STAT 110

Newton-Pepys Probability Lab

🎲 Interactive Monte Carlo Engine
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Total Trials
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Simulated Odds
0.00%
Exact Binomial
66.51%
MOST FAVORABLE
Bet A (Newton / Pepys)
Roll at least 1 six with 6 dice.
Exact: 66.51% Sim: --
RATIO 6:1
Bet B
Roll at least 2 sixes with 12 dice.
Exact: 61.87% Sim: --
RATIO 6:1
Bet C
Roll at least 3 sixes with 18 dice.
Exact: 59.73% Sim: --
The Newton-Pepys Intuitive Fallacy: In 1693, Samuel Pepys asked Isaac Newton which wager to take. Intuition tempts us to think 1/6 = 2/12 = 3/18 = equal odds. But expected value linearity doesn't apply directly to event probability thresholds!

Step 1: Complementary Event Strategy

Directly computing "at least k sixes" requires summing binomial probabilities. It is vastly easier to compute the complement: the probability of fewer than k sixes and subtract from 1.

P(At least k) = 1 - P(X < k)

Step 2: Bet A Breakdown (n=6, k=1)

The only complement is getting ZERO sixes in 6 rolls. Each die has a 5/6 chance of not showing a 6.

P(Zero 6s) = (5/6)⁶ = 0.3349
P(Bet A) = 1 - 0.3349 = 0.6651 (66.51%)

Step 3: Bet B Breakdown (n=12, k=2)

Complements are getting ZERO sixes OR EXACTLY ONE six in 12 rolls.

P(X=0) = (5/6)¹² ≈ 0.1122
P(X=1) = 12 × (1/6)¹ × (5/6)¹¹ ≈ 0.2692
P(Bet B) = 1 - (0.1122 + 0.2692) = 0.6187 (61.87%)

Step 4: Story Proof - Why odds decay as n increases

As n grows, the binomial variance relative to the mean decreases, narrowing the distribution around the mean μ = n/6. For Bet C (n=18, k=3), the distribution gets tighter around 3, putting more weight on exact values < 3 than Bet A puts on 0!

Active Wager: Bet A (6 dice / ≥1 six) Harvard Stat 110 Lab

Threshold probabilities in the saved dice lab

Read the explanation

The saved dice lab asks for at least one six in six independent fair dice. The easier complement is no sixes: each die misses with probability five sixths, so all six miss with five sixths to the sixth power. Subtracting from one gives about point six six five one. The bar shows about sixty-six point five percent. Expected number of sixes is one, but that expectation is not a guarantee that at least one appears. The second bet asks for at least two sixes in twelve dice. Subtract the probability of zero and exactly one from one. The result is about sixty-one point eight seven percent. The third asks for at least three in eighteen dice and gives about fifty-nine point seven three percent. All three bets have expected six-count equal to their target, yet their success probabilities differ. These bars use the same binomial complement function as the saved source. A custom two-dice threshold illustrates the whole distribution. Zero sixes has probability twenty-five thirty-sixths, exactly one has ten thirty-sixths, and two has one thirty-sixth. At least one therefore has eleven thirty-sixths, about thirty point five six percent. The source sums each binomial term below the threshold and subtracts it from one. Simulated trial frequencies fluctuate around the exact value; neither one lucky roll nor a short batch changes this calculation.

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