Wrapping my head around hypergeometric distributions

Question

I'm playing a game where I'm allowed to draw three cards without replacement and can select one option from the three. The objective is to walk away with the best possible outcome. Different suits are better than others: Clubs are great, Spades are good, Hearts are bad, and Diamonds are terrible. So basically: pick a Club if one is drawn, otherwise pick a Spade, otherwise pick a Heart, otherwise be sad.

If I understand correctly, the odds of pulling a great outcome (any Club) can be calculated with a hypergeometric cumulative distribution function: with a population of 52 cards, 13 desired cards, 3 draws, and anywhere between 1 and 3 desirable outcomes, I'm calculating the odds as ~58.65%.

How would I then calculate the odds of having to walk away with just a good outcome? That is to say, no Clubs are drawn but at least one Spade is drawn. What about the same for walking away with a bad outcome (No Clubs or Spades but at least one Heart)? And finally a terrible outcome (only Diamonds are drawn)?

Is this a reasonable approach to understanding this problem? If so, should I expect the four odds to add up to 100%?

Thanks for your help!

Answer 1

To have drawn $0$ clubs but at least $1$ spade, the probability is $$\frac{\binom{13}{0}\left(\binom{13}{1}\binom{26}{2}+\binom{13}{2}\binom{26}{1}+\binom{13}{3}\binom{26}{0}\right)}{\binom{52}{3}} = \frac{503}{1700}.$$ The way this is counted is to note that there are $\binom{13}{0}$ ways to choose $0$ of the $13$ clubs, and $\binom{13}{1}\binom{26}{2}$ ways to choose exactly $1$ spade and $2$ non-club/non-spades, etc.

Similarly, the probability of a bad outcome (best card drawn is a heart) is $$\frac{\binom{13}{0}\binom{13}{0} \left(\binom{13}{1}\binom{13}{2} + \binom{13}{2}\binom{13}{1} + \binom{13}{3}\binom{13}{0}\right)}{\binom{52}{3}} = \frac{89}{850}.$$

Finally, the probability of a terrible outcome (best card drawn is a diamond) is simply $$\frac{\binom{13}{0}\binom{13}{0}\binom{13}{0} \binom{13}{3}}{\binom{52}{3}} = \frac{11}{850}.$$

Note that in such a case, all cards drawn are diamonds. Since the probability of the best card drawn being clubs is $\frac{997}{1700}$ as you correctly calculated, the total probability does add to $1$:

$$\frac{997}{1700} + \frac{503}{1700} + \frac{89}{850} + \frac{11}{850} = 1.$$

The unifying counting principle here is to think of the number of ways to draw $c, s, h, d$ cards of each suit, where $0 \le c, s, h, d \le 3$ are integers satisfying $c + s + h + d = 3$. This is $$\binom{13}{c}\binom{13}{s}\binom{13}{h}\binom{13}{d}.$$ The corresponding probability of drawing exactly $c,s,h,d$ cards of each suit is simply $$\frac{\binom{13}{c}\binom{13}{s}\binom{13}{h}\binom{13}{d}}{\binom{52}{c+s+h+d}}.$$ So to get $0$ clubs and $1$, $2$, or $3$ spades, this corresponds to $$(c,s,h,d) \in \{(0,1,0,2), (0,1,1,1), (0,1,2,0), (0,2,0,1), (0,2,1,0), (0,3,0,0)\}.$$ But we took a shortcut and collapsed the hearts and diamonds categories, so we get $$(c,s,h+d) \in \{(0,1,2), (0,2,1), (0,3,0)\}.$$ You can use this counting principle to compute other probabilities and generalizations.