Forty-card frustration solitaire: why you almost always lose

1. The rules and the right question

With a 40-card Neapolitan deck, reveal cards one by one while calling 1, 2, …, 10, then restarting at 1: four complete cycles. Ace counts as 1, knave as 8, knight as 9 and king as 10. You lose at the first match between the card’s value and your call; you win only by completing all forty cards without a match.

Assume all orders of the 40 cards are equally likely. The result is striking: the winning probability is about 1.56216%, and the losing probability about 98.43784%. You do not always lose: you almost always lose. Here is how to calculate these numbers.

2. A winning arrangement exists

One counterexample proves that defeat is not inevitable:

Call:  1 2 3 4 5 6 7 8 9 10
Card:  2 3 4 5 6 7 8 9 10 1

Repeat the second row four times, once for each suit. All cards are used without a single match. Winning is possible; a random shuffle rarely produces a favourable arrangement.

3. The first estimate: useful, but not exact

Before seeing the deck, each fixed position has four dangerous cards out of forty. Its match probability is 4/40 = 1/10, and its no-match probability is 9/10.

If all forty trials were independent, we would have:

P(win) = (9/10)⁴⁰ ≈ 0.01478088 = 1.47809%

This is only an estimate: cards are not replaced. If the first card is a 2, only three twos remain among 39 cards when you call 2; if the first is a 7, four twos remain. You survived the first card in both cases, but the next risk differs. The correct product uses conditional probabilities, not forty identical factors of 9/10.

4. Why it is hard: four matches on average

Imagine revealing every card even after the first match. Let X be the total number of matches. Its mean is:

E(X) = 40 · (1/10) = 4

Expectations add even without independence. A mean of four does not imply four matches in every game: some have none, others many. Winning requires precisely X = 0. The mean helps explain the difficulty, but does not by itself determine P(X = 0).

5. Counting without double-counting

The exact calculation uses inclusion-exclusion: subtract arrangements with specified matches, then correct overlaps.

A small example explains the signs. With just three distinct cards 1, 2, 3 and calls 1, 2, 3, there are 3! = 6 orders. Each position matches in two orders, so subtract 3 · 2 = 6. Every pair of matches was subtracted twice, so add 3. Finally subtract the arrangement matching in all three positions:

Match-free orders = 6 − 6 + 3 − 1 = 2

They are 231 and 312. The same principle applies to forty cards, but now we must handle four cards of each value. The symbol n! means n · (n−1) · … · 1; by convention 0! = 1.

6. One value: where 1, 16, 72, 96, 24 come from

Consider the value 7. Four distinct sevens correspond to four positions where you call 7: 7, 17, 27, 37. How many ways can we specify j matches of this value?

Choose j of the four cards, choose j of the four positions, then pair cards with positions in j! ways. This gives C(4,j)² · j!, where C(4,j) counts choices of j objects from four.

For j = 2, six pairs of cards, six pairs of positions and two pairings give 6 · 6 · 2 = 72. We impose some matches, without requiring that they be the only ones.

7. Combining the ten values

Use x as a bookkeeping label for the number of specified matches:

Q(x) = 1 + 16x + 72x² + 96x³ + 24x⁴
Q(x)¹⁰ = r₀ + r₁x + r₂x² + … + r₄₀x⁴⁰

The coefficient of x² is 72 because one value permits 72 ways to specify two matches. Every value has the same counts and uses different cards and positions, so combine ten values by multiplying ten copies of Q.

Let r_k be the coefficient of x^k in Q(x)¹⁰. It counts compatible ways to specify k matches altogether. For example:

r₀ = 1
r₁ = 10 · 16 = 160
r₂ = 10 · 72 + C(10,2) · 16²
   = 720 + 11520 = 12240

Two matches can involve the same value or two different values, giving the two terms. Obtain all coefficients by multiplying by Q ten times; the final list contains r₀, …, r₄₀.

8. Reading the exact formula step by step

The forty physical cards are distinct despite repeated values, giving 40! equally likely orders. After fixing k compatible matches, the remaining 40 − k cards can be arranged in (40 − k)! ways. They may create additional matches; alternating signs correct for those.

W = Σ(k=0,…,40) (−1)ᵏ rₖ (40−k)!
p = W / 40!

The sum has 41 terms, from k = 0 to k = 40. The sign is positive for even k and negative for odd k. First compute the integer W counting winning orders, then divide by 40!. This avoids losing precision by subtracting nearly equal decimal quantities.

p = 0.0156215686393598…

Winning probability: 1.5621568639…%.
Losing probability: 98.4378431361…%.

This is a combinatorial calculation for forty cards, not a simulation and not the answer for a 52-card deck.

9. One win in 64 games is not a deadline

With independent shuffles, each new game has winning probability p. The mean number of games until the first win, including the winning game, is 1/p ≈ 64.014.

The probability of losing n consecutive games is (1 − p)ⁿ. For 50 games it is about 45.51%; for 100 it is still about 20.71%. No wonder a player may remember never winning.

A win does not become “due” after many losses. A fresh uniform shuffle starts with the same probabilities. The calculation depends on the rules and shuffling: it does not automatically apply if calls are changed according to previously revealed cards.

Mathematical reference: Doyle, Grinstead, Snell, Frustration solitaire. Calculation adapted here to ten values and four suits.