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?
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 ways.
So the total is 65,536*3=196,608.
But this includes straight flushes as well.
It's, no doubt, more complicated than this, but perhaps it's a start.
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: . 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: possible selections.
Hence, there are: . ways.
. . and: .