probability of a run in pinochle

[rjay]

i'd like to know the probability of attaining a run, i.e., an Ace, Ten, King, Queen, & Jack in the same suit from a deck of 48 cards in a single pinochle deck which consists of eight each of aces, tens, kings, queens, jacks and nines.
i thought the factors might be as follows:

40/48 -> any card but a nine
8/47 -> if 1st card is 10 of clubs, there are eight desirable clubs left, i.e., AC,KC,QC,JC (2 of each)
6/46 -> 6 desirable clubs left
4/45 -> 4 desirable clubs left
2/44 -> 2 clubs left
but i don't care what order the cards are received in, so i believe there would be 5! ways of arranging the five cards...which led me to think of the following probability:

40/48*8/47*6/46*4/45*2/44*5! ~= .00897

but i'm not confident with that result and think it should be better than that, maybe about 2 or 3 % instead of less than 1%.

btw, i don't care what the other cards are and there are 11 cards dealt to each player (4 players) and four cards in the "kitty"

any help would be greatly appreciated

robert

[jemidiah]

Since I'm strapped for time, I'll refer you to a previous pinochle question (http://www.vbforums.com/showthread.php?t=574151&highlight=pinochle). If it's not helpful I'll read your post and respond specifically, but later :).

[rjay]

thanks for the reply, but that doesn't help me on this problem

[jemidiah]

Doing it your way with factors is problematic, because you haven't figured in that you might have multiple tries to get a certain card. For instance, say you deal yourself a 10C and AC; you're waiting on any 1 of 6 cards, with chance 6/46. Say you then deal a 9; you now have a 6/45 chance of getting a good card. Your model doesn't figure in these permutations.

Perhaps a better way of finding this probability is by counting the number of ways you could possibly deal a "run". To do so, find an algorithm with definite choices that doesn't impose an order and uniquely finds each possible "run":

1. Choose a suit; there is a choice between 4 things.
2. Choose which Ace of that suit to use; choice between 2 things.
3. Choose which K, Q, J, and 10 to use; 4 choices between 2 things = 2^4 choices; 2^5 from both (2) and (3).
4. Put those 5 in your hand without choosing their order. You have 6 remaining spots to fill. This is your only choice. (i.e. this step can be done in 1 way)
5. Choose your remaining 6 cards from the deck of 48-5=43. I'm fuzzy on a couple details, so I won't finish this part now.

Details:
What's the "kitty"? When is it dealt?
Do you want the probability that, assuming 4 people get dealt hands, one of them gets a run, or do you want the probability that a single person gets a run? (Note that they're not just related by 1/4.)
Is it still a run if you have both face cards of the given suit? Ex: is AC AC KC QC JC 10C 9D 9H 9S AH QD a run or not?

However you finish the above, you'll get some number of ways to construct a hand with a run (unordered). To get the overall probability, you'll need the number of ways to construct a hand in general (unordered). This is probably quite simple, though it depends on the kitty and the other players. Most likely it'll use a binomial coefficient.



One important thing to note is that I've treated the two different cards of each suit for a given value as different. That is, I've implicitly indexed the ACs as AC1 and AC2. This model gets us around questions of whether or not, say, AC1 KC1 and AC2 KC1 are equivalent hands--they're not.

[rjay]

thanks again for the reply jedimiah.
i'm only interested of the probability of a single run in any one suit to one person and not the probability of another person getting a run.
The kitty consists of four cards and is of no consequence to the one player getting a run.
Although, that would bring up an interesting point, i.e., the probability of a person having four of the five cards required for a run and then what would the probability be of the fifth card being in the kitty?

thanks again,

robert

[jemidiah]

Having never played Pinochle, from your previous response I'll completely ignore the existence of the kitty. It could be a hundred cards for all I care--hopefully this is correct. What about this point,

"Is it still a run if you have both face cards of the given suit? Ex: is AC AC KC QC JC 10C 9D 9H 9S AH QD a run or not?"