Five cards are drawn without replacement from a 24 card Euchre deck, what is the probability that at least two cards have the same rank?

Clarification: A Euchre deck consists of just A-K-Q-J-10-9 of each suit, thus four cards in each of six ranks.

I'm confused on how to approach this problem, at first I thought it was fairly straightforward and used to combinations to determine the probability but realized that it is not the correct result. This is what I did

4C2 * 20C3 = 6,840, for getting two cards of the same rank in a hand of five
4C3 * 20C2 = 760, for getting three cards of the same rank in a hand of five
4C4 x 20C1 = 20, for getting four cards of he same rank in a hand of five

I then added them together which is 7,620 and divided by the combinations of Euchre hands 42,504 which is ≈ 17.93%

To test my math I created a Monte Carlo simulation with 5,000,000 iterations and the result was, any hand with at least two or more cards of the same rank occurred ≈ 85.54086% of the time which is nowhere near my previous result.

• Please describe a Euchre deck. – vadim123 Feb 23 '17 at 0:24
• It may be easier to compute the probability that no two cards have the same rank. – awkward Feb 23 '17 at 0:31

$\newcommand{\cbinom}{\operatorname{{}^{#1}\mathrm C_{#2}}}$ You can do it that way.

However, you don't want to just add those probabilities, but alternatively add and subtract. That is, use the Principle of Inclusion and Exclusion.

You want to include selection of a pair (value and two suits) and three other cards (not of that suit), exclude selections of two pair and a singleton, exclude triplet and two other cards, include triplet and pair, include quadruplet.

$$\frac{\cbinom 61\cbinom 42\cbinom{20}3-\cbinom 62(\cbinom 42)^2\cbinom {18}1-\cbinom{6}{1}\cbinom 43\cbinom{20}2+\cbinom61\cbinom 43\cbinom 51\cbinom 42+\cbinom 61\cbinom {20}1}{\cbinom{24}5}$$

I think I've gotten that okay, but I'm not going to check it too closely, because it is much easier to go through complementation.   The probability for at least two cards of the same value, is the probability for not all the cards being distinct.

$$\dfrac{\cbinom {24}5-\cbinom 65(\cbinom 41)^5}{\cbinom {24}5}$$

One way to look at the problem is this. Draw the cards sequentially. The first card you draw, the probability of having two cards of the same rank is obviously $0$. When you draw the second card from the remaining $23$ cards, you have $\frac{3}{23}$ probability of completing the pair. With probability $\frac{20}{23}$ you go on to draw the third card. This time, there are $6$ cards out of the remaining $22$ that match either of the cards already in your hand. Continuing in this fashion, you get:

$$\frac{3}{23} + \frac{20}{23} \Big( \frac{6}{22} + \frac{16}{22} \Big( \frac{9}{21} + \frac{12}{21} \cdot \frac{12}{20} \Big) \Big) = 0.8554488989 \enspace,$$

in good agreement with your simulation.