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I'm designing a game with a custom deck. The deck has 5 suits (hearts, spades, clubs, diamonds and crosses) and 5 ranks (1 to 5) for 25 cards total. I'm trying to calculate the odds of revealing 3 cards of the same suit if I draw 10 cards total.

In my head, that math would be: $$ \frac{{5 \choose 1}{5 \choose 3}{20 \choose 7}}{{25 \choose 10}} $$

Where ${5 \choose 1}$ is to pick a suit, ${5 \choose 3}$ is to pick any 3 cards from that suit, ${20 \choose 7}$ is to pick any random 7 cards from the remaining 4 suits, and ${25 \choose 10}$ is the total number of ways to pick 10 cards from a deck of 25. The problem is that my numerator ends up being $3,876,000$, and the denominator is $3,268,760$. Obviously I expect the answer to be very close to 100%, but in this case the probability is greater than 1 (which makes no sense, as you could draw 2 cards from each of all 5 suits and not have met the condition).

What am I doing wrong?

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  • $\begingroup$ What do you mean $3$ cards of the same suit. Selecting ${20 \choose 7}$ can include also another $3$ cards of the other same suit. $\endgroup$
    – Lion Heart
    Commented Jun 28 at 21:36
  • $\begingroup$ With this deck, and that hand, you have to worry about drawing three cards from each of two suits or even three suits. $\endgroup$
    – lulu
    Commented Jun 28 at 21:38
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    $\begingroup$ As you remark, the only way to fail to get three of a suit is if you get two each of every suit. Work out the probability of that. $\endgroup$
    – lulu
    Commented Jun 28 at 21:40
  • $\begingroup$ "Three cards" or "three or more cards"? Your $\binom{20}7$ suggests that you require "three". However, those such as @lulu and MrPuffer who reckon that you succeed unless you draw two cards of each suit seem to have taken you as meaning "three or more". Does SSSSHHHHDC count or not? $\endgroup$
    – Rosie F
    Commented Jun 29 at 14:24

2 Answers 2

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So the problem lies in the fact that you're overcounting, Why? You're clearly calculating ${5 \choose 1}$ combinations of suits and ${5 \choose 3}$ combinations of ranks. Where it goes wrong is when you're multiplying by ${20 \choose 7}$, because it adds the option to draw 3 of the same suit at least another time (from another suit clearly).

Let
H = Hearts
D = Diamonds
C = Clubs
S = Spades
X = Cross

Which means that HHHDDDCXSS and DDDHHHCXSS are both being counted, which is wrong as they're different permutations, but the same combination (assuming they represent the same rank in this example). Where HHH is the 3 cards you select first and then the rest gets selected at ${20 \choose 7}$, at the second option the DDD gets selected first, then the rest by ${20 \choose 7}$.

I hope this is a simplistic and extensive enough explanation of why your answer is false and why your probability exceeds 1.

The only way to not reveal 3 cards of the same suit is if your hand consists of 2 cards of every suit.

Which can be calculated as follows: ${5 \choose 5} {5 \choose 2}^5$ You're obviously choosing all 5 suits, and for every of the 5 suits you choose 2 cards.

Then $1 - \frac{{5 \choose 5} {5 \choose 2}^5}{25 \choose 10}$ is the probability you will reveal 3 cards of the same suit. Which should be about 97% chance.

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  • $\begingroup$ Is there a way to generalize this, or would the conditions change for every number of cards drawn? For example, obvious there’s no risk of drawing a second set of 3 when drawing less than 6 cards. Do I need to split my calculations into less than 6 and 6+? $\endgroup$
    – Ian Clayman
    Commented Jun 28 at 23:17
  • $\begingroup$ @N.F.Taussig Thanks for mentioning, completely missed it, I fixed it! $\endgroup$
    – MrPuffer
    Commented Jun 28 at 23:57
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    $\begingroup$ The calculation in the question also counts each hand with suites DDDDHHCCSS four times: once for each way you can choose one of the diamonds to make the remaining $7$ cards along with the HHCCSS cards. Any hand of the form DDDDDHHCCS is counted ten times. $\endgroup$
    – David K
    Commented Jun 29 at 3:25
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    $\begingroup$ @IanClayman The most generalized for is [Inclusion-Exclusion[(en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle), but often there are ways to simplify to calculations. For instance, in this case, as lulu pointed out in a comment to your question, you can instead calculate the probability on not getting three of the same suit. $\endgroup$
    – Acccumulation
    Commented Jun 29 at 19:13
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You asked in a comment whether it is possible to generalise this, though it is not quite clear in what way you would wish to do so. It is easy enough to calculate the number of ways and probability of any particular pattern: for example

  • to get $4$ cards from one suit and $2$ each from three suits and $0$ from the one remaining suit, there are ${5 \choose 4}^1{5 \choose 2}^3{5 \choose 0}^1 \frac{5!}{1!3!1!} = 100000$ ways

  • to get $2$ each from all five suits, there are ${5 \choose 2}^5 \frac{5!}{5!} = 100000$ ways too

and so on for the other patterns, each of which is a partition of the number of cards drawn into up to a number of parts no more than the number of suits, with each part no more than the number of cards per suit.

