Janine primes: a card solitaire

 

I had recently the idea of a card solitaire involving the first prime numbers. Call it “Janine primes”.
We play with a single traditional 52-card deck, all faces down.
Four piles will be used, they are empty at the beginning of the game.
We now place the first four cards of the deck on the table, face up and side by side – like in the hereunder example: 

(Sorry for the Yves Saint Laurent logo: the deck used here is one of the dozens belonging to my mother – a former great bridge player. She died last year, and I inherited her card decks. They still smell of the talc she used to make the cards slide smoothly. Farewell Janine!-)

The goal:
To get rid of all the cards, step by step, by placing them on the existing piles.

The game:
If two cards sum up to a prime number, you can cover them with the next two cards of the deck – face up.

We see hereunder that the player covered the above Queen and Ace of clubs (as 12+1=13) with two 9s:

We go on like that – and decide here to cover the leftmost pair 10 and 9 as its sum is 19 (we could have chosen the combination 10+7=17 instead); we get this: 

We will illustrate now the second possibility to advance in the game: the player can cover a single card if he wants – provided this card shows a prime number (2, 3, 5, 7, Jack, King). The above 7 of diamonds was chosen here – and covered hereunder with the 3 of clubs:

This 3 (a genuine "prime card") was itself covered at the next step by the 4 of clubs:

 The above central pair 1 and 4 (total 5) was covered hereunder with the pair 2 and Queen:

The above combination Jack and Queen (11+12=23) was covered hereunder by the 5 of spades and another Queen:

We will illustrate here a (bad) strategic decision: the solitaire player decided to cover the leftmost pair 5 and 2 (total 7) and got that – which unfortunately ends the game:

Indeed, no prime card is visible above, and no combination of two cards sums up to a prime.

Here is what the player should have done – we rewind to the position he messed up: 

Instead of covering the 5 of spades and the 2 of diamonds, he should have covered the 2 of diamonds and the 9 (sum 11). Why? Because whatever the values of the future appearing cards, this would leave a spare prime combination for the next step: the above 5 and Queen of spades sum up to 17.

We see hereunder that the same two appearing cards (the 6 of clubs and the Queen of diamonds) even offer two possibilities to continue the game (the possible prime pairs being now 5+6 and 5+12).

The “Math-Fun” question now:

What are the odds to win the “Janine primes” solitaire? (a game is won when all cards of the deck are placed somewhere in the four piles; in case you end with a single card to place – the 52^nd of the initial deck –, and there is no visible single “prime card” to cover, you lose).

Here is a small sketch to help computing those odds:

The upper row reads: Ace + 2 is a prime number / Ace + 4 also / Ace + 6 also, etc.

The next row reads: 4 + Ace is a prime number / 4 + 3 also / 4 + 7 also; etc.

The last sum of the last row is Queen + Jack = 12 + 11 = 23 (a prime number).

The hereunder variants might be fun to explore:

1) forget the “single” prime card rule: at each step you must cover exactly two cards (summing up to a prime). This drastically lowers the odds to win (I guess);

2) start with two piles instead of four; if you are blocked, start the 3^rd pile. And if you are stuck again, start the 4^th pile, the 5^th, the 6^th, etc. You are sure to empty the 52-card original deck but the less piles you use, the better.

3) combine the above variants 1 and 2;

4) play with exactly four piles and allow the prime sums of three cards (in addition of two or one single prime card).

Sorry if this is old hat.
Best,
É.