[Math] With a standard deck of 52 cards, how many different 25 card spreads are available

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With a 52 card standard deck, if you need to deal 25 cards, how many different spreads are possible?

For example, with a 2 card spread:

Ace of Diamonds + King of Spades

Would count as one combination. However, reversing it

King of Spades + Ace of Diamonds

Would NOT count, because the order/shuffle not the goal.

With a three card spread:

Ace of Diamonds + King of Spades + Jack of Hearts
Ace of Diamonds + King of Spades + Five of Clubs

Both examples would count as one combination. Changing the order around however like this:

King of Spades  + Ace of Diamonds + Jack of Hearts
Ace of Diamonds + Five of Clubs   + King of Spades

Neither of these would count, because they are just variations of the examples above.

Backstory: I am just curious how many different possible scribbage spreads there are, and what is the math to figuring it out.

Best Answer

The problem boils down to the following operation:

Put out the 52 cards in a row.
Keep the first 25 cards in the row.
Throw away the remaining 27 cards.

In how many ways can you do this? You can put out the 52 cards in 52! ways, assuming that the order matters. But the order should not matter, as you say, so there is a lot of double counting here. The only thing you care about is that you keep the first 25 cards, no matter what order they are in, and throw away the remaining 27 cards, again, irrespective of their order. So for each given manner of how the 52 cards are arranged, you can rearrange the first 25 cards in 25! ways and the remaining 27 cards in 27! ways, and you’ll still have the same effective spread. Therefore, for a given spread, there are $$25!\times 27!$$ occasions of double counting.

In order to avoid all these redundancies, therefore, one should divide the total number of orderings, 52!, by the number of multiplicities per spread. The total number of spreads is thus: $$\frac{52!}{25!\times 27!}=\binom{52}{25}=477\mathord,551\mathord,179\mathord,875\mathord,952.$$

Related Solutions

[Math] How many different ways are there to get a 21 in Black Jack

This Python code by @niemmi on Stack Overflow says that there are 186,184 ways.

import operator
from math import factorial

def bcoef(n, k):
    return factorial(n) / (factorial(k) * factorial(n - k))

def combinations(limit, start, used):
    if limit == 0:
        # For each face value figure out how many combinations can be used
        combs = (bcoef(4 if i != 10 else 16, x) for i, x in enumerate(used))
        res = reduce(operator.mul, combs, 1)
        return res

    res = 0
    for i in range(start, 12):
        if i > limit:
            break

        index = i if i != 11 else 1

        if used[index] < 4:
            used[index] += 1
            res += combinations(limit - i, i, used)
            used[index] -= 1

    return res

print combinations(21, 1, [0] * 11) # 186184

[Math] Number of $13$-card hands satisfying the stated conditions

a) There are only $4$ ways that all $13$ cards can have the same suit because there are only four possible suits (we can have all $13$ be hearts, spades, diamonds, or clubs).

b) If we know that all $4$ aces are part of the $13$ cards, then we are really only choosing $9$ cards out of $48$. We are choosing $4$ of the $4$ aces, and then there are $48$ remaining cards, from which we are choosing $9$. So, the answer is ${4\choose 4}{48\choose 9}$.

c) There are $13$ possible numbers. The first card we choose can take any number ($13$ options). The second card we choose can take any number except for that of the first number $(12$ options), and so on. The answer is $13!$ because there are $13 \cdot 12 \cdot 11 \cdot \cdots 1$ ways to do this.

d) There are $13$ total spades in the deck, and we want exactly $7$ of them. So, there are ${13 \choose 7}$ ways to select the spades. We still have six more cards that can be anything but spades, though. There are ${39 \choose 6}$ ways to select them. Therefore, our final answer is ${13 \choose 7} \cdot {39 \choose 6}.$

e) Now, we want at least seven of the cards to be spades. So, our answer becomes the number of ways to have exactly $7$ spades plus the number of ways to have $8$ spades, all the way up to $13$. Mathematically, this is given by ${13 \choose 7} \cdot {39 \choose 6} + {13 \choose 8} \cdot {39 \choose 5} + {13 \choose 9} \cdot {39 \choose 4} + {13 \choose 10} \cdot {39 \choose 3} + \ldots$.

Best Answer

Related Solutions

[Math] How many different ways are there to get a 21 in Black Jack

[Math] Number of $13$-card hands satisfying the stated conditions

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