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Question
QUESTION: A person draws 6 cards from a deck of 52 cards. What is the probability that he
has a royal
flush? To obtain a royal flush, he must have ace, king, queen, jack, and ten of
the same suit.
ANSWER: C(a,b) = a choose b = a!/((a-b)!b!).
He draws six cards, so the number of ways this could be done is
C(52,6).
There are only 4 ways to get a royal flush, but for each of these, there are 47 ways to get the other card. This would mean the answer is 4*47/C(52,6) = 4*3*2*1*47*6*5*4*3*2*1/(52*51*50*49*48*47) =
4*3*2*6*5*4*3*2/(52*51*5*49*48).
---------- FOLLOW-UP ----------
QUESTION: please can u explain why 4! is used instead of 4.
Now that I am almost at the bottom, you are right in this case -
it should be a 4 since there are only 4 ways to get a royal flush.
I notice at the start it is only a 4, but for some reason since
we are talking probability, I neglected that since factorials
show up so much and started doing 4!. Note that on that day I
answer 19 questions in a row and was starting to get warn out.
Maybe I should have taken a break ...
-----------------------------------------------------------
Where this does work is as follows:
If you look at the chance of drawing an ace out of a standard deck
that has 52 cards, you get 4/52 = 1/13.
If you look at the chance of getting two aces in 2 cards drawn,
you get 4*3 / 52*51.
If you look at the chance of getting all four aces in 4 cards,
the chance would be 4*3*2*1/(52*51*60*4).
The reason that it is 4*3*2*1 is after the first ace (4 chances
out of 52 = 4/52), there are only 3 left (out of of 51 cards), so
you multiply by 3/51. For the chance of another ace, there are
only 2 aces left, out of 50 cards, so the answer would be
multiplied by 2/50. For another ace, there is only 1 way
to get it out of 49 cards. The chance of the last ace is
therefore 1/49.
So in order to draw 4 aces from this deck, the probability would
be 4!/(52*51*50*49). NOte that the value in the denominator is
really 52!/48!, so the final answer is 4!48!/52!.
------------------------------------------------------------------
Again, now that I think about it, I'm sorry, but I think I gave
out the wrong formula. In fact, I know it's wrong. It should be
C(a,b) = a!(b-a)!/b!. That way the (b-a)! cancels with several
terms in the b! on the bottom.
I think I made another mistake. I must have been out of it
wnen I answered this question. There are only 4 ways to get
a flush, so it should be a 4 in this case and it was at the
start. I missed that when I was quickly checking it and thank
you for correcting me.
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Scott A Wilson
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