So we are working on finding the probability of getting a hand with four aces in a 5-card poker game. To do this, we have to count the total number of possible poker hands. This turns out to be possible using the combination formula:

\[{C}{(}{n}{,}\hspace{0.33em}{r}{)}\hspace{0.33em}{=}\hspace{0.33em}\frac{n!}{{r}{!(}{n}\hspace{0.33em}{-}\hspace{0.33em}{r}{)!}}\]And in this case,

\[{C}{(}{52}{,}\hspace{0.33em}{5}{)}\hspace{0.33em}{=}\hspace{0.33em}\frac{52!}{{5}{!(}{52}\hspace{0.33em}{-}\hspace{0.33em}{5}{)!}}\hspace{0.33em}{=}\hspace{0.33em}\frac{52!}{5!(47)!}\]But I left my last post not calculating this, and cautioning you to not calculate this before doing some simplification. This is because 52! and 47! are HUGE numbers and because of the limitations of many calculators, you will get inaccurate results.

Well how do we simplify \[\frac{52!}{5!(47)!}\]?

Well notice that 52! = 52×51×50×49×48×47! . In other words, you can always start counting down when writing an expanded factorial but the remaining numbers are just the factorial of where you stopped. So now you can cancel the 47! in the numerator and the denominator,

\[\begin{array}{l}{\frac{52!}{5!(47)!}\hspace{0.33em}{=}\hspace{0.33em}\frac{{52}\hspace{0.33em}\times\hspace{0.33em}{51}\hspace{0.33em}\times\hspace{0.33em}{50}\hspace{0.33em}\times\hspace{0.33em}{49}\hspace{0.33em}\times\hspace{0.33em}{48}\hspace{0.33em}\times\hspace{0.33em}{47}{!}}{{5}{!}\hspace{0.33em}\times\hspace{0.33em}{47}{!}}\hspace{0.33em}{=}\hspace{0.33em}}\\{\frac{{52}\hspace{0.33em}\times\hspace{0.33em}{51}\hspace{0.33em}\times\hspace{0.33em}{50}\hspace{0.33em}\times\hspace{0.33em}{49}\hspace{0.33em}\times\hspace{0.33em}{48}}{5!}\hspace{0.33em}{=}\hspace{0.33em}{2}{,}{598}{,}{960}}\end{array}\]That’s a lot of hands! Well that number is the denominator in our generic probability formula:

Probability = \[\frac{{\mathrm{Number}}\hspace{0.33em}{\mathrm{of}}\hspace{0.33em}{\mathrm{favorable}}\hspace{0.33em}{\mathrm{outcomes}}}{{\mathrm{Total}}\hspace{0.33em}{\mathrm{number}}\hspace{0.33em}{\mathrm{of}}\hspace{0.33em}{\mathrm{possible}}\hspace{0.33em}{\mathrm{outcome}}{s}}\]

So what is the numerator? Well if you want all four aces in your hand, that leaves just one more card. There are 48 other cards that can be in your hand that can be the fifth card. So there are 48 possible ways to have four aces in your hand of five cards. So the probability of having four aces is

\[\frac{48}{2,598,960}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{54,145}\hspace{0.33em}{=}\hspace{0.33em}{0}{.}{00001847}\]A very small probability! You may need to brush up on your bluffing skills.