Chinese Poker - Expectation of Royalties (Win 25$)

[cardcore]

Hi,

this is kind of a math question and math isn't my strength [img]/images/graemlins/laugh.gif[/img] would be nice if you give it a shot.

I'm looking for the Probability that my Opponent has (lets say its headsup) a Royalty in his 13-Card Chinese Poker Hand. I need that probability in terms of points. Lets say my actual CP Hand has an expectation of -2 (points), how much worse does the probability that he has a royalty make my hand?

We are playing 2-4 (2 for normal win, 4 for scoop) point system and following royalties:

Royal Flush: 10 points
Straight Flush: 4 points
Quads: 3 points
Full House in Middle Hand: 1 point
Trips in Front Hand: 3 points

natural royalties (instawins)
Six Pair - 6 points
Flush Flush Flush - 3 Points
Straight Straight Straight - 6 Points
13 Card Straight - 13 Points
13 Card Straight Flush - 50 Points


It doesnt have to be exact.

Thank you!
Best answer gets 25$ on PokerStars [img]/images/graemlins/grin.gif[/img]

[jay_shark]

The royal flush can only happen from the middle or the back . Given 5 random cards , the probability it contains the royal flush is 4/52c5.

a)4/52c5 + 4/52c5 - 4/52c5*3/47c5 but we can exclude the times that both events occur(the negative term) since this quantity is very small 4/52c5*2 and so we multiply this number by 10 .

b)Since it doesn't have to be exact , the probability is a bit less than 4*9/52c5*2 which omits the royal flush case . Also , remember to multiply this number by 4 .

c) 48*13/52c5*2 and we multiply this number by 3.
d) 13c2*4c3*4c2*2/52c5 . The probability that the back hand is better than a fh is simply:

4*2/52c5 + 4*9*2/52c5 + 48*13*2/52c5 . Take this number and multiply it by the probability of hitting a full house from the middle hand which should give you about 0.000000736 . Notice that this is even more rare than hitting a royal flush on either hand.

e) 4c3*13/52c3.We wish to determine the approximate probability that the back and middle hands are better . The only other hands we have omitted thus far are the flushes and straights .

straights: 10200 =4*4*4*4*4 -40 (sf)
flush: 5108 = 13c5*4-40(sf)

Add all the hands better than trips and you should get 4c3*13/52c3[(10200+5108+3744 +624 +40 )/49c5]^2 =0.0000002515.... Multiply this number by 3 .

Let me emphasize that these are rough estimates . There are tons of cases to consider but I'm neglecting them since they won't change the outcome very much .

Thats it for now , I will complete the rest later .

[jay_shark]

Part Two :


a)6 pair from 13 hands : There are 13c6 ways to select which pairs . once we've selected the 6 pairs , there is now an odd card out . 13c6*(4c2)^6*28 = 2,241,727,488 . This number divided by 52c13 is 0.0035302... . Again multiply this number by 6 .

b)13c5*13c5*13c3*4c3*3 We multiply by 4c3 since there are 4 ways to select 3 suits from 4 .Also we may have something like 10 cards in spades and 3 cards in diamonds . So we have to add an additional 13c5*8c5*13c3*4c2 + 13c5*8c3*13c5*4c2 . In total we have 6 364 873 944 different ways this may occur . The probability is close to 1% if you take that number above divided by 52c13 . I'm omitting the 13 card straight flushes as well as the occasional time that one of the 5 card hands is a straight flush .

c) I have to think of an easy way to approximate c .

d)(4^13 -4)/52c13=0.0001056. Multiply this number by 13 .
e) 4/52c13 . If you want it in points , multiply it by 50 .

a,d and e are exact answers but b is pretty close.

[cardcore]

thank you very much for the effort, that's awesome

PM me your stars sn and I ship you teh money [img]/images/graemlins/laugh.gif[/img]