TU Wien:Statistik und Wahrscheinlichkeitstheorie UE (Levajkovic)/Übungen 2022W/HW01.1

Poker game

A deck of 52 cards has 13 ranks (2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A) and 4 suits A poker hand is a set of 5 cards randomly chosen from a deck of 52 cards.

Problem a)

A full house in poker is a hand where three cards share one rank and two cards share another rank. How many ways are there to get a full-house? What is the probability of getting a full-house?

Problem b)

A royal flush in poker is a hand with ten, jack, queen, king, ace in a single suit. What is the probability of getting a royal flush?

Lösungsvorschlag von Friday

--Friday Sa 30 Jan 2021 13:59:00 CET

Solution a)

First we need to calculate all possible hands there are. Simplified we want to count the number of subsets, while ignoring the order. The number of all combinations can be calculated with:

$${\binom {n}{k}}={\binom {52}{5}}=2\,598\,960$$

Next, we calculate how many hands result in a full-house. We can think of it as choosing a rank from which we draw 3 cards and then choosing a different rank from which we draw 2 cards. (There are always 4 cards in each rank)

$${\binom {13}{1}}*{\binom {4}{3}}*{\binom {12}{1}}*{\binom {4}{2}}=3\,744$$

Now, all we have to do is to devide the desired outcomes by number of possible outcomes.

$${\frac {3\,744}{2\,598\,960}}\approx 0,00144=1,44*10^{-3}\quad \Box$$

Solution b)

Since a royal flush fills a full hand and all cards must be of the same suit and there are only ${\textstyle 4}$ suits, it is obvious that there are only ${\textstyle 4}$ hands that result in a royal flush.

$${\frac {4}{2\,598\,960}}\approx 0.00000154=1.54*10^{-6}\quad \Box$$