23 November 2007

strategy in Tammet's solitaire

This was written a few months ago; I'm just getting around to posting it now for some reason.

Recently I read Daniel Tammet's book Born on a Blue Day: Inside the Extraordinary Mind of an Autistic Savant. Tammet is a savant with Asperger's syndrome; he's very good as a mental calculator; he also is synaesthetic. In his particular case, he experiences numbers and words with certain colors, which explains the title of the book. He also sometimes sees numbers as shapes, to which he attributes his calculational ability; there's an interesting point where he explains that he sees multiplication of numbers by seeing the "shapes" that he associates them come together, and the space in between is the shape of the product. This seems to imply that each prime number would have to have some basic sort of shape.

At one point he mentions a solitaire game to be played with cards that he invented. The game works as follows:

Some of the time I played a simple form of solitaire of my own creationo, with a deck of cards in which each card was given a numerical value: ace as 1, jack 11, queen 12, king 13, while the numbers on the other cards determined their values. the object of the game is to keep as many cards as possible from the deck. At the start, the deck is shuffled and then four cards ar played onto a pile. If, after the first card, the total value of the cards in the pile is at any point a prime number, then those cars are lost....
The player now decides whether to risk putting another card onto the pile or to start a new pile from scratch. If the player decides not to risk a new card on the pile, then the cards from the pile are safe and are retained. If the player plays a new card and the total reaches a prime number at any point, then the whole pile of cards is lost and a new pile is started. The game ends when all fifty-two cards in the deck have been played into piles, some lost and some successfully held. The player counts up the total number of safely retained cards to work out his final score.

Presumably the reason for not declaring the pile dead if the first card is prime is because the chances of the number on any given card being prime are just too high -- six out of the first thirteen natural numbers are prime! The first natural question to ask is the probability that that initial four-card pile is allowed to live. It wouldn't be hard to compute this exactly, but I figured I'd get a better feel for the game by playing with actual cards. A fifty-two card deck naturally breaks up into thirteen such piles; I went through a deck of cards ten times, seeing 130 piles, and got fifty-three piles which were allowed to live -- about 41% of them.

I then ran the following MAPLE code:

S := 0;
for i from 1 to 13 do
for j from 1 to 13 do
for k from 1 to 13 do
for l from 1 to 13 do
if not(isprime(i+j) or isprime(i+j+k) or isprime(i+j+k+l)) then S := S+1; fi;
od; od; od; od;

This loops through all the quadruplets (i, j, k, l) with each member between 1 and 13; 1 is added to the accumulator S for each quadruplet which corresponds to a good pile. There are 10026 possible "good" piles, it turns out, out of a total of 28,561; that's about 35%. Going through the deck ten times as I did, you'd expect 46 good piles; I got 53, which isn't that far off given the small sample size.

(Note that implicit here is the assumption that each quadruplet is equally likely; in fact, those which contain repetitions are less likely when you're dealing with an actual deck of cards.)

Incidentally, if 13 is replaced by some other number n, then the probability of a "good" quadruplet increases as n is increased. If n=2 it's 1/16 -- the only good quadruplet is (2, 2, 2, 2). If n = 10 we get 0.3074. If n = 20 there are 64575 good quadruplets out of 160000, and the probability of a good quadruplet is 0.4035. If n=52 (this corresponds to each card in the deck having its own number) the probability of a "good" quadruplet is up around 48%. I suspect the probability approaches one as n gets large, but this code runs quite slowly as n gets larger. There are faster ways to do the computation, but they take more coding.

Now, let's say we're dealt four cards, and their sum is k. Do we want a fifth? As Tammet points out, you want a fifth card if there aren't that many primes in the next thirteen numbers. My instinct is that you only want to "hit" (somehow the blackjack terminology seems natural here) if the sum has probability less than one-fifth of being prime -- that is, if there are two or less primes in the numbers from k+1 to k+13. k can be at most 52; it turns out that you should only hit on 44 or 45, in this strategy.

But this is actually the correct strategy for maximizing the number of cards you get on a single trick. (We're trading in having four cards with certainty to having five cards with at least a 4/5 probability.) The object of the game, as stated, is to maximize the number of cards you get going through the whole deck. with the naive strategy of always staying after the fourth card, we get to keep on average 35% of the cards.

So let's say I have a four-card pile going, with sum k. Considering those four cards and the next one, I expect to keep 4.35 of them if I stay; I expect to keep 5m/13 of them if I hit, where m is the number of composites between k+1 and k+13. So hitting is the right move exactly when 4.35 > 5m/13, or m > 11.31; I only want to hit if the numbers from k+1 to k+13 contain zero or one primes. But k is at most 52. The sequence of primes begins

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67...

and so I should never hit. I wonder if Tammet is aware of this?

Obviously, though, not hitting makes the game less interesting...

Daniel Tammet has a website and a blog

1 comment:

Charles Tye said...

Doesn't the strategy you outline assume replacement of the cards?

Without replacement, it would be quite different with the end-game being more and more predictable.

The last card to be played is known with certainty. Say your pile is 15 and the last card is a 3. Then it would be obvious to hit.

Isn't the fact that the cards are not replaced the whole point of the game?