Can You Win At Italian Solitaire?

Riddler Classic

From Nicola Paciolla comes a game that you can win (or more likely lose?) all by yourself:

There’s a version of solitaire played in southern Italy with a deck of 40 Neapolitan cards, with four suits numbered from 1 to 10. The deck is shuffled and then cards are turned over one at a time. Flipping over the first card you say “one,” the second card “two” and the third card “three.” You repeat this, saying “one” for the fourth card, “two” for the fifth card and “three” for the sixth card. You continue your way through the deck, until you at last say “one” for the 40th card.

If at any point the number you say matches the value of the card you flip over, you lose.

What is your probability of winning the game?

Solution

In order to win:

1. The four 1s must appear only when you say "two" or "three".
2. The four 2s must appear only when you say "one" or "three".
3. The four 3s must appear only when you say "one" or "two".

WLOG, suppose the four 1s and four 2s locations in the shuffled deck have been predetermined. Where can the four 3s be located without losing?

• There are 14 times you will say "one". However, some have already been allocated for 2s.
• There are 13 times you will say "two". However, some have already been allocated for 1s.
• There are five distributions of the four 3s:
  1. None will be when you say "one", all four will be when you say "two". There is only one arrangement.
  2. One will be when you say "one", three will be when you say "two". There are four arrangements.
  3. Two will be when you say "one", two will be when you say "two". There are six arrangements.
  4. Three will be when you say "one", one will be when you say "two". There are four arrangements.
  5. All four will be when you say "one", none will be when you say "two". There is only one arrangement.

The sum of those five distributions yields the total number of ways given 1s and 2s locations have been predetermined.

However, the 2s follow the distribution between when you say "one" and "three", which affects 3s locations.

Similarly, the 1s follow the distribution between when you say "two" and "three", which affects 2s locations, which in turn affects 3s locations.

I ran code and counted the number of ways to place the four 1s, four 2s, and four 3s such that none occurs when the corresponding word is spoken. That yielded 149,967,879,125,350.

Answer

There are $28!$ ways to order the remaining numbers in the deck. There are $40!$ decks.

The probability of winning is thus:

$$\dfrac{149967879125350 \cdot 28!}{40!} \approx 5.604 \cdot 10 ^{-5}$$

Rohan Lewis

2023.02.20

Code can be found here.