Mathy Things

Twenty four the basic Game

Twenty four is played with a standard 52 card deck of playing cards. The aces are have the value 1, jacks eleven, queens twelve and kings 13.
The basic game proceeds by having 4 cards turned face up. The first player that can achieve the number 24 exactly using all the cards and only allowed operations (such as addition, subtraction, multiplication, division, and parentheses) wins the hand.

Should I keep trying

The first point to not is that not every combination of cards has a solution. So I went through the process of writing an algorithm to check if a specific combination has a solution. Like most people I did this by having a computer try every possible set of operations in every possible order. I don't program much and have only recently been introduced to javascript, but my algorithm appears to match others on the internet. However, mine differs in a key way. Mine does not tell its solution. It is handy to check to see if you should keep trying. Here is my version of the checker. Then we can go onto some of the interesting math behind the game.

Analysis

Since not every combination has a solution, this immediately makes several questions pop into my mind:

• What is the probability that a hand has a solution?
• Given a specific combination how many ways are there to perform operations on them?
• Do certain cards guarantee that there will be a solution?
• Do certain cards guarantee that there will not be a solution?
• What would happen if we used a different number as a target?
• what would happen if we used a different number of cards?
• What would happen if we used a broader range of numbers

Some of these questions we can tackle directly, others we may only be able to approximate solutions.

How many card combinations are there really

To know how often there is a solution, we take the number of combinations that have solutions and divide by the total number of card combinations. Easy Right?
Maybe not.
First we have to decide how to count. To do that it might be worth while to introduce some mathy definitions. From Wikipedia:

In mathematics, a combination is a way of selecting items from a collection, such that (unlike permutations) the order of selection does not matter. In smaller cases it is possible to count the number of combinations. For example, given three fruits, say an apple, an orange and a pear, there are three combinations of two that can be drawn from this set: an apple and a pear; an apple and an orange; or a pear and an orange.

That was all assuming all the cards are unique, but in the game, suit doesn't matter. We don't want to repeat a computation if the only thing that changes is the suit. At the same time, we need to know how many ways we can get the same face values by only mixing up the suits, to know how frequently a hand will show up in a game.

We broke the set of all face value combinations into types, based on whether a number was repeated. The first type, there was only one number. Because this uses all four suits, there is only one combination that gives four kings or aces or threes. In contrast, when there are four distinct face values, each could be any suit. So there are 4*4*4*4 combinations that contain the numbers A,2,3,10.

The table breaks down the ways we can get combinations by focusing on how many distinct face values are shown. The weight factor accounts for the different ways these can be combinations can occur based on altering the suits or the numbers involved.

If done correctly the total (which is the product of the weights) will match our previous total for the number of four card hands. And it does. This doesn't guarantee we did everything right, but if it was different we would know we did something wrong.

How many possible solutions do we need to check

There may be some underlying structure that allows us to determine which combinations have solutions without actually solving them, but I don't know it. The fastest way for me to solve this problem was to try all the possible ways of combining the four numbers using three operators. Take some time and try to figure out a way to go through all the possible ways of using the +,-,*./ operators on four numbers. Two key goals are to make sure you get all possible results and you don't repeat a lot of computations.

Each operation takes two numbers and returns one. We will think of each of these operations as a step. How many ordered pairs can we get from our set of four numbers. Four for the first choice and three for the second, 12 possible ordered pairs. But for the addition and multiplication operator order doesn't matter, which means 6 unordered pairs. 12*2+6*2=36 possible first steps.

Step two is similar with less numbers. Now there are three choices for the first number and 2 choices for the second, thus six ordered pairs and half as many unordered pairs 6*2+3*2=18
Step three follows the same pattern. Now there are two choices for the first number and 1 choice for the second, that means 2 ordered pairs and 1 unordered pairs 2*2+1*2=6
The total number of ways 36*18*6=3780
This is not the most efficient way to get all possible results, but it is simple and generalizable and a small number of computations for a computer to do.
I tried to do a quick test, but rounding errors caused the test to be inconclusive. Later we might go back and explore this problem further.

Computer Explorations:

My algorithm has been implements and some of the results tabulated. Of the 1820 types of hands 1361 have at least one solution, this is a success rate of only 75%. But many of the hands that had no solution were rare. If we look at all possible hands 217781 of 270725 have solutions. This is a success rate of 80%. So we expect one hand in five to have no solution.

Target Value: Maybe we should be playing 25 or 27

Call the number we are trying to reach the target. I assumed that 24 was chosen as a target because it was special in some way. It may still be, but the way I thought it would be special was that it would have a higher probability of success than other targets. I decided to test this hypothesis and found all the integers that could be created with our standard deck of cards and four operations using four card hands. The following table gives the results. They are not what I would have expected.

Relative Difficulty and altering the deck

We can try to gauge the difficulty of a particular combination. My first approach was to count how many solutions my algorithm could produce for a given hand. In general my assumption was that the more solutions there were, the more likely I would be to find them. The amount of time it takes me to solve is only loosely correlated with the rankings. This might not be an effective measure of difficulty.

If we alter the deck, either by removing or adding cards we change the likelyhood of getting a hand that has no solution and the distribution of the difficulty of the hands, but do we change it in a significant way.
If we restricted ourselves to one suit, we could only get combinations where every number was different. If we restricted to only one color, no number could show up more than twice. We could remove the face cards. These are all relatively simple alternations. We could could create a deck that had two aces, a king, a jack and three sevens removed. This removes the symmetry involved in our calculations and they become more challenging.