IMO, Just don't chop. People play so terribly.

Or you can run the scumbag route and only chop in sb.

getting AA in holdem is 1 in 221 i think. what are the odds of getting AAxxx in 5 card omaha? what are the odds some has AAxx when we block an A?

What's crazy is that in holdem there's like 200 hand (169) combos
In Omaha there's over 16k insane

I think question one is very similar to the one I answered above (we of course need bip! to verify the answer there)

Question two is just pokerstove no? I'm assuming you lost here and are pissed about it lol...if you are actually curious, 92.2%, 3.5%, 4.2% respectively.

No, I wasn't involved in hand 2.

Assume that you are starting with two suited broadway cards. What are your chances of making a royal flush by the river?

So I haven't quit the thread - I am actually working on a "all you ever wanted to know about BBJ odds" post. This much I know already - it is a hell of a lot of work! Holy ****. Been going for 6 hours and I don't think I am even half way to all of the corner cases for BBJ situations.

One teaser:
Odds of making quads using your pocket pair... an being 4th place at showdown in a 9 handed game..?!?! (not sure how the casino would divvy up the BBJ in this situation):

Boards:
52*51*50*49*48 / 5! = 2,598,960 boards
QJTXY - 543XY where X and Y each pair the boards = 4 suits * 3 higher pair combos * 3 lower pair combos * 8 ranks of the suited run * 3 pair-pair combos = 864 boards / 2,598,960 possible boards

Hole cards =
1 * 1 * 1 * 1 * 39*38/2 * 37*36/2 * 35*34/2 * 33*32/2 * 31*30/2 / 5! =
1 / 262,047,913

Put together - the odds of your BBJ quads being 4th place are:

864 / 2,598,960 * 1 / 262,047,913 =
864 / 681,052,043,970,480
~= 1 in 788,254,680,521

There 50*49*48*47*46 / 5! boards possible for your hand = 2,118,760 boards.

Of those, we need to figure out how many include the other 3 cards of your straight flush:

1*1*1*47*46/2! = 1,081 boards do the trick.

1,081 / 2,118,760 = 1 / 1,960 so one in 1,960

To get the exact number of Omaha starting hands:

52*51*50*49 / 4! = 270,725

In hold' em
52*51 / 2! = 1,326

Now, there is no appreciable difference between many of those "different" hands because the only difference is suit.. but there are the exact numbers

For holdem (yes you are correct), but here is the math:

4 / 52 * 3 / 51 = 1 / 221

Alternatively, there are 52*51 / 2 (ranked by order) starting hands = 1326
Out of those, there are 6 combinations of AA - AA , AA , AA , AA , AA , AA

6/1326 = 1/221

In Omaha, to get any AAxx (excluding AAAx and AAAA):

52*51*50*49/4! starting combos = 270,725

AAxx = 6 combos of AA * 48*47/ 2 = 6768 / 270,725 ~= 1 in 40 chance of a clean AA.

what are the odds some has AAxx when we block an A?

Heads up this would be (for clean AA, i.e. no AAAx):
3 combos of AA * 45*44/ 2 = 2970 / 270,725 ~= 1 in 91 chance

9 handed:
(48*47*46*45/4! * 44*43*42*41/4! * 40*39*38*37/4! * 36*35*34*33/4! * 32*31*30*29/4! * 28*27*26*25/4! *
24*23*22*21/4! * 20*19*18*17/4! )/ 8! ordered combinations of villains hands

(3*2*45*44/(2!*2!) * 44*43*42*41/4! * 40*39*38*37/4! * 36*35*34*33/4! * 32*31*30*29/4! * 28*27*26*25/4! *
24*23*22*21/4! * 20*19*18*17/4!) / 7! ordered combinations where one Villain has a clean AAxx

Take B / A from above (a lot cancels out) an leaves us:

(3*2*45*44 /4 / 7!) / (48*47*46*45/ 4! / 8!) =( 8 * 6 * 45 * 11) / (2 * 47 * 46 * 45) = 528 / 4324 ~= 12.2% ~= 1 in 8.2

This is incorrect (although the conclusion about whether completing the blinds for BBJ may be correct.. the math for BBJ odds is wrong). I will post a complete BBJ guide in the coming days.

To be fair, a couple of my early posts in the thread are also incorrect (i.e. a lot of times I said 90X more likely 10 handed for some stuff, but it is actually often 45X more likely)

With unpaired broadway cards:

You have AB, you need a board that is AABxx or ABBxx or AAAxx or BBBxx. I will ignore the straight flush possibility as I am assuming you have to use both cards (?yes). So if you are unsuited there is no chance for SF, but if suited there is a small chance (which we will ignore) to make your straight flush hand.

Anyways, there are 50*49*48*47*46 / 5! boards = 2,118,760 possible boards to come (if the cards were ordered by priority after all 5 dealt).

We need to know how many satisfy your FH+. I assume you have to use both cards:
First, some corner cases:
AABBB (where you have overfull and quads.. don't want to count it twice):
3*2*3*2*1/(2*6) = 3 boards
ABBBx (underfull and quads)
3*3*2*1*44/6 = 132 boards
AAABB = 3 boards
AAABx = 132 boards
AABBx = 3*2*3*2*44/(2*2) = 396 boards

normal cases
AABxx = 3*2*3*44*43 / (2*2) = 8,514 boards
ABBxx = 8,514 boards
AAAxx = 3*2*1*44*43/(6*2) = 946 boards
BBBxx = 946 boards

So, odds of making a FH or quads with unpaired broadway =
(3+132+3+132+396+8514+8514+946+946) / 2,118,760
= 19,586 / 2,118,760 ~= 1 in 108

(.. will come back to this for pocket pair in a second)

What about my question, cause I have another one ready for you?

