Re: combinatorial question
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Date: 03/18/05
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Date: 18 Mar 2005 09:36:27 -0800
As is often the case, calculate the probability of that which you do
not want and then subtract from 1.
Let us take the cards A (hearts), 8 (hearts), Q (clubs) as a starting
point.
To show the gaps this can be written as [A]1234567[8]9(10)J[Q]K
There are 4 factors to be multiplied.
(1) The factor 13.
Since you say that the ace can be both high and low, any card can be at
the beginning of the list.
(2) A factor for the number of subgroupings.
To show the grouping, [A]234567[8]9(10)J[Q]K can be written as 742.
The various groupings are
13
11, 2
10, 3
9, 4
9, 2, 2
8, 5
8, 3, 2
7, 6
7, 4, 2
7, 3, 3
7, 2, 2, 2
6, 5, 2
6, 4, 3
5, 5, 3
5, 4, 4
4, 4, 3, 2
4, 3, 3, 3
3, 3, 3, 2 2
3, 2, 2, 2, 2, 2
For 742 the number of subgroupings is 3!/(1!1!1!). Instead of
[A]234567[8]9(10)J[Q]K you could have [A]234[5]6789(10)J[Q]K.
(3) A factor for making up 10 cards. With the chosen example there are
3×3=9 cards (the cards must be of the same denomination as those
already there) to choose from, and you have to choose 10-3=7. The
number of possibilities is combination(9,7).
You can forget about the grouping 76 and others in which you have only
2 denominations, since you need 8 more cards and you can only select
from 6 cards.
(4) A factor to take into account that there are different suits.
There are 4^3 possibilities in the chosen example.
There is another factor that one has to divide by, and that is for
cyclicality. However, for 13 cards in a suit this factor will always
be 1. Should you decide not to include, say, the king, and hence have
12 denominations, this factor will have to be considered.
The easy part is that there are combination(52,10) ways in which 10
cards can be selected from 52 cards.
Multiply the factors for the groupings, add, divide by
combination(52,10) and subtract from 1.
rbmadden@ualr.edu wrote:
> Ten cards are selected without replacement from a deck of 52. What
are
> the chances that at least two of the cards will be in sequence. A
> sequence is defined to be at least two cards with no gaps and may or
> may be in the same suit. Also an Ace counts as a low card(so an A23
> would be a sequence) and as a high card(so an AK would also be a
> sequence). Again, the cards in sequence do not all have to be in the
> same suit. I can solve the problem(using Feller's theory of runs
> concepts) when the cards must be of the same suit, but not requiring
> the same suit in the sequence makes the problem(IMO) much more
> difficult.
> Any help would be appreciated>
>
> Burt
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