Royal Flush in holdem or double quads in Omaha
07-20-2013
, 01:21 PM
07-20-2013
, 06:03 PM
Regardless, two quads in omaha is much, much rarer, and you should propbet your friend if possible.
07-21-2013
, 04:03 PM
07-27-2013
, 05:58 PM
In a set of 13 there are 78 distinct groups of 2. ([13*12] / 2 = 78). To wit, since there are 13 different ranks in a deck of cards, there are 78 distinct varieties of “two-pair”. (Though doing so with larger numbers is typically impractical, the numbers with which this question deals are small enough to allow you to verify this in a matter of seconds. AAKK-AA22 = 12. KKQQ-KK22 = 11. QQJJ-QQ22 = 10 . . .)
In a set of 4 there are 6 distinct groups of 2. ([4*3] / 2 = 6). Therefor, each pair can occur in 6 distinct ways, which means each two-pair can occur 36 ways. (Once again, this can be verified with almost no effort: Spades/hearts, spades/diamonds, spades/clubs, hearts/diamonds, hearts/clubs, diamonds/clubs = 6 combinations)
In a set of 52 there are 270,725 groups of 4. ([52*51*50*49] / 24 = 270,725)
Frequency with which you're dealt two pair in Omaha: ~ 1 in 96.4
*
There are 1,712,304 distinct final Omaha boards. ([48*47*46*45*44] / 120), of which 44 will give you quads twice.
Frequency with which your two pair will quad up twice: 1 in 38,916
Frequency with which you'll end up with quads over quads in your hand (assuming you routinely travel all the way to the river): ~ 3,751,500 (96.4 * 38,916)
*
Translation: sixhigh nailed it.
In a set of 13 there are 78 distinct groups of 2. ([13*12] / 2 = 78). To wit, since there are 13 different ranks in a deck of cards, there are 78 distinct varieties of “two-pair”. (Though doing so with larger numbers is typically impractical, the numbers with which this question deals are small enough to allow you to verify this in a matter of seconds. AAKK-AA22 = 12. KKQQ-KK22 = 11. QQJJ-QQ22 = 10 . . .)
In a set of 4 there are 6 distinct groups of 2. ([4*3] / 2 = 6). Therefor, each pair can occur in 6 distinct ways, which means each two-pair can occur 36 ways. (Once again, this can be verified with almost no effort: Spades/hearts, spades/diamonds, spades/clubs, hearts/diamonds, hearts/clubs, diamonds/clubs = 6 combinations)
In a set of 52 there are 270,725 groups of 4. ([52*51*50*49] / 24 = 270,725)
Frequency with which you're dealt two pair in Omaha: ~ 1 in 96.4
*
There are 1,712,304 distinct final Omaha boards. ([48*47*46*45*44] / 120), of which 44 will give you quads twice.
Frequency with which your two pair will quad up twice: 38,916
Frequency with which you'll end up with quads over quads in your hand (assuming you always travel all the way to the river): ~ 3,751,500 (96.4 * 38,916)
Conversely, since the frequency with which you'll hold two to a Royal in your hand is ~ 1 in 33, while the frequency with which you'll get there (assuming you're dealt such a starting hand) is 1 in 1,960, you'll make a Royal by the river ~ 1 in 65k tries.
In other words, the Royal is in excess of 50 times more likely. (The disparity is even greater if you count the times when you use only one, or even zero, cards from your hand).
*
Translation: sixhigh nailed it.
*
By the way, a hearty LOL @
Grrrr . . . FML . . . I need to take a cr*p. But I'm “feeling a little too lazy” to actually stand up and walk all the way to the bathroom. Then again, this is a vinyl chair, and since I'm going to take a shower before I leave the house, maybe I'll just sh*t in my pants.
“Feeling a little too lazy” . . . Seriously ?
It took me all of fifteen seconds to do the math for the double quads in Omaha, and had I not already known the figure for the Royal in Holdem I'm guessing another fifteen seconds would have been enough to figure it out. As such, even allowing for the fact that I'm a math savant (oh wait . . . I'm not . . . dammit) it eludes me how anyone who's familiar with how to do such calculations could need more than a minute.
No offense intended, Rusty. You're probably a nice guy. Moreover, embarrassing strangers isn't how I get my thrills. (I much prefer to do it to those close to me
). But you just stepped right into that one.