Part two of ruken

Paired hole cards, what are odds of FH by river? I will assume that as long as your two cards can make a FH+, it does not matter if the board plays 4 cards or even 5 cards.. meaning I count TT with JJJT9 board.. even though you only play 1 card because you could still make a FH playing both (Omaha style)

(Note: this contradicts how I answered part 1 - really need clarification of what you are asking here)

(Very important thing for Ruken to clarify - does the FH have to be of a certain rank? i.e. are Jacks full of Twos good enough for your high hand promotion? I will assume yes...)

Again, 50*49*48*47*46 / 5! boards = 2,118,760 possible boards

We need any of the following:
(say your pair is of rank A)
ABBBB = 2*4*3*2*1*12 ranks of B/4! = 24 boards
ABBBx = 2*4*3*2*44*12 ranks of B/3! = 4224 boards
ABBCC = 2*4*3*4*3*12 ranks of B * 11 ranks of C / (2!*2!*2!) = 4752 boards
ABBCD = 2*4*3*4*4*12 ranks of B * 11 ranks of C * 10 ranks of D / (2! * 2!) = 126,720 boards
AAxxx = 2*1*48*47*46/(2!*3!) = 17,296 boards
BBBxx = 4*3*2*44*43*12 ranks for B/(3!*2!) = 45,408 boards

(24 + 4,224 + 4,752 + 126,720 + 17,296 + 45,408) / 2,118,760
= 198,424 / 2,118,760 = 1 in 10.7

You by definition have [KK-22]. I am going to igore any cases of 1 outer quads beating your higher set.. I think you are more asking about the odds of someone flopping a higher set than you.

The question implies the board is not paired. So I will also go with that.

We have hand PP, first the odds we flop a set:
Flop combos = 50*49*48 / 3! = 19,600
Flops with our card = P x y = 2 * 4 * 4 * 12 ranks for x * 11 ranks for y / 2! orders x-y ranks = 2,112

Our odds to flop a set (nothing more) are 2,112 / 19,600 = ~1 in 9.28

Let's assume 9 handed. There are 47*46*45*44*43*42*41*40*39*38*37*36*35*34*33*32 / (2^8 * 8!) opponent holdings possible.

The odds 2 other villains have sets:
3*2 * 3*2 * 43*42*41*40*39*38*37*36*35*34*33*32 / (2^8 * 2! * 6!)..
18 * 56 / (47*46*45*44) = 1,008 / 4,280,760
~= 0.0002355
~1 in 4,246.8

The odds only 1 villain has a set:
3*2 * 45*44*43*42*41*40*39*38*37*36*35*34*33*32 * 2 ranks / (2^8 * 7!) = 8*2*3*2 / (47*46) - 1,008 / 4,280,760 = 96/2,162 - (discount the already counted 3 sets on flop) 1,008 / 4,280,760
~=0.044168
~1 in 22.64 chances

So if you have a pocket pair, chances in isolation here:
You flop a set: 1 / 9.28
You got outflopped by another set
(with bottom set): 1 / 22.52
(with middle set): 1 / 44.8
You get a PP next hand [KK-22]: 1 / 18.42
You flop another set: 1 / 9.28
You got outflopped again by another set
(with bottom set): 1 / 22.52
(with middle set): 1/44.8

All together (I will assume you were bottom set each time):
~= 1 in 804,500

Thx, your assumptions were correct. I did some quick math and this should not happen to me in my lifetime ever again.

(Assuming live cash games)

Indeed - Very tiny chance you ever see this happen again.

...You would need to play full time for over a century to have ~50% chance of this happening again.

Yeah I rarely see set over set otf. The only time I seen it I flopped top set 77 over bottom set 55. He talked about that hand for my whole 10 hour session that night lol.

LOL just saw this. That was my hand Ibelieveinchipkelly was at my table. I had checkraised all in got both to call and the guy with the dominated flush draw went runner runner straight which of course also required the 3rd player to hit his set since he was holding blockers to the straight.

I'm guessing he posted it because I had just posted in this thread the same night the hand took place.

And truthfully I believe its a Murphy's law thing and the odds of the top pair top kicker w/ nut flush draw winning are inversely proportional to villain's stack size

odds of getting quads twice in a row ( I had back to back pocket pairs) Live at my local casino

I am going to assume you mean "getting pocket pair, making quads by river, next hand getting pocket pair, making quads by river again".

Odds of PP = 52/52 * 3/51 = 1/17
Number of boards:
50*49*48*47*46 / 5! = 2,118,760

Number of boards that give you quads:
2*1*48*47*46 / (2! * 3!) = 17,296

Odds of quads with your PP = 17,926 / 2,118,760 = 2 / 245

Odds of the entire sequence happening on your next two hands:
(1/17 * 2/245)^2 = 4 / 17,347,225
~= 1 in 4,336,806

In my first response to this, I overlooked the 5 card omaha... woops.

Here we go again:

52*51*50*49*48 / 5! starting hands = 2,598,960


# of those that are AAxxx (without AAAxx or AAAAx)
4*3*48*47*46 / (2! * 3!) = 103,776

103,776 / 2,598,960 ~= 1 in 25

Odds ILCD will get AA on LATB in 3 hours assuming 26 hands per hour?

Chances of AA in 1 hand = 1 / 221

So over 78 hands, chances of AA:
0 times = (220/221)^78 = 70.2%
more than 0 = 1 - (220/221)^78 = 29.8%


And if we got very specific, the odds of having AA exactly:
0: 70.2%
1: 78* (1/221)*(220/221)^77 = 24.9%
2: 78*77/2 * (1/221)^2 * (220/221)^76 = 4.35%
.. (I will fill all in up to 78 times in a row later)

at a 9 handed table, odds of someone having KKs or AAs when you have QQs.