You can then turn these into probabilities by dividing by ${25 \choose 10}$. As lulu has commented and MrPuffer has calculated, the answer to your particular question is the complement of $2$ cards from each of the five suits, i.e. $1-\frac{100000}{3268760} \approx 0.97$.

Using R, you can find the number of ways and probabilities for each possible pattern (and this can easily be adapted for other examples to help you generalise) with

library(partitions)
suits        <- 5
cardspersuit <- 5
cardsdrawn   <- 10
restrictparts <- restrictedparts(cardsdrawn, suits)
limitparts <- restrictparts[, restrictparts[1,] <= cardspersuit] 
ways <- numeric(ncol(limitparts))
for (n in 1:ncol(limitparts)){
    part      <- limitparts[,n]  
    tablepart <- table(part)
    ways[n]   <- prod(choose(cardspersuit, part)) * 
                 factorial(suits) /  prod(factorial(tablepart))
    names(ways)[n] <- paste(part, collapse="+")
    }                
cbind(ways, prob=ways / choose(suits*cardspersuit, cardsdrawn))

giving

             ways         prob
5+5+0+0+0      10 3.059264e-06
5+4+1+0+0    1500 4.588896e-04
5+3+2+0+0    6000 1.835558e-03
4+4+2+0+0    7500 2.294448e-03
4+3+3+0+0   15000 4.588896e-03
5+3+1+1+0   15000 4.588896e-03
4+4+1+1+0   18750 5.736120e-03
5+2+2+1+0   30000 9.177792e-03
4+3+2+1+0  300000 9.177792e-02
3+3+3+1+0  100000 3.059264e-02
4+2+2+2+0  100000 3.059264e-02
3+3+2+2+0  300000 9.177792e-02
5+2+1+1+1   25000 7.648160e-03
4+3+1+1+1  125000 3.824080e-02
4+2+2+1+1  375000 1.147224e-01
3+3+2+1+1  750000 2.294448e-01
3+2+2+2+1 1000000 3.059264e-01
2+2+2+2+2  100000 3.059264e-02

If you changed the number of cards drawn to $5$ you would instead get

           ways         prob
5+0+0+0+0     5 9.410879e-05
4+1+0+0+0   500 9.410879e-03
3+2+0+0+0  2000 3.764352e-02
3+1+1+0+0  7500 1.411632e-01
2+2+1+0+0 15000 2.823264e-01
2+1+1+1+0 25000 4.705439e-01
1+1+1+1+1  3125 5.881799e-02

and the probability of drawing at least three cards from a single suit would become $\frac{10005}{53130}\approx 0.19$.

Alternatively, consider the shape of a Bridge hand, so $13$ cards drawn from a pack of $4$ suits and $13$ cards per suit:

                 ways         prob
13+0+0+0            4 6.299078e-12
12+1+0+0         2028 3.193633e-09
11+2+0+0        73008 1.149708e-07
10+3+0+0       981552 1.545718e-06
9+4+0+0       6134700 9.660739e-06
8+5+0+0      19876428 3.130079e-05
7+6+0+0      35335872 5.564585e-05
11+1+1+0       158184 2.491033e-07
10+2+1+0      6960096 1.096055e-05
9+3+1+0      63800880 1.004717e-04
8+4+1+0     287103960 4.521226e-04
7+5+1+0     689049504 1.085094e-03
6+6+1+0     459366336 7.233961e-04
9+2+2+0      52200720 8.220410e-05
8+3+2+0     689049504 1.085094e-03
7+4+2+0    2296831680 3.616981e-03
6+5+2+0    4134297024 6.510565e-03
7+3+3+0    1684343232 2.652452e-03
6+4+3+0    8421716160 1.326226e-02
5+5+3+0    5684658408 8.952027e-03
5+4+4+0    7895358900 1.243337e-02
10+1+1+1      2513368 3.957975e-06
9+2+1+1     113101560 1.781089e-04
8+3+1+1     746470296 1.175519e-03
7+4+1+1    2488234320 3.918396e-03
6+5+1+1    4478821776 7.053112e-03
8+2+2+1    1221496848 1.923576e-03
7+3+2+1   11943524736 1.880830e-02
6+4+2+1   29858811840 4.702075e-02
5+5+2+1   20154697992 3.173900e-02
6+3+3+1   21896462016 3.448188e-02
5+4+3+1   82111732560 1.293071e-01
4+4+4+1   19007345500 2.993219e-02
7+2+2+2    3257324928 5.129536e-03
6+3+2+2   35830574208 5.642490e-02
5+4+2+2   67182326640 1.057967e-01
5+3+3+2   98534079072 1.551685e-01
4+4+3+2  136852887600 2.155118e-01
4+3+3+3   66905856160 1.053613e-01

allowing you to say that the most likely shape is $4+4+3+2$, then $5+3+3+2$, then $5+4+3+1$, then $5+4+2+2$, and then $4+3+3+3$, while the probability of a hand having its longest suit with $4$ cards is about $0.35$, $5$ cards about $0.44$, $6$ cards about $0.17$, and $7$ or more about $0.04$.

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