# Diamond Rush: Back to... um

It is summer in the Pse School Of Logicians, and everyone studying there is dying for a break (they absolutely hate school, and can do anything to get out of it). And because the teachers there are nice, the students there are offered a chance to be off school for as long as they like, given that the students there are perfect logicians.

All students are off school starting from the first of July, but they are each invited to the gates at two in the afternoon every day. There, the principal would show everyone decks of cards with each and every student's name on it. These are not normal decks, but extra special decks with two sets of only diamonds, like this: $$\rm A\diamondsuit, Q\diamondsuit, K\diamondsuit, J\diamondsuit, 10\diamondsuit, ..., 2\diamondsuit$$ $$\rm A\diamondsuit, Q\diamondsuit, K\diamondsuit, J\diamondsuit, 10\diamondsuit, ..., 2\diamondsuit$$

Each student will then pick a card randomly with a uniform distribution, then the card is returned to the deck. Then, if they did not pick an ace, they are asked to decrement any number of the rest of the cards by the value of the card: $$\rm Q = 13, K = 12, J = 11, 10, ..., 2$$ That is, for example, if the card is a five and the rest of the cards are all tens, they can turn five of the tens into nines, or one into an eight and three into sevens, or even turn one into a five and leave the rest unchanged. They can decrement the card they picked. Their deck will be shown and they can choose which cards to decrement.

However, they cannot decrement a two (because it is the lowest card). If at any point, a student's deck becomes all twos, they will be sent back to school the next morning. If their card is 'too big' (for example, they picked a ten and all other cards are twos), their card would be returned immediately and they cannot do anything with their deck.

If they do pick an ace, they can add one to a non-ace card in their deck. That is, their deck will be shown, and they pick a non-ace card, add one to its value then return it back to the deck. If at any point, a student's deck becomes all aces, they can be off school every day for the rest of their life.

Question: What is the probability a student gets to be off school for the rest of their life, that is, they make their deck all aces before they make their deck all twos?

• How do you make aces. Is it one up after Q? Commented Apr 19 at 21:08
• Reassigning the numerical values of Q and K doesn't seem to change anything.
– Bass
Commented Apr 21 at 6:56
• @FlorianF Yes, you can make an ace by adding one to a queen. Commented Apr 21 at 8:48
• What happens if the game does not end? I.e. there are only 2's and a single 3. Commented Apr 21 at 10:18

0% or very very very close.

because

There is a huge bias towards lowering the total value of your deck.

The best strategy (without proof) seems to be to increment the largest card and decrement the smallest ones. This maximizes the probability of making aces and doesn't change the average number of points you remove when you don't pick an ace. This strategy quickly gets you to the state where there are only aces and twos and one card you try to push towards becoming an ace.
But even if you reach the point where half of the deck is aces, you still must fight an uphill random walk where you equally likely increment by 1 or decrement by 2 or more. It is almost certain you will progress down and lose more aces rather than getting a new ace.

So you are inexorably pushed toward losing all points and ending up with only twos. It is virtually impossible to get all aces. One caveat is that you might end up with all twos and a single three. In that situation the game stalls. Whatever you pick you cannot do anything so you reached a dead end. But the question is: "what is the probability they make their deck all aces before they make their deck all twos". If the game stalls you certainly won't make all aces. So it doesnt' count as success.

In conclusion, I don't know the exact value but the probability to win that game is "astronomically" small and is therefore not worth being played. Unless you can skip school while playing the game.

There is an easily calculable lower bound on the probability: in the case that aces are selected consecutively each day, and through best play, the player chooses the card to increment that will result in additional aces soonest (thereby increasing the probability of aces being drawn on subsequent days). As noted by another answer, this is achieved by incrementing the greatest non-ace value each time. The number of increments required is equal to 2 (the number of suits) times the sum of differences below the value of an ace across each suit; that is, 1 + 2 + 3 + ... + 12, which has a familiar formula: n*(n+1)/2 = 156 in this case.

So to calculate the probability of an ace being drawn the first 156 times, which must necessarily be less likely than all possible events that lead to the deck being full of aces, we know there will be 156 terms multiplied together. In each term the denominator will be 26, the total number of cards in the deck. The first term will be 2/26, because there are 2 aces in the deck initially. After the first ace is drawn and a king converted to an ace, there will be 3 aces and so the second term will be 3/26. The same logic applies to the third term (4/26) but the fourth term will also be 4/26 because the next highest card needs to be incremented twice before yielding an ace. It soon becomes evident that we can 'collapse' the numerators in the following pattern: 2^1 * 3^1 * 4^2 * 5^2 * 6^3 * 7^3 * 8^4 * 9^4 * 10^5 * 11^5 * 12^6 * 13^6 * 14^7 * 15^7 * 16^8 * 17^8 * 18^9 * 19^9 * 20^10 * 21^10 * 22^11 * 23^11 * 24^12 * 25^12. The denominator is 26^156, of course.

I make that out to be 1.45694338E−34. Of course this is only a lower bound, and the actual probability will be slightly higher, as there are a few more (read: infinitely many) ways a full deck of aces could be achieved besides drawing only aces from the start. But the way to calculate the true probability of these permutations is beyond my skill in statistics, so I leave that up to someone else.