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View Full Version : Wild cards and 5K vs Royal calculations- please correct!

Easy E
04-26-2003, 10:53 AM
Some of this doesn't "feel" right- please straighten out the idiot:

5K with 4 deuces wild= each card rank &gt; 8*7*6*5*4/5! times 12 ranks (can't have 5 twos) = (6720/120)*12 = 672 five of a kind

Royal flush = each suit = 9*8*7*6*5/5! time 4 suits= 15120/120 *4 = 504 royal flushes

Now, what are the hands with 1 rank card and 4 deuces- are they royal flushes or 5K or both?
There are 20 hands that meet this criteria. Do we subtract them out of both, or just the more common hand (i.e five of a kind, so it would be 652 vs. 504)?

b) With bug (joker) for 5 card hands
[NOTE- I believe I am misusing the !factorial notation here. 7!= 7*6*5*4*3*2*1 normally. I think the proper notation should be 7!5.... and the correct uniqueness equation would be (7!5)/5!, or 21]

5K- each rank = 1 chance (5!/5!) * 13 chances = 13 five of a kind
Royal- each suit &gt; 6!/5! * 4 = 24 royal flushes

c) With 2 jokers:
5K =&gt; 6!/5! * 13 = 78 five of a kind
Royal = 7!/5! * 4 = 21* 4= 84 royals

d) With 3 wilds: 2 jokers, 2 of hearts wild (for easiest approximation; would depend on the card- if one-eyed jacks and suicide king are the wildcards, the calculations should change?)

5K&gt; 7!/5! * 13 = 273
Royal&gt; 8!/5! * 4 = 224 royals

Someone braver than me can try the other card rankings. I think four of a kinds would have to be calculated in this manner:

Natural, non-deuce kicker +
3 ranks, 1 wild, one kicker +
2 ranks, 2 wild, one kicker +
1 rank, 3 wild, one kicker not a straight flush kicker
= sum of one rank's 4 of a kind hand * 12 (again, using 2's as wild, four twos wouldn't be possible)

So,
Natural= 4*3*2*1*40/5!= 8 of a rank
3+1+k = 4*3*2*4*40/5! = 32 of rank
2+2+k= 4*3*4*3*40/5! = 48
1+3+k = 4*4*3*2*40/5! = 32
= 120 *12 ranks
= 1440 4 of a kinds?

(in a five card hand, there are 624 non-wild 4 of a kind hands = 48 for each rank * 13 ranks. I'm not sure why factorials don't work for that calculation- I must be thinking incorrectly about this)

Thanks
Easy E

BruceZ
04-26-2003, 10:32 PM
Now, what are the hands with 1 rank card and 4 deuces- are they royal flushes or 5K or both?
There are 20 hands that meet this criteria. Do we subtract them out of both, or just the more common hand (i.e five of a kind, so it would be 652 vs. 504)?

Subtract the 20 from the lower ranking hand, since you will count these as the higher ranking hand. Normally 5-of-a-kind is higher ranking than any straight flush including a royal flush. It depends on your rules. The only standard version of poker that allows 5-of-a-kind would be draw played with a joker, and then the joker only counts as an ace or a bug to complete straights and flushes, so you can only make 5 aces, and this is the highest hand.

[NOTE- I believe I am misusing the !factorial notation here. 7!= 7*6*5*4*3*2*1 normally. I think the proper notation should be 7!5.... and the correct uniqueness equation would be (7!5)/5!, or 21]

5K- each rank = 1 chance (5!/5!) * 13 chances = 13 five of a kind
Royal- each suit &gt; 6!/5! * 4 = 24 royal flushes

You want C(7,5) = 7!/(5!2!) = 21. This is combinations of 7 things taken 5 at a time. P(7,5) = 7!/5!, this is permuatations of 7 things taken 5 at a time which counts order. C(7,5) = P(7,5)/2!

