} [q]=\sum _{ n=1 }^{ \infty }{ n\prod _{ 1\le i\le n } a_{ i } }$ for some sequence $a_{ i }\in \left\{ p,1-p \right\}$ . Without loss of generality, assume $p\le 1-p$ .

For the case of $p < 1-p$ , we have that

$\sum _{ n=1 }^{ \infty } np^{ n }\le \mathbf{E} [q]=\sum _{ n=1 }^{ \infty }{ n\prod _{ 1\le i\le n } a_{ i } } \le \sum _{ n=1 }^{ \infty } n(1-p)^{ n }$

Since $p, 1-p < 1$ , both sides converge and we have $\mathbf{E} [q]$ is finite.

For the case of $p=1-p=\frac{1}{2}$ , we have that

$\mathbf{E} [q]=\sum _{ n=1 }^{ \infty } \frac { n }{ 2^{ n } } =2$

Hence, $\mathbf{E} [q]$ is finite.

It's kind of amazing, how for the case of a fair coin, we can generate a probability of $\sqrt{0.5}$ with only an expected of 2 throws

@Julian Poon – That's correct.

This is not precisely your question, but I explored a version of this question for a class years ago. Here's some of the work.

Eli Ross Staff - 3 years, 10 months ago

The other Brilliant problem that you proposed is here!

Plinio SD - 3 years, 6 months ago

Yup, that's the problem. Which apparently I proposed myself.

Ivan Koswara - 3 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$