Re: Probability of a "Super" Royal Flush

nice one lads: it's pretty unlikely then! thanks aswell for the
workings - I think it's all pretty much cemented in my head ok.

If there were 10 of us sitting round a table playing one hand of
betholdem what'd be the chance that during that game one of the 10 of
us got that set of cards!!? (this game involves 5 cards being in the
middle of the table and shared by all 10 players and then each player
getting dealt 2 cards). I know it must be more likely then the below
but how much more likely.

Any help greatly appreciated
J



Janzon wrote:

WhytheQ skrev:

Can anyone please show me how to calculate the following probability
The chance of being dealt, IN SPADES, the following:

8: 9: 10: J: Q: K: Ace (i.e 8 of spades through to Ace of spades)

Any help geatly appreciated,
Jason

(...if workings could be included that'd be much appreciated, as I'm
trying my best to understand some basic probability)

Assuming you are familiar with the "number of combinations" function
(also called "choose"):

How many ways are there to get the above hand, not caring about the
order of the cards? Answer: 1

How many ways are there to deal a seven card hand, not caring about the
order of the cards? Answer: choose(52,7).

Hence the probability is 1/choose(52,7) = (52-7)! 7! / 52! = 1/133784560

.

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