Can you find the luckiest coin?

A FiveThirtyEight Riddler puzzle.

By Vamshi Jandhyala in mathematics Riddler

February 18, 2022

Riddler Express

It’s time for a random number duel! You and I will both use random number generators, which should give you random real numbers between $0$ and $1$. Whoever’s number is greater wins the duel!

There’s just one problem. I’ve hacked your random number generator. Instead of giving you a random number between $0$ and $1$, it gives you a random number between $0.1$ and $0.8$.

What are your chances of winning the duel?


The random generator that I have generates numbers according to $X \sim \mathcal{U}[a,b]$ where $0\leq a \leq b\leq1$ and the random generator that you have (in the general case) generates numbers according to $Y\sim\mathcal{U}[0,1]$. We need the probability, $\mathbb{P}[X< Y]$. This is given by the area of the trapezium $EFGH$ divided by the area of the rectangle $EFIJ$. We have $|EF| = |IJ| = b-a$, $|FG|=b$, $|EH|=a$ and $|EJ|=1$ in the figure below

$$ \mathbb{P}[X< Y] = \frac{\frac{1}{2}(b-a)(b+a)}{(b-a)} = \frac{b+a}{2}. $$

In our particular case $a=0.1$ and $b=0.8$, therefore the probability of me winning is $\textbf{0.45}$.

Computational validation

The probability of me winning the duel as per the simulation below is $\textbf{0.45}$ which validates the result we got earlier.

from numpy.random import uniform

def prob_win(p1l, p1h, p2l, p2h, runs = 1000000):
    total_wins = 0
    for _ in range(runs):
        x, y = uniform(p1l, p1h), uniform(p2l, p2h)
        if x > y:
            total_wins += 1
    return total_wins/runs


Riddler Classic

I have in my possession 1 million fair coins. Before you ask, these are not legal tender. Among these, I want to find the “luckiest” coin.

I first flip all 1 million coins simultaneously (I’m great at multitasking like that), discarding any coins that come up tails. I flip all the coins that come up heads a second time, and I again discard any of these coins that come up tails. I repeat this process, over and over again. If at any point I am left with one coin, I declare that to be the “luckiest” coin.

But getting to one coin is no sure thing. For example, I might find myself with two coins, flip both of them and have both come up tails. Then I would have zero coins, never having had exactly one coin.

What is the probability that I will — at some point — have exactly one “luckiest” coin?


Let $\mathbb{P}[N]$ be the probabillity that we will have exactly one “luckiest” coin starting with $N$ coins. We have the following recurrence relation:

$$ \begin{align*} \mathbb{P}[N] &= \sum_{i=1}^N \mathbb{P}[i] \frac{N \choose i}{2^N} \\
\implies \mathbb{P}[N](1 - \frac{1}{2^N}) &= \sum_{i=1}^{N-1} \mathbb{P}[i] \frac{N \choose i}{2^N} \\
\implies \mathbb{P}[N] &= \frac{1}{2^N-1}\sum_{i=1}^{N-1} \mathbb{P}[i] {N \choose i} \\
\end{align*} $$

because the probability of ending up with $i$ heads when you flip $N$ coins is $\frac{N \choose i}{2^N}$ and then it is equivalent to starting the game with $i$ coins. When $N=1$, $\mathbb{P}[N]=1$.

Computational Solution

From the code below, we get $\mathbb{P}[100] \approx \textbf{0.7214}$. Assuming convergence, $\mathbb{P}[1000000] \approx \textbf{0.7214}$

from functools import lru_cache
from math import comb

def prob_luckiest_coin(n):
    def prob(n):
        if n == 1:
            return 1
            total_prob = 0
            for i in range(1, n):
                total_prob += prob(i)*comb(n,i)
        return total_prob/(2**n-1)
    return prob(n)