Editor’s Note: this is a chapter from David’s forthcoming short book Probability for 12 Year Olds (And Maybe You
Probability books typically spend some time showing you how to calculate what they call “permutations and combinations,” after which they show you how to use that knowledge to answer probability questions. Not me. I’m going to reverse that order. We’re going to look at some problems that are solved via the counting of combinations before showing how to actually count them. That’s because I want to be kind to your brain. When taught in the normal matter it’s likely that you would be distracted by the calculating technique you just learned rather than focusing on the logic.
Say you flip six coins (say one after another although it doesn’t really matter if it was at the same time) and want to know the chances of getting exactly three heads and three tails. It would be simple if I had specified which flips I needed to be heads and which were to be the tails. If I wanted HHHTTT the answer would simply be (1/2) (1/2) (1/2) (1/2) (1/2) (1/2) or 1/64. But that is just one of the ways to flip three of each. Those three heads can be in any of those six positions. And each of those configurations, say THTTHH, has that same probability of 1/64. Thus, we need to count all the different ways we could place those heads into three of those slots. How many different ways can we pick three out of six?
The answer is 20. Count them one by one if you like. But there’s a shortcut to getting this answer which will be explained soon. Thus, the answer to the question is 20/64 since there are 20 equally likely, mutually exclusive ways of getting three heads, each with a probability of 1/64
Had I asked about the probability of getting exactly two heads out of six flips we would need to count the number of ways the two heads could be placed in those possible six positions. That would be 15. And the answer would now be 15/64 since each configuration would again be 1/64 when we are considering six coin flips.
What if instead it was six throws of a die. And we wanted to know the probability of getting exactly three fours. Same basic technique. Putting the three fours in the first three positions followed by three non fours results in a probability of (1/6) (1/6) (1/6) (5/6) (5/6) (5/6) or 125/46,656. But those three fours can occur anywhere in the six possible positions and I have already told you that there are 20 ways to do that. So the answer is 2,500/46,656. Had I wanted exactly two fours I would multiply 125/46,656 times the 15 ways you can choose two out of six, just like in the coin flip example. 1,875/46,656
You have just learned the logic behind what they call the “binomial distribution” even though you have not been told the explicit formula. And I contend that after learning it this way you are much more likely to remember it five years from now compared to those who learned the formula first.
What about being dealt all the same suit on a poker machine. This is one of those questions that can be done a few different ways. For instance, as we have already done earlier, we can multiply fractions to figure out the probability of a flush in a specific suit, and multiply that answer by four.
(13/52) (12/51) (11/50) (10/49) (9/48) (4).
But a lot of math types do it differently. They use the combination formula to count up how many different flushes there are. (If even one card is different it’s a different combination.) Specifically, they are counting how many ways to choose five cards from 13. That, as you’ll soon see is 1,287. So, there are 5,148 total combos (since there are four suits). Now they invoke a number that you are perhaps familiar with. The total number of possible five card poker hands. Which is 2,598,960. (You’ll know why soon.) Thus, making the pat flush probability 5,148 out of 2,598,960, or about one in 505. (We’re lumping in straight flushes and royal flushes for simplicity’s sake.)
What about a pat straight? Here, slightly different thinking is required. There are ten different straights going from five high to ace high. (In this case, we won’t be lumping in straight flushes which includes the ace-high straight flush which is called the royal flush.) And there are 1,020 of each of them. Why 1,020? Because given a specific straight, there is a choice of four suits for each of the five cards and (4) (4) (4) (4) (4) = 1,024 different “straights.” Except four of them are straight flushes. Leaving 1,020. And since there are 10 different straights, there are 10,200 straights out of 2,598,960. About one in 255.
To calculate the chances of being dealt a full house one way would be to figure the chances of a specific one, say aces full of kings, and multiply times the number of possible full houses. You can be dealt three aces four different ways. That should be obvious. Somewhat less obvious is the fact that two kings can be dealt six different ways. That means there are 24 different full houses that are called aces full of kings. But there are all those other choices of three of one kind and two of the other. I have 13 choices for the trips and then 12 choices for the pair. So there are 156 different kind of full houses, each of which can be formed in 24 different ways. Thus, there are (156) (24) = 3,744 different full houses. Putting that over the total number of five card hands gives us a chance of about one in 694 of being dealt a pat “boat.”
