Who Wants To Be A Riddler Millionaire?

Riddler Express

On the $1 million question with choices $A,B,C,D$, you are $70\%$ sure the answer is $B$, with the other three equally plausible. You use the $50{:}50$, which removes two wrong answers and always keeps the correct one. $B$ survives. How confident are you now that $B$ is correct?

Solution

Let $L$ be the event that $B$ is one of the two answers the $50{:}50$ leaves. The lifeline always keeps the correct answer and one of the three wrong answers at random. If $B$ is correct (prior $0.7$) it certainly survives, so $\Pr(L\mid B) = 1$. If $B$ is wrong (prior $0.3$), it survives only as the lone wrong answer kept, one of three, so $\Pr(L\mid \text{not }B) = \tfrac13$. By Bayes, $$\Pr(B \mid L) = \frac{1\cdot 0.7}{1\cdot 0.7 + \tfrac13\cdot 0.3} = \frac{0.7}{0.8} = \boxed{\tfrac78} = 0.875.$$ Seeing $B$ survive nudges your confidence from $70\%$ up to $87.5\%$, because a wrong $B$ would often have been removed.

The computation

Simulate the lifeline: place the correct answer ($B$ with probability $0.7$, else one of the others), keep it plus a random wrong answer, and among the trials where $B$ survives, measure how often $B$ was right.

import numpy as np
rng = np.random.default_rng(0); runs = 4_000_000
survived = correct = 0
for _ in range(runs):
    truth = 'B' if rng.random() < 0.7 else rng.choice(['A', 'C', 'D'])
    wrong = [c for c in 'ABCD' if c != truth]
    kept = {truth, rng.choice(wrong)}        # 50:50 keeps truth + one wrong
    if 'B' in kept:
        survived += 1; correct += (truth == 'B')
print(correct / survived)                    # ~0.875

Riddler Classic

The classic birthday problem needs only 23 people for better-than-even odds that two share a birthday. How many people do you need for better-than-even odds that at least three share a birthday? (Ignore leap years.)

Solution

Work with the complement: no birthday is shared by three or more people, meaning every day is used at most twice. Among $n$ people, the number of birthday assignments with no day used three or more times, divided by $365^n$, gives $\Pr(\text{no triple})$. There is no tidy closed form, but the count is built up day by day: process the $365$ days one at a time, and for each decide whether $0$, $1$, or $2$ of the remaining people land on it, in $\binom{n-k}{j}$ ways.

Running this for growing $n$, the probability of at least three sharing first passes one half at $$n = \boxed{88}, \qquad \Pr(\ge 3 \text{ share}) \approx 0.511.$$ So $88$ people suffice, well above the $23$ of the two-person version, since coincidences of three are much rarer.

The computation

For each $n$, count the assignments with every day used at most twice (a dynamic program over the days), divide by $365^n$ for the no-triple probability, and report the first $n$ whose complement exceeds one half.

from fractions import Fraction as F
from math import comb
def no_triple(n, D=365):
    dp = [0] * (n + 1); dp[0] = 1            # dp[k]: ways for k people
    for _ in range(D):                       # add one day at a time
        nd = [0] * (n + 1)
        for k in range(n + 1):
            if dp[k]:
                for j in (0, 1, 2):          # 0, 1 or 2 people on this day
                    if k + j <= n:
                        nd[k + j] += dp[k] * comb(n - k, j)
        dp = nd
    return F(dp[n], D**n)
n = 80
while 1 - no_triple(n) <= F(1, 2):
    n += 1
print(n)                                      # 88