# Thread: Question from someone via email

1. ## Question from someone via email

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?

2. Originally Posted by MathGuru
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?
Oh, a sorted question makes this one tough. Would this follow a hyper geometric distribution?

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.

4. There are some real issues with this one.
Suppose we allow multiple denominations.
That is say 2246667Q is a sorted hand of eight.
Then there are 120055 possible hands.

5. There are some real issues with this one.
That's exactly what I was thinking, but I gave my 2 cents anyway. Probably worthless, but...........

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