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Given a deck of cards and N players, once I have drawn a card, how can I determine my odds of having the highest card?

I want to know what the formula is for calculating a players odds of having the highest card among players once they have each drawn a card but each player only knows the card he has drawn, the number of other players, and the cards that were in the deck before he drew his card.

For example, lets assume a fresh deck and 4 players. If I draw a 7. What are the odds that I will have the highest card among all players?


I intend to use this in some code I am writing as a demonstration, and I just want to make sure my logic is correct.

So far I figure that I need the total number of cards in the deck minus 1 (for the card I drew), lets call it C (so for a fresh deck that would be 52, but its still a variable)

Then I need to determine how many cards in the deck are lower than my card, lets call it L.

  • So for the first other player I need to do L / C
  • And for the second other player I need to do L - 1 / C - 1
  • And for the third other player I need to do L - 2 / C - 2

And I need to multiply those results together (L / C) x (L - 1 / C - 1) x (L - 2 / C - 2) to get the odds.

Am I right? Also, I am sure there is probably a better way to do this if it is right.

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  • $\begingroup$ Not sure the rules are clear. Are all the cards ranked or are there suits? Phrased differently, are ties possible? $\endgroup$ – lulu Jan 10 '18 at 21:28
  • $\begingroup$ @lulu I believe the solution works either way; the only question is whether you choose to include "ties" in the definition of $L$ or not. $\endgroup$ – user326210 Jan 10 '18 at 21:29
  • $\begingroup$ @user326210 Yes, that sounds right. $\endgroup$ – lulu Jan 10 '18 at 21:33
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Yes, that's exactly right.

Here's another way to compute it using combinations. Imagine you take out a card, then spread out the $C$ remaining cards in a row. Your $N$ friends each pick their own card from the row of cards. If that row contains $L$ cards that are lower than yours, then there are ${L \choose N}$ ways of choosing N cards lower than yours, out of ${C \choose N}$ total ways of choosing any N different cards from this list.

That's the same as:

$$\frac{L!}{N! (L-N)!} \div \frac{C!}{N! (C-N)!} = \frac{L!}{(L-N)!}\div \frac{C!}{(C-N)!} = \frac{L\times(L-1)\times \ldots (L-N+1)}{C\times(C-1)\times\ldots\times(C-N+1)}$$

which is your answer exactly.

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  • $\begingroup$ Wow, Thanks. I am just curious, does this kind of function have a name? One that identifies this kind of formula? $\endgroup$ – Inbar Rose Jan 10 '18 at 22:59
  • $\begingroup$ That's a good question. I'm not sure. There's terminology for "rising and falling factorials" for example, but that's more of a notation than a combinatoric formula. $\endgroup$ – user326210 Jan 10 '18 at 23:49
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Yes, that works. It's also equivalent to $\dfrac{\binom{L}{3}}{\binom{C}{3}}$

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