In a set of 4 there are 6 distinct groups of 2. ([4*3] / 2 = 6). Therefor, each pair can occur in 6 distinct ways, which means each two-pair can occur 36 ways. (Once again, this can be verified with almost no effort: Spades/hearts, spades/diamonds, spades/clubs, hearts/diamonds, hearts/clubs, diamonds/clubs = 6 combinations)
In a set of 52 there are 270,725 groups of 4. ([52*51*50*49] / 24 = 270,725)
Frequency with which you're dealt two pair in Omaha: ~ 1 in 96.4
*
There are 1,712,304 distinct final Omaha boards. ([48*47*46*45*44] / 120), of which 44 will give you quads twice.
Frequency with which your two pair will quad up twice: 1 in 38,916
Frequency with which you'll end up with quads over quads in your hand (assuming you routinely travel all the way to the river): ~ 3,751,500 (96.4 * 38,916)
*
Translation: sixhigh nailed it.
In a set of 13 there are 78 distinct groups of 2. ([13*12] / 2 = 78). To wit, since there are 13 different ranks in a deck of cards, there are 78 distinct varieties of “two-pair”. (Though doing so with larger numbers is typically impractical, the numbers with which this question deals are small enough to allow you to verify this in a matter of seconds. AAKK-AA22 = 12. KKQQ-KK22 = 11. QQJJ-QQ22 = 10 . . .)
In a set of 4 there are 6 distinct groups of 2. ([4*3] / 2 = 6). Therefor, each pair can occur in 6 distinct ways, which means each two-pair can occur 36 ways. (Once again, this can be verified with almost no effort: Spades/hearts, spades/diamonds, spades/clubs, hearts/diamonds, hearts/clubs, diamonds/clubs = 6 combinations)
In a set of 52 there are 270,725 groups of 4. ([52*51*50*49] / 24 = 270,725)
Frequency with which you're dealt two pair in Omaha: ~ 1 in 96.4
*
There are 1,712,304 distinct final Omaha boards. ([48*47*46*45*44] / 120), of which 44 will give you quads twice.
Frequency with which your two pair will quad up twice: 38,916
Frequency with which you'll end up with quads over quads in your hand (assuming you always travel all the way to the river): ~ 3,751,500 (96.4 * 38,916)
Conversely, since the frequency with which you'll hold two to a Royal in your hand is ~ 1 in 33, while the frequency with which you'll get there (assuming you're dealt such a starting hand) is 1 in 1,960, you'll make a Royal by the river ~ 1 in 65k tries.
In other words, the Royal is in excess of 50 times more likely. (The disparity is even greater if you count the times when you use only one, or even zero, cards from your hand).
*
Translation: sixhigh nailed it.
*
By the way, a hearty LOL @
Quote:
I'm not gonna lie, I'm feeling a little too lazy to actually work out the numbers.
“Feeling a little too lazy” . . . Seriously ?
It took me all of fifteen seconds to do the math for the double quads in Omaha, and had I not already known the figure for the Royal in Holdem I'm guessing another fifteen seconds would have been enough to figure it out. As such, even allowing for the fact that I'm a math savant (oh wait . . . I'm not . . . dammit) it eludes me how anyone who's familiar with how to do such calculations could need more than a minute.
No offense intended, Rusty. You're probably a nice guy. Moreover, embarrassing strangers isn't how I get my thrills. (I much prefer to do it to those close to me
). But you just stepped right into that one.
07-27-2013
, 06:35 PM
07-30-2013
, 05:33 AM
And the odds of getting a Royal in Texas Hold'em are greater than 1 in 65k overall. OP didn't ask that both cards were needed.
With no restrictions, the probability defaults to 7 card stud:Combinations of royals: 4*47*46/2 = 4324
Total Combinations = C(52,7) = 52*51*50*49*48*47*46/(7*6*5*4*3*2*1) = 133784560
Probability of making a Royal flush by the river in hold 'em: 4324/133784560 = 1 in 30,940
Now probabilities of making a royal flush with using both hole cards:
Cominations of suited broadways:
4*C(5,2) = 40
Boards to complete the Royal:
1*C(47,2) = 47*46/2 = 1081
Total Hole card choices:
C(52,2) = 1326
Total board choices:
C(50,5) = 2118760
Probability of Royal flush using both hole cards:
40*1081/1326/2118760 = 1 in 64,974.
This post did take me more than 15 seconds, fwiw, but I hope it's a bit clearer at least.