deuces wild questions

On the INITIAL draw, what are the chances of getting:

(a) a deuce

(b) 2 deuces

(c) 3 deuces

(d) 4 deuces

thanks in advance

Thanks JB,

I only had the figures correct for the 4 deuces

I have a follow up question, In the initial deal, what is the probability of getting (all ways):

(a) one to a royal,

(b) two to a royal,

(c) three to a royal,

(d) four to a royal

(e) a royal (even though i know this one, may as well put it here too)

Quote: ksdjdj

Thanks JB,

I only had the figures correct for the 4 deuces

I have a follow up question, In the initial deal, what is the probability of getting (all ways):

(a) one to a royal,

(b) two to a royal,

(c) three to a royal,

(d) four to a royal

(e) a royal (even though i know this one, may as well put it here too)

Here are my answers:

Hand	Ways	Frequency
zero to a royal	201,376	1 in 12.91
one to a royal	1,731,200	1 in 1.50
two to a royal	622,200	1 in 4.18
three to a royal	43,240	1 in 60.11
four to a royal	940	1 in 2,765
a royal	4	1 in 649,740
TOTAL	2,598,960

Quote: ChesterDog

Here are my answers:

...

I agree. This wasn't as easy as the deuces question, since a hand can have up to four "one to a royal" combinations and up to two "2 to a royal" combinations and other mixtures where a lower count was dwarfed by a higher count in a different suit.

Thank you for all the help JB and ChesterDog,

would these formula(s) be correct, for the situations provided below (only guessing from the information provided)


(a) "Going for a Royal Flush, discarding 4 cards (keeping 1 of any card for a royal lets say a '10 of diamonds' as an example)

'=COMBIN(4;4)*COMBIN(44;1)/COMBIN(47;4)'

or about 0.02467% chance of happening

(b) "Going for 4ok deuces, discarding 3 cards (keeping 2 deuces)

'=COMBIN(3;3)*COMBIN(45;1)/COMBIN(47;3)'

or about 0.27752% chance of happening

nb: i am using OpenOffice, so it replaces "," with ";" when using the combin function

again i am only guessing here, so will be happy to be corrected if the above is wrong

edit: i think i only got the "(b) example" correct, after having a think about it?

Drawing 1 to a royal (ie: Hold Ten of Diamonds, discard 4 others)....your chance of getting a royal are significantly less than 0.02%.

I don't know how these fancy math people do it (well, I "get" it, but I've never liked permutations and combinations). Holding 1 to a royal.....

the first of four cards you draw can be a JQK or A (4 cards). And there are 47 cards remaining. So the chance of that first card being "good" is 1/47.

Chance the second of four cards is good is dependant on the card before it (if the previous card is no good, then it doesn't matter what this next card is). There are 3 remaining good cards out of a possible 46 cards.

Do the same for the next 2 cards, and you get:


4/47 * 3/46 * 2/45 * 1/44

=

(4 * 3 * 2 * 1) / (47 * 46 * 45 * 44)

=

4! / (47! / 43!)

thanks RS

Quote: ksdjdj

On the INITIAL draw, what are the chances of getting:


(d) 4 deuces

thanks in advance

standard answer is 50/50 :-)

I was fortunate just a couple days ago. Had only $40 to play on y way home from work. Needed 2400 pts for president tier at Stations. Decided to play -ev 5 play 5c DW,'

Dealt 4 2's for 5 hands. cleared my tier, and cashed out $500.

Went over to my regular FPDW and caught 4 2's again for $250, Net for the day +$650.