Standard deck of cards, full straight flush probability question

Assume a standard $52$ card deck (no jokers) well shuffled. I was wondering how many cards would need to be randomly drawn from it (without replacement) on average to get about a $50$% chance of getting all $13$ cards in exactly one suit. For example, getting all $13$ of the diamond cards. How do you solve for this type of problem?

For clarification, a "winning" draw would be when one of the four suits is completely filled (all $13$ ranks of that suit are in the drawn cards). Any of the $4$ suits can be filled then the # of cards in the draw would be # of interest. There will be no cases where more than $1$ suit is completely filled in the hand. So for example, if the first $13$ cards drawn are all hearts, then hand stops at $13$ cards. If $12$ cards of each suit are drawn ($48$ cards total), then the hand would stop at the $49$th card draw (thus completing one of the suits in full).

• If iI'd firstly compute the chance of getting all 13 of diamonds cards at drawing $i$ cards for some $i$-s then I'd use the Inclusion-Exclusion Principle. – Leonhardt von M Mar 30 '15 at 20:29
• I will try to write a computer simulation to see if I can get a solution to this but I was curious about how to do it using probability techniques. – David Mar 31 '15 at 1:12
• For those of you interested, my simulation program just picks random cards (no repeats) for each hand until one complete suit is full (all $13$ cards are gotten). I then run it for 1 million decisions to make sure I get a good average. I get $45.33$ cards per decision on average which is $3$ decisions every $136$ cards (on average). – David Mar 31 '15 at 2:23
• Actually a faster simulation might be to start with a full deck and just remove $1$ card at a time until no suit has all $13$ cards present then run that simulation $1$ million (or more) times to see if the same result of $45.33$ cards is achieved. – David Mar 31 '15 at 19:27

If you are looking for all the diamonds, it might be easier to draw from the bottom of the pack and look at the probability of not getting a diamond.

0: $1 \gt \frac12$

1: $\frac{39}{52} \gt \frac12$

2: $\frac{39}{52} \times \frac{38}{51} \gt \frac12$

3: $\frac{39}{52} \times \frac{38}{51} \times \frac{37}{50} \lt \frac12$

So if you draw $52-3=49$ cards then the probability of having all the diamonds is less than $\frac12$ but if you draw $52-2=50$ cards, the probability of having all the diamonds is more than $\frac12$.

If your question is how many cards do you need to draw until your probability of having all $13$ of at least one suit, then you can do something similar, again drawing from the bottom of the pack and looking at the probability of not having all four suits. I think the probabilities may be as follows, though you will want to check

Suits   0   1           2           3           4
Cards
0       1   0           0           0           0
1       0   1           0           0           0
2       0   0.235294    0.764706    0           0
3       0   0.051765    0.550588    0.397647    0
4       0   0.010564    0.299640    0.584298    0.105498
5       0   0.001981    0.145918    0.588355    0.263745
6       0   0.000337    0.066841    0.506340    0.426482
7       0   0.000051    0.029347    0.401023    0.569578


If so then if you draw $52−7=45$ cards then the probability of having all of any suit is less than $\frac12$ but if you draw $52−6=46$ cards, the probability of having all of any suit is more than $\frac12$.

I can't think of a straight up combinatorial argument without splitting it into too many cases, so I'll use the inclusion-exclusion principle.

Number the suits $1,2,3,4$ (this is for commodity of notation). Let:

$$A_k=\text{the draw contains 13 cards of suits k}$$

Then the desired event is $A_1 \cup A_2 \cup A_3 \cup A_4$

Let $n$ be the size of your draw. Then there are ${52 \choose n}$ possible draws.

Then $P(A_1)={{52-13 \choose n-13} \over {52 \choose n}}$, since you need to choose $n-13$ cards which are not of suit 1.

Similarly: $P(A_1 \cap A_2)={{52-26 \choose n-26} \over {52 \choose n}}$

$P(A_1 \cap A_2 \cap A_3)={{52-39 \choose n-39} \over {52 \choose n}}$

The others are similar. Compute them to obtain an expression which depends on $n$ and try out different values.

Obviously some of the probabilities might be $0$ depending on $n$.

I ran a computer simulation of $1$ million decisions and got about $45.33$ cards on average are needed to get a winning hand (exactly $1$ of the $4$ suits all filled). It might be fun to play a game with a naive person betting them even money (dollar for dollar) that they cannot get all $13$ cards of one suit drawing from $3/4$ of the deck ($39$ cards). The answer is somewhat surprising because worst case is $49$ cards. That is, if you draw $49$ cards you are guaranteed to fill at least one suit completely. $45.33$ is not much less than $49$ so the average case is close to the worst case and "far away" from the best case of $13$.

• My calculations would suggest a mean of about $45.33492$. Does your simulation give the median (I think $46$) and mode (I think $47$) as well as the mean? – Henry Mar 31 '15 at 7:24
• My simulation program is only about $20$ lines of code and only gives the average # of cards needed. So for example, if I run $1$ million decisions, the card count will be $45.33$ million (approximately). I could get $1$ billion decisions but I strongly suspect the result will be about the same, $45.33$. – David Mar 31 '15 at 12:41
• Computer simulation is a great tool because in math it seems like you first have to choose which method to use before you can work an answer. In computer simulation, it seems more straightforward. For this particular problem, I would probably struggle with it using math techniques but with computer simulation it is easy for me. Also, instead of just using random numbers to simulate card draws, I could simulate all $45$ or $46$ card draws and have the computer count them up and get an exact probability but knowing the approx # of cards needed is what I really wanted and my simulation did that. – David Mar 31 '15 at 12:52