Subject: Re: Probabilities of decks of cards
Date: Thu, 17 Apr 1997 13:46:25 -0400
From: Alexander Bogomolny
Just to make sure we use the same notations: (n m) stands for the binomial coefficient n!/(m!(n-m)!). This is the number of ways to select m objects out of n. Also, to get a better understanding of this kind of problems, I would solve a simplified, more manageable analogue. For example use a deck of two suits with 3,4, or 5 cards each. You would gain a lot of insight by just having a whole sample in front of you.
Now, your basic formula p = e/s is correct. Personally, I would make a distinction between an event e and the number #e of elementary events (elements) it contains in which case the formula would look a little different p = #e/#s.
I understand in all 3 problems we deal with a selection of 4 cards out of 52. Therefore, #s = (52 4). Using this formula implies that you did away with a particular order of cards at the outset.
There are 13 = (13 1) distinct units. Therefore, #e = 13.
There are (12 4) ways to select 4 out of 12 face cards. Therefore #e = (12 4).
That's right: p = 1 - p(no two are alike). The first card is selected arbitrarily. There are 52 choices. The second card may be selected out of 48 cards that have a different unit. The third one is selected out of 44, and the fourth out of 40 remaing cards. In all you have 52*48*44*40=44*13*12*11*10. The four selected cards may be reshuffled in 4! ways, Therefore, #e = 44*13*12*11*10*4!. Which is the same as 44*(13 4). This shows we might have computed the numbers differently:
There are (13 4) ways to select 4 different units out of 13. For every selection there are 4 ways to pick a suit.