A perfect hand, with some combinatorial tricks

Question: You deal the first 13 cards from a well shuffled, 52-card deck. What are the chances the 13 cards you deal are of the same suit?

There’s a couple of ways to do this – both incorporate useful techniques for solving these kind of problems.

First Answer – simple probability

The first way is just to look at the first 13 cards dealt and work out how many ways there are to deal a single-suit hand, and divide this by the number of ways there are to deal any hand.

There are 52 ways you can deal a first card. There are then 12 ways you can deal a second card if we restrict it to the suit of the first card. So we now have 52 * 12 permutations. With the third card, we’re up to 52 * 12 * 11 permutations. So we can see that the ways to deal a single-suit 13 card hand is:

= 52 * 12!

Using the same logic, we can see that:

Ways to deal any 13 card hand = 52 * 51 * 50 * … * 40. This equals:

= 52! / 39!

(Hopefully this is clear – the remaining sequence to finish making 52! is 39!).

So the probability of dealing a single-suit 13 card hand:

= (52 * 12!) / (52! / 39!) 
= 12! * 39! / 51!

Is there a way to work this out without blowing up our calculator? We can go old school and get out our factorial log tables.

The log of the answer equals the log of (12! * 39! / 51!). If you remember your log rules, this is the same as log(12!) + log (39!) – log(51!). If you don’t have log of factorial tables you can type that equation into google and it magically gives you the answer. So you get:

= log(12!) + log (39!) - log(51!) 
= -11.200723

So the log of the probability is -11.200723. Now let’s take the antilog to get our answer:

= 10 ^ -11.200723
= 6.299078 * 10 ^ -12

This is the probability of dealing a perfect deck. Take the reciprocal to get the odds to 1 against:

1 in 158,753,387,100

Holy cow – that’s what in Australia we call between Buckley’s and none.

Second Answer – let’s use some combinatorial tricks!

The first thing you have to know is: if I have a group of things of size n, how many ways are there to select a subset of size r? The anser is: n! / (r! * (n – r)!).

So for example if I had 5 balls, I could arrange 2 balls out of the 5 in: 5! / (2! * 3!) = 15 ways.

As a side note – and I cover this more at the end of this post – I’m treating this as an order doesn’t matter, repetitions not allowed arrangement.

How does this help us in this problem? Let’s think of the number of ways that I can select 13 cards from this deck. The order in which the cards are selected doesn’t matter. The answer, following our example above would be 52! / (13! * 39!). Out of all of these possibilities, how many of these give us 13 cards a particular chosen suit? The answer is only 1. The reason is that there’s only 13 cards of the chosen suit, and the arrangement I’m looking for has all of them. And since order doesn’t matter, there’s only one arrangement that has all of these 13 particular cards.

So dividing the number of arrangements with 13 cards being our chosen suit by the total number of arrangements gives us: 1 / (52! / (13! * 39!))

Now we have to multiply this by 4, as we have 4 suits.

= 4 * 1 / (52! / (13! * 39!))
= 4 * 13! * 39! / 52!

We can see that this becomes our original answer above:

12! * 39! / 51!

Some useful combinatorial definitions

It’s probably a good idea at this point to formalize a little the definitions of permutations and combinations.

Permutations are when the order of the arrangement does matter – the order is a factor in making the arrangement unique.

Combinations are when we want to get the number of arrangements without regard to order. So we can treat the arrangements as buckets we can shake up and it’s still the same arrangement.

Within these, we can also allow repetitions or not. In other words, in a given arrangement, can a particular item be used more than once?

Some examples will help:

Combinations

Repetitions allowed: Let’s say you’re ordering a sandwich at Subway and can choose 3 meats out of a possible 5. If you love salami, you could chose salami, salami and salami for your meats. And we don’t really care what order we get the meats in the sandwich – for example, bologna, salami, bologna is the same as salami, bologna, bologna.

No Repetitions: Let’s say Subway had only one of each of the meats left – so you couldn’t choose the same meat twice. Another good example is a lottery where you need to match 6 numbers drawn from a pool of 40. Once a number is drawn it doesn’t go back in to the pool.

Permutations

Repetitions: the classic example is that of a combination lock (which should really be called a permutation lock!). Order does matter, and repetitions are allowed as your combination could be 7777.

No Repetitions: How many ways are there to get the first 3 positions in a 100 person race? Order matters, and you can’t have the same person in more than one position.

The formulas:

Permutation, repetition: n ^ r
Permutation, no repetition: n! / (n-r)!
Combination, repetition: (n + r – 1)! / (r! * (n – 1)!)
Combination, no repetition: n! / (r! * (n-r)!)

It’s interesting to note that in the above, the combination no repetition case is just the permutation no repetition case divided by r!. You can see this by thinking that for arrangements with 3 elements, there will be 3! more ways to arrange them if we care about order as in the permutation case than in the combination case.

The combination no repetition case is probably the most common and comes up in card-dealing problems and the like. It is sometimes called the “n choose r” case – or nCr, and also known as the “Binomial Coefficient”.