Question

Question: What is the probability that a hand of \(13\) cards contains no pairs?

Short answer

Answer

The probability that a hand of \(13\) cards contains no pairs is \( = \frac{{{4^{12}} \times 12! \times 39!}}{{51!}}\).

Step-by-step solution

Given data

A hand of \(13\) cards contains no pairs.

Concept of Combination

The number of ways to choose \(r\) numbers from \(n\) numbers is \(C(n,r) = \frac{{n!}}{{(n - r)!r!}}\).

Calculation to find the probability that contains no pairs

If we select first card, then we have same type of three cards. To select the second card we have to avoid other three cards.

Consider first card in your hand is an ace, you have to avoid other three aces in the deck.

For the second card, we have cards and of them will not match with the first card.

Then, the probability of the second card will not pair with the first card \(\frac{{48}}{{51}}\).

The probability of the third card will not pair with either of the first two cards is \(\frac{{44}}{{50}}\).

The probability of the fourth card by avoiding nine cards from \(49\) is \(\frac{{40}}{{49}}\).

This process will continue till \(13\) card.

The probability of \({13^{{\rm{th }}}}\) card is \(\frac{4}{{40}}\).

The probability that a hand of 13 cards contains no pairs is \(\frac{{48}}{{51}} \times \frac{{44}}{{50}} \times \frac{{40}}{{49}} \times \frac{{36}}{{48}} \times \frac{{32}}{{47}} \times \frac{{28}}{{46}} \times \frac{{24}}{{45}} \times \frac{{20}}{{44}} \times \frac{{16}}{{43}} \times \frac{{12}}{{42}} \times \frac{8}{{41}} \times \frac{4}{{40}} = \frac{{{4^{12}} \times 12! \times 39!}}{{51!}}\).

The probability that a hand of 13 cards contains no pairs is \(\frac{{{4^{12}} \times 12! \times 39!}}{{51!}}\).