[Math] French playing cards, probability to have a four (of a kind)

probability

We have a "French playing cards" composed of $52$ cards. Each kind of cards (Ace, King, Queen, …) is composed of $4$ cards : There are $4$ Ace, $4$ Jack, etc. We pull $5$ cards among the $52$ cards.

We are a four (of a kind) if among the $5$ cards pulled, there are the $4$ of a same value ($4$ cards Ace, or King, or 2, or 3……).

What is the probability to have a four (of a kind) if the first card pulled is an Ace, and the fifth card pulled is a King ?

I hope you understand, it's very difficult for me to translate that. Thank you for help in advance.

Best Answer

Here is a hint:

Given that you have drawn one ace and one king, you will need either to draw the remaining three aces, or draw the remaining three kings, from the remaining $50$ cards.

Can you continue from this point?

(Spoiler answer)

There are two winning ways to draw the remaining three cards, and ${50 \choose 3}$ total ways to draw the remaining three cards, so the probability is $2/{50 \choose 3}$.

Related Solutions

[Math] Playing Cards probability problem

probability of getting a king on your first draw is 4/52.

probability of getting a queen after a king on your second draw 4/51

jack after a king and a queen 4/50

So the probability of getting a king, then a jack, and then a queen is (4*4*4)/(50*51*52).

There are six orders that give you a king, a queen and a jack.

KQJ, KJQ, QKJ, QJK, JKQ, JQK

So, we take the original probability and multiply it by 6.

6*(4*4*4)/(50*51*52) = 16/5525 $\approx$ .29%

[Math] the probability that 5 randomly chosen cards in a deck add up to 40 or greater

Normal approximation:

Consider a single card draw from the deck:

E(X)=7

Var(X)=14

With the sum of 5 cards (drawn without replacement):

$E(X)=7 \times 5$

$\text{Var}(X)=14\times 5\times \frac{52-5}{52-1}$

(the last term being the 'finite population correction', which applies as much to the variance of the sum as it does to the variance of the mean).

The approach to normality in this situation is reasonably rapid.

This suggests the sum on 5 cards might be be roughly approximated by a normal distribution with mean $35$ and variance $70\times\frac{47}{51}\approx 64.5098$. Using a continuity correction this gives an approximate probability of totalling at least 40 on 5 cards of:

$$1-\Phi(\frac{39.5-35}{\sqrt{70\times\frac{_{47}}{^{51}}}})\approx 0.288$$

Simulation (in R) of ten million five-card draws indicates the probability is
about $0.293$ (with s.e. $\approx 1.4\times 10^{-4}$):

 res=replicate(10000000,sum(sample(rep(1:13,4),5,replace=FALSE)));mean(res>=40)  
 [1] 0.2927447

As a check on the earlier calculation of the variance, the standard deviation of of those simulated sums was 8.0316; the previous calculation gives 8.0318.

Edit: Here's a comparison of the empirical cdf of the simulated data with the above normal approximation; they're pretty close:

enter image description here

More extensive simulations are consistent with the other two answers based on complete enumeration:

enter image description here

Best Answer

Related Solutions

[Math] Playing Cards probability problem

[Math] the probability that 5 randomly chosen cards in a deck add up to 40 or greater

Related Question