How many cards for a Flush?

Let $E_{a,b,c,d}$ be the expected number of additional cards that have to be drawn to obtain a flush, given that $a$ Spades, $b$ Hearts, $c$ Diamonds and $d$ Clubs have already been drawn. Conditioning on the outcome of the next draw, we see that $E_{a,b,c,d} \; = \; 1 + \frac{(13-a)E_{a+1,b,c,d} + (13-b)E_{a,b+1,c,d} + (13-c)E_{a,b,c+1,d} + (13-d)E_{a,b,c,d+1}}{52-(a+b+c+d)} \hspace{2cm} 0 \le a,b,c,d \le 4$ and we also have that $E_{a,b,c,d} = 0$ when any one of $a,b,c,d$ is equal to $5$ or more. It is a simple matter to plug this recursive relation into a computer programme, and obtain the expected number of draws to obtain a flush as $E_{0,0,0,0} \; = \; 12.3598016$ making the answer $\boxed{12}$ .

Trying a little harder, we note that, if we define $F_{a,b,c,d} \; = \; \binom{52-a-b-c-d}{13-a\;\;13-b\;\;13-c\;\;13-d}E_{a,b,c,d}$ where $\binom{p+q+r+s}{p\;\;q\;\;r\;\;s}$ is the multinomial coefficient $\binom{p+q+r+s}{p\;\;q\;\;r\;\;s} \; = \; \frac{(p+q+r+s)!}{p!\,q!\,r!\,s!}$ then we have the recurrence relation $F_{a,b,c,d} \; = \; \binom{52-a-b-c-d}{13-a\;\;13-b\;\;13-c\;\;13-d} + F_{a+1,b,c,d} + F_{a,b+1,c,d} + F_{a,b,c+1,d} + F_{a,b,c,d+1}$ for $0 \le a,b,c,d \le 4$ , and hence $F_{0,0,0,0} \; =\; \sum_{a,b,c,d=0}^4 \binom{52-a-b-c-d}{13-a\;\;13-b\;\;13-c\;\;13-d}N_{a,b,c,d}$ where is $N_{a,b,c,d}$ is the number of ways of getting from $(0,0,0,0)$ to $(a,b,c,d)$ by taking steps of size $1$ in any one of the components at any one time. In other words, $N_{a,b,c,d} \; = \; \binom{a+b+c+d}{a\;\;b\;\;c\;\;d}$ and hence $F_{0.0.0.0} \; = \; \sum_{a,b,c,d=0}^4 \binom{52-a-b-c-d}{13-a\;\;13-b\;\;13-c\;\;13-d}\binom{a+b+c+d}{a\;\;b\;\;c\;\;d} \; = \; 663038316898820188679631600000$ To obtain $E_{0,0,0,0}$ , we need to divide this by $\binom{52}{13\;\;13\;\;13\;\;13} \; = \; 53644737765488792839237440000$ obtaining $E_{0,0,0,0} \; = \; \frac{1173956396185}{94981815404} \; = \; 12.35980162$

Yet another approach! For any $0 \le a,b,c,d \le 13$ , the probability $p_{a,b,c,d} \; = \; \frac{\binom{a+b+c+d}{a\;\;b\;\;c\;\;d} \, \binom{52-a-b-c-d}{13-a\;\;13-b\;\;13-c\;\;13-d}}{\binom{52}{13\;\;13\;\;13\;\;13}}$ is the probability that, when a pack is dealt, the first $a+b+c+d$ cards consist of $a$ Spades, $b$ Hearts, $c$ Diamonds and $d$ Clubs. If $N$ is the number of cards that have to be dealt to obtain a flush, then $P[N \ge k+1]$ is the probability that there is no flush in the first $k$ cards, and hence $P[N \ge k+1] \; = \; \left\{ \begin{array}{lll} \sum_{{0 \le a,b,c,d \le 4} \atop {a+b+c+d=k}}p_{a,b,c,d} & \hspace{1cm} & 0 \le k \le 16 \\ 0 & & k \ge 17 \end{array}\right.$ and hence $E[N] \; = \; \sum_{k \ge 0}P[N \ge k+1] \; = \; \sum_{k=0}^{16} \sum_{{0 \le a,b,c,d \le 4} \atop {a+b+c+d=k}}p_{a,b,c,d} \; = \; \sum_{a,b,c,d =0}^4 p_{a,b,c,d} \; = \; E_{0,0,0,0}$ retrieving the previous result.