C(7,5)*4 = 84.

c) With 2 jokers:
5K =&gt; 6!/5! * 13 = 78 five of a kind
Royal = 7!/5! * 4 = 21* 4= 84 royals

C(6,5)*13 = 78
C(7,5)*4 = 84

d) With 3 wilds: 2 jokers, 2 of hearts wild (for easiest approximation; would depend on the card- if one-eyed jacks and suicide king are the wildcards, the calculations should change?)

5K&gt; 7!/5! * 13 = 273
Royal&gt; 8!/5! * 4 = 224 royals

C(7,5)*13 = 273
C(8,5)*4 = 224

Someone braver than me can try the other card rankings. I think four of a kinds would have to be calculated in this manner:

Natural, non-deuce kicker +
3 ranks, 1 wild, one kicker +
2 ranks, 2 wild, one kicker +
1 rank, 3 wild, one kicker not a straight flush kicker
= sum of one rank's 4 of a kind hand * 12 (again, using 2's as wild, four twos wouldn't be possible)

So,
Natural= 4*3*2*1*40/5!= 8 of a rank
3+1+k = 4*3*2*4*40/5! = 32 of rank
2+2+k= 4*3*4*3*40/5! = 48
1+3+k = 4*4*3*2*40/5! = 32
= 120 *12 ranks
= 1440 4 of a kinds?

(in a five card hand, there are 624 non-wild 4 of a kind hands = 48 for each rank * 13 ranks. I'm not sure why factorials don't work for that calculation- I must be thinking incorrectly about this)

There are only 44 naturals of each rank since there is only one way to pick the 4 cards, and 44 non-deuce kickers. You don't have to divide by factorials when multiplying combinations, only when converting permutations to combinations.

Natural= 44 of a rank
3+1+k = 4*4*44 = 704 of a rank
2+2+k= 6*6*44 = 1584 of a rank
1+3+k = 4*4*44 = 704 of a rank
= 3036 *12 ranks = 36,432

Or you could just compute C(8,4) minus the 1 hand with 4 deuces to get [C(8,4)-1]*44 = 69*44 = 3036 of a rank. 3036*12 = 36,432 as above.

Easy E
04-27-2003, 09:52 AM
[NOTE- I believe I am misusing the !factorial notation here. 7!= 7*6*5*4*3*2*1 normally. I think the proper notation should be 7!5.... and the correct uniqueness equation would be (7!5)/5!, or 21]

You want C(7,5) = 7!/(5!2!) = 21. This is combinations of 7 things taken 5 at a time. P(7,5) = 7!/5!, this is permuatations of 7 things taken 5 at a time which counts order. C(7,5) = P(7,5)/2!

C(7,5)*4 = 84.

Bruce, I don't understand the combinations equation. Doesn't C(7,5) = 7!/(5!2!) translate to (7*6*5*4*3*2*1)/[(5*4*3*2*1)*(2*1)] ?? if so, what is the 2! for?

I KNEW the 4K equations looked wrong- proves I can't subtract properly either (duhhhh)

BruceZ
04-27-2003, 10:16 AM
P(7,5) = 7!/5!, this is permuatations of 7 things taken 5 at a time which counts order. C(7,5) = P(7,5)/2!

P(7,5) = 7!/2! and C(7,5) = P(7,5)/5!

BruceZ
04-27-2003, 10:19 AM
Bruce, I don't understand the combinations equation. Doesn't C(7,5) = 7!/(5!2!) translate to (7*6*5*4*3*2*1)/[(5*4*3*2*1)*(2*1)] ?? if so, what is the 2! for?

Yes. The 2! cancels the 2*1 in the numerator, so you just have 7*6*5*4*3/5! = 21. You evaluated all these correctly, just used the wrong notation, or at least one I have never seen. You are apparently using 7!5 for
P(7,5). The 7*6*5*4*3 is P(7,5) = 7!/2! and has 5 numbers. We divide by 5! to remove order and change
P(7,5) into C(7,5).