De Mere's Dice Bet and the Five Pumpkins

Probability · Easy · Free problem

Part 1. You must choose which of two games gives you the better chance to win:

  • Game 1: roll a single die four times; you win if you get at least one six.
  • Game 2: roll a pair of dice twenty-four times; you win if you get at least one double-six.

Which game should you pick?

Part 2. You have five pumpkins. When you weigh them two at a time, the ten pairwise weights come out as

1, 22, 23, \ldots, 30$ pounds (each value exactly once). What is the total weight of all five pumpkins?

Hints

  1. For the dice games, compute the probability of NO success per game -- the trials are independent so it factors cleanly.
  2. Compare
    - (5/6)^4$ against
    - (35/36)^{24}$ directly.
  3. For the pumpkins, sum all ten pairwise weights and note each pumpkin is counted in exactly four pairs, so the total is (sum of pairwise weights)/4.

Worked Solution

How to Think About It: Both parts are about reframing. For the dice, compute the probability of the COMPLEMENT (no success) -- it factors into independent trials. For the pumpkins, notice that summing all the pairwise weights counts each pumpkin a fixed number of times.

Quick Estimate: Game 1: roughly $4 \times \tfrac{1}{6} \approx 0.67$ naively, but with overlap a bit over $0.5$. Game 2: roughly 4 \times \tfrac{1}{36} \approx 0.67$ naively, but the smaller per-trial chance and overlap pull it to just under $0.5$. So Game 1 looks better. Pumpkins: the ten pairwise sums add to 55$; since each pumpkin appears in $4$ pairs, the total is 55/4$.

Formal Solution:

Part 1. $P(\text{Game 1 win}) = 1 - (5/6)^4 = 1 - 0.4823 = 0.5177$. $P(\text{Game 2 win}) = 1 - (35/36)^{24} = 1 - 0.5086 = 0.4914$. Since $0.5177 > 0.4914$, choose Game 1. (This is the historical de Mere problem.)

Part 2. Label the pumpkin weights $w_1, \ldots, w_5$. Summing all $\binom{5}{2} = 10$ pairwise weights counts each pumpkin in exactly $4$ pairs: $\sum_{\text{pairs}} (w_i + w_j) = 4\sum_i w_i = 21 + 22 + \cdots + 30 = 255.$ Therefore the total weight is $\sum_i w_i = \dfrac{255}{4} = 63.75$ pounds.

Answer: Part 1: Game 1 is better ($0.518$ vs $0.491$). Part 2: the pumpkins weigh $63.75$ pounds in total.

Intuition

Part 1 is the famous Chevalier de Mere bet that helped launch probability theory: the two games look nearly identical by the naive 'multiply the count by the per-trial chance' heuristic, yet they fall on opposite sides of

/2$. The lesson is to always work with the complement for 'at least one' events and never trust the linear approximation near probability
/2$ -- the exact geometric calculation is what separates the two.

Part 2 is a counting-symmetry classic: instead of solving for individual weights, sum the constraints and exploit that each unknown appears the same number of times. The trick generalizes -- with $n$ objects, each appears in $n-1$ pairs, so the grand total is (sum of all pairwise sums)$/(n-1)$. This 'sum the equations to collapse the unknowns' move is everywhere in linear algebra and in quick mental-math interview problems.

Open the full interactive solver →