Hi everyone,
I'm not sure if this is the right forum but it seems like the right place.
I wanted to see the probability of something happening and I haven't looked at probability in a while, so if someone could quickly check my work I'd be appreciative. Assume a 52 card deck of cards that has had one of the royal flushes removed. If I select 20 cards, the odds of a royal flush being in there can be calculated thus:
3*c(42,15)/c(47,20)
This comes to around 3%
I figure this is because there are 3 potential royal flushes (I pulled one), and 47 cards total.
Am I doing the math right here? I left my textbooks on another continent.
Thanks,
D

Hi there.
Not spent all that much time thinking about this, so the following may have a mistake (these types of problems are known to cause even those with lots of experience to jump in too quickly).

So I believe that you are currently double counting, not taking into account the fact that your hand of 20 cards could contain not just one royal flush, but two or three. The best way to solve such problems is using the Sieve method, or as it is also known inclusion-exclusion. Suppose you have two shopping lists, and that some of them have duplicated objects (so both you and I want to buy toothpaste, etc.), then the total number of distinct items is equal to the length of your list, plus the length of mine, minus the number of duplicated objects. In mathematical notation, given sets A and B,

|A U B|= |A|+ |B|- |A n B|,

where U and n are union and intersection respectively. For three sets the formula is

|A U B U C|= |A|+ |B|+ |C|- |A n B|- |A n C|- |B n C|+ |A n B n C|.

Suppose you have removed the royal flush in spades from the deck, and let

A = {lists of 20 cards (of 47) which contain the flush in Hearts},
B = {lists of 20 cards (of 47) which contain the flush in Clubs},
C = {lists of 20 cards (of 47) which contain the flush in Diamonds},

Then you want to count exactly |A U B U C|, which is the set of all lists that have at least one royal flush. Now

|A|= |B|= |C|= C(42,15).
|A n B|= |A n C|= |B n C|= C(37, 10).
|A n B n C|= C(32,5).

After dividing by the total number of ways to choose the 20 cards (which you already noted was C(47,20) ), one has

P = ( 3 * C(42,15) - 3 * C(37,10) + C(32,5) ) / C(47,20)

Now after all that, if you were only interested in the percentage than these calculations have made little difference, as your answer comes out as 3.032%, whilst this answer gives 3.021%...!

I hope that is correct, and moreover I hope that it helps.

Owen

Thanks Owen!
I'll keep that in mind. Though it didn't make a lot of difference in the final percentage, I do like seeing the logic in probability math.
-D

Just returning to say that I checked the calculations and they are correct (I was reluctant to be too certain given that it was late at night in the UK)!

Indeed, the calculations make little difference to the percentage, the reason being that the probability that you should have two or even all three of the remaining royal flushes in your packet of 20 is so low as to be negligible.

Owen

I won't lie, the look on my face if all three of the royal flushes should show up in those 20 cards would be tremendously entertaining.

Interesting problem. Here's owens's math run through Wolfram|Alpha:
http://www.wolframalpha.com/input/?i=%28......se+20%29

This makes me think of one of my favorite online tutorials about binomial coefficients:
http://web.archive.org/web/20120929011708/http://mindyourdecisions.com/blog/2012/09/17/monday-puzzle-buying-fresh-fruit/?

Correct me if I'm wrong, but wouldn't you calculate the probability of getting 1, 2, or 3 complete royal flushes in 20 cards pulled from a group of 47 like this:
http://tinyurl.com/c2t42gu

By this calculation, you should get at least 1 complete royal flush around 1 out of 5 times, which seems to make more sense than about 1 out of 33 times.

Ignore my previous post...I realized my error.