# Probability, Poker Hands and Mathematical Creativity

Finding the probability of poker hands can be entertaining. But learning how to manipulate probability can result in an invaluable tool for everyday life.

It is now time to evaluate the results of some recent homework problems. In two previous articles some probability questions concerning five-card poker hands were posted. To assist in assessing the accuracy of one’s work and understanding of the process here are the solutions.

**What would be the probability of a five-card poker hand with no matches?**

Since probability is defined as possible successes divided by total possible outcomes the numerator (possible successes) for this question would be the product of 52 x 48 x 44 x 40 x 36 divided by 5 factorial (mathematically symbolized by “!” and indicating 5 x 4 x 3 x 2 x 1). The denominator (total possible outcomes) would be 52 x 51 x 50 x 49 x 48 also divided by 5(!). The result would be o.5071 or 50.7%.

**How about the chances of exactly two cards matching?**

While the denominator is unchanged the numerator would become 52 x 3 (the two matching cards) divided by 2(!) or 78 multiplied by 48 x 44 x 40 (the three non-matching ones) divided by 3(!) which is 14,080. The product of 78 x 14,080 (1,098,240 is then divided by the denominator the result is 0.4226 or 42.26%.

**What would be the probability of a full house (three matching cards and two matching cards)?**

As is the case with all five-card hands the denominator (total outcomes) is the same. The three matching cards would be 52 x 3 x 2 divided by 3(!) (52). The two matching cards are found by 48 x 3 divided by 2(!). (72). The product of 52 x 72 divided by the denominator gives a result of 0.0014 or 0.14%.

**Four matching cards?**

The numerator would be 52 x 3 x 2 x 1 divided by 4(!) (13). Multiply that by 48 (1(!) is equal to 1) and then divide by the denominator and the probability is remote—0.00024 or 0.024%.

**Time for some mathematical creativity**

Here is a question I would present to my pre-calculus students on their probability exam. It began with two ludicrous situations. “If a person were given one card from a standard deck of cards the probability of it matching is zero since there is no other card. The probability of not matching would be 100%. Likewise if one received 14 cards the likelihood of a match would now be 100% and a non-match zero.” Then prior work would be referenced. “As previous calculations have found if five cards are drawn the chances of no matches is 50.7% and one match 42.3%.” Next, the question was stated. “How many cards must be drawn from the deck to ensure a higher probability of no matches than that of a match?”

**The three bears solve for probability**

One card was too little; 14 too much. What number would be just right? Five cards gave 50.7% to 42.3% which was close. The obvious next step would be either six or seven cards. Being very linear in nature most of my students would try six. For no matches the numerator would be 52 x 48 x 44 x 40 x 36 x 32 which would then be divided by 6(!). The numerator would be 52 x 51 x 50 x 49 x 48 x 47 also reduced by 6(!). The probability would be 34.5% a significant drop.

And what would be the chances of exactly two of the cards matching? The numerator would be the product of 52 x 3 divided by 2(!) which is 78 and 48 x 44 x 40 x 36 divided by 4(!) which equals 129,888. Multiplying those two numbers together gives 10,131,264. Dividing by the previous denominator the probability becomes 49.8%. Thus when receiving six cards one is more likely to have a match than not (49.8% versus 34.5%). These relative probabilities represent a reversal of the values from five-card hands. Not surprisingly, when receiving a sixth card the likelihood of three or four matching cards increases as well. But not every traditional poker hand would follow the same trajectory.

**Finding the exception**

What would be the chances of having a six-card flush? In order for all six cards to be the same suit the numerator would become 52 x 12 x 11 x 10 x 9 x8 divided by 6(!). The resulting 20,592 divided by the denominator gives a probability of 0.01%. Comparing the likelihood of a flush in five cards (0.198%) to six (0.01) it can be said that the sixth card makes acquiring a flush 19.8 times more difficult.

**Yet another homework assignment**

Working with the concept of a six-card hand, which is more likely to occur: a hand containing two sets of three matching cards or one with four matching and two matching? Stay tuned.