While playing cards what is the probability of being dealt a hand of 8 cards that is already sorted if we ignore suits and just sort numerically, either high to low, or low to high?

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- Dec 20th 2007, 11:51 AM #1

- Dec 20th 2007, 12:06 PM #2

- Dec 20th 2007, 12:47 PM #3
I may be misinterpreting and way off base, but this is sort of like a straight only with 8 cards instead of 5. Are there restrictions?. Are we allowed, say, J,Q,K,A,2,3,4,5?.

Or just numbered cards, 2,3,4,5,6,7,8,9, for example?.

The number of consecutive groups of 8 cards is 3. If we use A,2,3,...,10.

In each sequence of 8 we can choose each card in 4 ways, this gives $\displaystyle 4^{8}=65,536$ ways.

So the total is 65,536*3=196,608.

But this includes straight flushes as well.

$\displaystyle \frac{196,608}{C(52,8)}$.

It's, no doubt, more complicated than this, but perhaps it's a start.

- Dec 20th 2007, 01:15 PM #4

- Dec 20th 2007, 01:25 PM #5

- Dec 20th 2007, 07:09 PM #6

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Hello, MathGuru!

What is the probability of being dealt a hand of 8 cards that is already sorted

if we ignore suits and just sort numerically, either high to low, or low to high?

Without stretching the interpretation of the problem,

. . I assume we want an "eight-card straight " dealt in numerical order.

There are: .$\displaystyle {52\choose8} \:=\: 4,515,258,175$ possible eight-card hands.

Let's assume*increasing*order . . .

There are 7 possible straights . . . from A-to-8 to 7-to-A.

For each straight, there are: $\displaystyle 4^8$ possible selections.

Hence, there are: .$\displaystyle 7 \times 4^8 \:=\:458,752$ ways.

Therefore: .$\displaystyle P(\text{8-card straight, increasing order}) \;=\;\frac{458,752}{4,515,228,900} \;=\;\frac{16,384}{161,258,175} \;\approx\;0.0001 $

. . and: .$\displaystyle P(\text{increasing or decreasing}) \;\approx\;0.0002$