Two players pick cards from standard 52 card deck without replacement...

Question

I am struggling with this interview prep question... SOS

Two players pick cards from standard 52 card deck without replacement: 1st player picks a card, then 2nd, then again 1st, then 2nd etc. They stop once somebody picks a king (of any suit), the player who picks the king wins. What is the probability that the first player wins? the second player?

Which player has more chances of winning: the first or the second? It is sufficient to set up the correct formula for the probabilities, no need to evaluate it numerically. The last question can be answered without numerical evaluation, by analysis of the formulas

What is the probability that the first player wins? the second player? couldn't be far of from 50% for both... right?

Which player has more chances of winning: the first or the second? Intuition tells me the first person but I'm not sure if that follows or how to setup a formula -- EDIT: for this I'm thinking certainly the first player because they have one more opportunity to win by choosing a king first

Answer 1

In a given round of two draws, you start with $n$ cards of which $4$ are kings. The first player wins with probability $\frac 4n$. The second player wins with probability $\frac {n-4}n\cdot \frac 4{n-1}=\frac 4n\cdot\frac {n-4}{n-1}$ because they need the first player not to draw a king and there is one less card in the pack for their draw. As the last factor is less than $1$, on each round the first player has a greater chance to win than the second, so the first player has a greater chance to win overall.

I made a spreadsheet to compute the probability. I find the first player wins about $51.98\%$ of the time.

Answer 2

Let $E_k$ be the event that the kings are all in the last $k$ cards. Then a win in turn $52-k$ is the event $E_{k+1}\setminus E_k$. Since $E_k\subset E_{k+1}$, we have $\mathsf P(E_{k+1}\setminus E_k)=\mathsf P(E_{k+1})-\mathsf P(E_k)$, so the probability for the first player to win is

$$ \sum_{j=2}^{26}\left(\mathsf P(E_{2j})-\mathsf P(E_{2j-1})\right)=\sum_{k=4}^{52}(-1)^k\mathsf P(E_k)=\sum_{k=4}^{52}(-1)^k\frac{\binom k4}{\binom{52}4}=\frac{433}{833}\approx51.98\%\;. $$

You can tell that this is greater than the second players win probability without computing the sum by noting that each summand $\mathsf P(E_{2j})-\mathsf P(E_{2j-1})$ is greater than the corresponding summand $\mathsf P(E_{2j-1})-\mathsf P(E_{2j-2})$ for the second player.

Answer 3

The first player winning at the first draw: $$\frac{4}{52}=\frac{{4\choose 1}}{{52\choose 1}}$$ The first player winning at the second draw: $$\frac{4}{52}+\frac{48}{52}\cdot \frac{47}{51}\cdot \frac{4}{50} =\frac{{4\choose 1}}{{52\choose 1}}+\frac{{48\choose 2}\cdot {4\choose 1}}{{52\choose 2}\cdot {50\choose 1}}.$$ The first player winning at the third draw: $$\frac{4}{52}+\frac{48}{52}\cdot \frac{47}{51}\cdot \frac{4}{50} +\frac{48}{52}\cdot \frac{47}{51}\cdot \frac{46}{50}\cdot \frac{45}{49}\cdot \frac{4}{48}=\\ \frac{{4\choose 1}}{{52\choose 1}}+\frac{{48\choose 2}\cdot {4\choose 1}}{{52\choose 2}\cdot {50\choose 1}}+\frac{{48\choose 4}\cdot {4\choose 1}}{{52\choose 4}\cdot {48\choose 1}}.$$ The first player winning in general: $$\sum_{n=0}^{24} \frac{{48\choose 2n}\cdot {4\choose 1}}{{52\choose 2n}\cdot {52-2n\choose 1}}=\sum_{n=0}^{24} \frac{48!\cdot (52-2n)!\cdot 4}{(48-2n)!\cdot 52!\cdot (52-2n)}=\\ \sum_{n=0}^{24} \frac{(49-2n)(50-2n)(51-2n)}{13\cdot 51\cdot 50\cdot 49}=\\ \frac{1}{1624350}\cdot \sum_{n=0}^{24} (-8n^3+600n^2-14998n+124950)=\frac{433}{833}\approx 0.5198.$$

Answer 4

Previous answers are correct, but here is an alternate way to think about the problem.

Consider the sequence of 52 cards. Define a slot to be an odd location followed by an even location; thus, we have exactly 26 (non-overlapping) slots. Let us partition the $52!$ ways of arranging the deck into three categories: Type $A$ will be those where the first king appears in an odd location, and the card following the first king (i.e., the other card in the same slot) is a non-king.

Type $B$ will be those where the first king appears in an even location.

Type $C$ will be those where the first king appears in an odd location, and the card following the first king (i.e., the other card in the same slot) is also a king.

Note that $A$ and $C$ are wins for player 1, and $B$ is a win for player 2.

We have $|A|+|B|+|C|=52!$. It is also not hard to see that $|A|=|B|$ because we have a bijection given by swapping the cards in the slot that contains the first king. Since it is clear that $|C| \neq 0$, this already tells us that player 1 wins more of the time.

One can then go ahead and compute $|C|$. It seems tedious but not particularly difficult.

Answer 5

Consider arrangements of the deck. If Player 2 wins, putting the top card (not a King or else Player 1 would win) on bottom gives an arrangement where Player 1 wins and the bottom card is not a King. This is a bijection.

So, the difference between the probability that Player 1 wins and the probability that Player 2 wins is the probability that Player 1 wins and the bottom card is a King. This can be expressed in terms of Player 1 winning a similar game with $3$ Kings in $51$ cards (times the chance of a King as the fifty-second card). Proceed recursively until an easily solvable case like $1$ King in $49$ cards.

Express the probability that Player 1 wins as $\frac{1}{2}$ plus half the difference.

$$\frac{1}{2} + \frac{1}{2}\cdot\frac{4}{52}\left(\frac{1}{2}+\frac{1}{2}\cdot\frac{3}{51}\left(\frac{1}{2}+\frac{1}{2}\cdot\frac{2}{50}\cdot\frac{25}{49}\right)\right)=\frac{433}{833}$$

(It is also possible to take $0$ Kings in $48$ cards as a base case where Player 1 always wins since the fourty-ninth card is actually a King.)