Re: Probabilities
- From: Ray Vickson <RGVickson@xxxxxxx>
- Date: Sun, 2 Mar 2008 09:39:41 -0800 (PST)
On Mar 2, 9:00 am, andreas <anpa...@xxxxxxxxx> wrote:
Hi.
I am currently trying to understand how to calculate probabilities and so far I've done some progress on the simple areas.
However I'm quite stumped on a problem that was presented to me;
I have a normal card deck with 52 cards.
I randomly take 5 cards from the deck.
What is the probability that the 5 cards is a pair and three of a kind? (ex. 4,4,A,A,A)
My take so far:
The probability formula is if I understood it; P=G/M where P is the probability, G is the number of outcomes for the event and M the total outcomes.
That means that M is the 5 cards from the 52.
I can get that by using the combination formula; C=n!/(r!*(n-r)!)
so if n=52 and r=5, that would result in 2598960 outcomes of selecting 5 cards from 52.
That was the easy part, finding out G what I don't understand even where to begin.
There's 4 of each type in the deck and that confuses me. :/
Thanks for any pointers or advice.
One way is to compute the probability for a specific pair and triple
(such as your 44AAA) then multiply be the number of different such
choices, because for each way of choosing the pair and the triple, the
probability is the same. The pair can be any one of the 13, then the
triple any one of the remaining 12, for a total of (13)(12) = 156, so
P{pair,triple} = 156*P{44AAA}. Now to get P{44AAA}, imagine laying out
all 52 cards randomly in a row; there are 52! such possible rows. We
look only at the first 5 cards and ask for the number of rows that
have two 4's and three A's in the first 5 positions. That would be
your G, while the 52! would be your M. Note that we are looking at
permutations, not combinations, so the order matters. How many
permutations give the first five cards as 44AAA /in that order/? The
first card can be any of the 4's, then the second any of the remaining
4's, the third any of the A's, the fourth any of the remaining A's,
etc., for a total of (4)(3)(4)(3)(2)*N, where N = 47! is the number of
permutations of all the other cards. Remember: we are looking at
permutations, so the orders 4spade,4club and 4club,4spade are counted
separately. The other orders such as 4A4AA, 4AA4A, 4AAA4, AAA44,
A4AA4, A44A4, .... have the same number, so all that needs to be done
now is to figure out the correct number of distinct possibilities. One
way is just to enumerate all of them, but there may be fancier ways,
depending on your background knowledge.
R.G. Vickson
.
- References:
- Probabilities
- From: andreas
- Probabilities
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