How do you work out the probability of winning rummy in the first hand of cards?

Question

presumimg 2 people are playing, you get given 7 cards to start with and they are the winning ones!

Answer 1

I assume you are playing seven card rummy with one deck. There are different ways to achieve a winning rummy hand:

Set of 3 and set of 4: For any given rank, there are 4 sets of 3 and 1 set of 4 so this is just 4*13*12 = 624

Run of 3 and set of 4: Every run will exclude a set in exactly 3 ranks so this is just 4*12*10 = 480

Set of 3 and run of 4: Every run will partially exclude 4 ranks so this is 4*11*(9*4 + 4) = 1760

Run of 4 and run of 3:
Case 1 -- different suits is just 4*12*3*11 = 1584
Case 2 -- same suit the first run excludes possibilities for the second depending what rank it begins with. This is
4*(7+7+6+5+5+5+5+5+5+6+7+7) = 280 The numbers in parentheses are the number of runs of 4 for each possible run of 3 beginning with ace low and ending with queen low.
Total = 1864

Grand total = 624 + 480 + 1760 + 1864 = 4728 possible rummies.

There are 52C7 = 133784560 rummy hands so the probability is

about .00003534 or about 1 in 28000.

Answer 2

Presuming you get dealt 7 cards the probability of getting a designated set of 7 cards based on removal and not replacing is about 1 in 90000000 this is making the (hopefully) unlikely assumption you are playing on your own. You would then need to take into account how may possible independent successful hand there are.

But I believe that the results would vary depending on how many people you dealt to...so who is playing?

Answer 3

p=1/no of players

Answer 4

Too many variables