Problem of drawing poker

Question

Randomly shuffle a deck of poker cards (52) and continually draw the card without replacement. On average how many cards you need to draw in order to get a card which has a same value as the cards you've drawn? That is to say, if the card you've drawn is 1->2->3->4, then you would stop at the next step if you draw one of the 1,2,3 or 4.

This is a problem I came upwith when I was playing desk card games...

I thought "on average" it could be seen as $E(\sum_{i=1}^{52}I_i)$ of which $I_i$ means the i-th card has already been seen (0) or not (1). Then $E(\sum_{i=1}^{52}I_i) = \sum_{i=1}^{52}P_i$ of which $P_i$ is the prob that the i-th card is a new card (not seen before). But apparently the $P_i$ depends on $i$....so what can I do next?

=============== The original post is "stop when you draw a card with a same face value as the first card". I think that's just splitting the remaining deck (52-1=51 cards) by 3 cards, i.e. we have (51-3)/4 cards per subdeck. Then the answer could be (51-3)/4 + 1 + 1. But not so sure...

The background comes from a Chinese desk card game. The game aims to kill other characters (lower down the life points). One special hero has a special skill, i.e. when he is almost dying (life point goes from 1 to 0 due to the attack) he could draw a card from the deck and collect it, if the card has a same face value compared to the cards he collected by this method (i.e. draw a card when life point goes from 1 to 0) then he dies. I am just wondering on average how many cards he could collect (or how many times you need to "attack" him).

Answer 1

This answer is for the game where you stop once some value is repeated for the first time. In the words of the OP "if the card you've drawn is 1->2->3->4, then you would stop at the next step if you draw one of the 1,2,3 or 4.".

Model the deck by dependent random variables $X_1,\ldots,X_{52}$ that each take values in the set $\{1,\ldots,13\}$.

Let $N$ denote a random variable modeling the rank of the first repeated value. Note that $P(N>n) = 0$ when $n\geq 14$ by the pigeonhole principle.

Fix $n\in \{1,\ldots, 13\}$. Note that $$\begin{aligned} P(N>n) &= P(X_1,\ldots,X_n \text{ are all distinct}) \\&= \sum_{v_1\neq\ldots\neq v_n \in \{1,\ldots,13\} } P(X_1=v_1,\ldots,X_n=v_n) \\&\stackrel{(1)}{=} P(X_1=1,\ldots,X_n=n) \times |\{(v_1,\ldots,v_n) \in \{1,\ldots,13\}^n: v_1,\ldots,v_n \text{ are all distinct}\}| \\&= P(X_n=n|X_{n-1}=n-1,\ldots, X_1=1)\times \ldots\times P(X_2=2|X_1=1)P(X_1=1) \frac{13!}{(13-n)!} \\&= \frac{13!}{52!} 4^n \frac{(52-n)!}{(13-n)!}. \end{aligned}$$

Equality (1) holds by symmetry.

Finally, $E[N] = \sum_{n=0}^{13} P(N>n) = 1+ \sum_{n=1}^{13} \frac{13!}{52!} 4^n \frac{(52-n)!}{(13-n)!} = \frac{226087256246}{39688347475} \approx 5.696$.

This value is confirmed by the following simulation.

import numpy as np 

def find(arr):
    seen = set()
    for (i,el) in enumerate(arr):
        if el in seen:
            return i+1
        else:
            seen.add(el)

res = []
deck = np.repeat(np.arange(1,14),4)
for _ in range(10000):
    np.random.shuffle(deck)
    res.append(find(deck))
print(np.mean(res))

---

The same approach applies for the other game where you stop once the first value is repeated.

Note that $P(N>0)=P(N>1)=1$, $P(N>n)=0$ if $n\geq 50$ and for $n\in \{2,\ldots, 49 \}$, by similar techniques as above $$\begin{aligned} P(N>n)&=13 P(X_2,\ldots,X_n \text{ all distinct from }1 | X_1=1) P(X_1=1) \\&= 4\times 13 \times \frac{48!}{52!} \frac{(52-n)!}{(48-n+1)!}. \end{aligned}$$

Thus $E[N] = 2 + \sum_{n=2}^{49} 4\times 13 \times \frac{48!}{52!} \frac{(53-n)!}{(50-n)!} = 14.$

This value is confirmed by the following simulation.

import numpy as np

def find2(arr):
    val = arr[0]
    for (i,el) in enumerate(arr[1:]):
        if el == val:
            return i+2

res = []
deck = np.repeat(np.arange(1,14),4)
for _ in range(10000):
    np.random.shuffle(deck)
    res.append(find2(deck))
print(np.mean(res))