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.

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3 Answers 3


The problem boils down to the following operation:

  1. Put out the 52 cards in a row.
  2. Keep the first 25 cards in the row.
  3. 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: $$\frac{52!}{25!\times 27!}=\binom{52}{25}=477\mathord,551\mathord,179\mathord,875\mathord,952.$$

  • $\begingroup$ Less formally, but probably clearer, it's 52 × 51 × ... × . $\endgroup$
    – chepner
    Commented Aug 14, 2015 at 13:52

That's the binomial coefficient $$\binom{52}{25}=\frac{52!}{25!27!}=477551179875952.$$ For background, see https://en.wikipedia.org/wiki/Combination


I may be misunderstanding, but it looks like an $n$-card spread is just a collection of $n$ cards in which the order doesn't matter. If that is the case, the number of 25-card spreads can be calculated by counting the number of ways to choose 25 items from a collection of 52 items. We write $\binom{52}{25}$ to indicate this calculation, which is $\frac{52!}{25!(52-25)!}=477551179875952$.


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