This will not be an exact solution, but more of an educated guess. But maybe it is still valuable for people to get a feeling for probability. The minimum number of cards you need to draw for a flush is obviously 5. The maximum number is 17. Because if you haven't hit a flush with 16 cards, it means you have 4 cards of each color. Therefore the 17th card has to complete one flush. Now the average of 5 and 17 is 11. But since the distribution of probability in a situation like this (drawing something with a probability lower than 1/2 from a large set) is typically shifted towards the right, 12 is a good guess. Especially considering the possible choices.

Here is some quick python code implementing Mark brilliant solution :

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from functools import lru_cache

@lru_cache(maxsize=1296)
def F(a,b,c,d):
    if (a==5 or b==5 or c==5 or d==5):
        return 0
    return 1 + ((13-a)*F(a+1,b,c,d)+(13-b)*F(a,b+1,c,d)+(13-c)*F(a,b,c+1,d)+(13-d)*F(a,b,c,d+1))/(52-a-b-c-d)

print(F(0,0,0,0))

I wrote a bit of code in processing to simulate drawing cards and ran it 100,000,000 times to get 12.359815. With this method the number of cards needed for a flush, the number of suits and the number of cards in a suit are each variables so it could be worked out for other variations. I know there's undoubtedly better and more elegant solutions in both code and concept but this worked for me. I should probably have a go at Python though.

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/* 
  The Code:
    -Sets all cards to not have been picked
    -Picks a random card that hasn't been picked
    -Checks if there's a flush and stops if there is
    -Repeats then averages all results
*/


//number of repeats
int countNumber = 10000;

//number of cards to make a flush
int flushNumber = 5;

//number of suits
int suitNumber = 4;

//number of cards in each suit
int valNumber = 13;


int cardNumber = suitNumber * valNumber;
boolean cards [][];


void setup()
{
  cards = new boolean[suitNumber][valNumber]; 
  int total = 0;

  for (int i =0; i<countNumber; i++)
  {
     total += countCards();
  }

  print(float(total)/float(countNumber));
}

int countCards()
{  
  int count = 0;

  refreshCards();

  while(count < cardNumber)
  {
    count ++;
    if(pickCard()) return count;
  }

  return -1;
}

boolean pickCard()
{
  while(true)
  {
    int suit = int(random(cards.length));
    int num = int(random(cards[0].length));
    if (cards[suit][num] == false) 
    {
      cards[suit][num] = true;
      return flushCheck();
    }
  }
}


boolean flushCheck()
{
  boolean verdict = false;

  for (int i = 0; i<cards.length; i++)
  {
    int val = 0;
    for(int j = 0; j<cards[0].length; j++)
    {
      if (cards[i][j]) val ++;
      if (val >= flushNumber) verdict = true;
    }
  }

  return verdict;  
}



void refreshCards()
{
  for (int i = 0; i<cards.length; i++)
    {
      for (int j = 0; j<cards[0].length; j++)
      {
        cards[i][j] = false;
      }
    }
}

Not a solution

Let me define some questions as,

$Q_5$ What is the probablity of forming exactly one flush on your 5th pick?

$Q_6$ What is the probablity of forming exactly one flush on your 6th pick?

..

$Q_{17}$ What is the probablity of forming exactly one flush on your 17th pick? (since 17th pick guarantees the flush)

And the answers of these questions as $A_5,A_6,...,A_{17}$ , respectively

So finally our answer will be $5 \times A_5 + 6 \times A_6 + ... + 17 \times A_{17} \approx12$

I found that

.. ..

But I could not find a formula for $A_{10}$ to $A_{17}$