In the thread, you ask an odds question, and I do my best to answer it with an explanation of the math behind it.

i.e.
What chance does Lloyd Christmas have with Mary Swanson?

(about tree fiddy?)


More serious question:
What are the chances of getting a suited connector dealt to you?

Answer:
For hole card combinations, there are 52 * 51 / 2 orders they can come = 1326 combos

To be a suited connector, there are per suit:
AK.. down to 32.. = 12 combinations * 4 suits = 48 combos

48 / 1326 =
8 / 221 = ~3.6% of hands.. maybe once an hour live.

Awesome, thank you. Definitely taking advantage of this

Odds of flopping an oesd with a one gapper?

Odds of AQo VS my AK of diamonds on a 6d 7c 2d Jh board.
And a Q landing on the river.

Odds of royal flush over straight flush using both hole cards from both hands. Odds for flopping it and turning it, rivering it?

Odds of having quads twice in an hour (after all board cards are revealed) assuming 30 hands/hr?

I will help out... These 3 everyone should know for live

- odds and percentage of flopping a flush draw with 2 suited cards
- of flopping a draw with connectors and gappers
- of flopping a set

Math not essential.. Just percent/ Answer is fine

This will be helpful stuff for the peeps!

Odds of making contact with aliens OR aliens making contact with us,
within my lifetime (I am 39 yrs old now, and expect to die when 78)
Thx.

2 outs (the two non diamond queens) out of 44 unknown cards:

2/42 = 1/21
~4.76%

... tough beat on that one

More of a short stack tourney question. In a play money online poker tournament (I know I know...) Anyway I had ~10 bb on the button with 44.

I tried to pokerstove my shove there against 2 random players and pokerstove seemed to be having issues. It gave less equity to V1 than V2. What sort of equity 3 handed on the button do you have with lower pocket pairs? (4-8). Follow up, do you shove in that spot?

78 is a good goal to make it to... sex is likely over after that so why go longer.

About the aliens... we will not have the technology to contact any life-capable planet in your lifetime, so the only possibility is them contacting us. In which case, the alien race would be far more advanced than us, and would IMO not reveal themselves... kind of to remind themselves of their long forgotten blissful innocence.

So 100% - they are among us now.

Odds of any straight flush over straight flush flopped.

Ex:

KQss vs 78ss

Flop JsTs9s

Set over set

Sent from my SCH-I535 using 2+2 Forums

Odds of me nailing the cute dealer that has been showing escalating interest in me?

Odds of that being a bad idea?

And a non fun one, odds of running quads vs a royal.

1. ~10% at min. 1/10 if your at a full table. (She's gonna sleep w atleast one of ya) If youre charming and have a big package odds increase dramatically.
2 . 1/69 of it ever being a bad idea. Do it and fist pump when finishing.
3. Not fun enough to want to figure it out.
(Bip) take over!

I will get to all of these.. just will go in order of easiest first.

I will work this one from the flop backwards.

Total number of possible flops:
(52*51*50) cards / (3*2*1) order = 22,100 flops

Number of flops that include 3 coordinated straight flush cards:
4 suits * [345 - TJQ] = 4 * 8 = 32

(32 / 22,100) flop odds... save for later

Now, let's assume full ring with 10 handed play.

V1 and V2 get exact cards needed:
1 / (49*48/2) * 1 / (47*46/2)


Number of ways V1 and V2 can be in a 10 handed game:
10 * 9 = 90

So 32 / 22,100 * 90 * 4 / (49*48*47*46) =11,520 / 112,379,030,400

~= 1 in 9,755,124

(anyone is welcome to check the math.. no guarantees! )

you sir, are one sick individual...I love it

This one is more difficult to be exact on, but I will give it a go...

A3 and AQ can flop open ended on very specific boards:
456 (A3) and 9TJ (AQ). Of course, we should be wary of the fool's side of the draw (A3), but the AQ on a JT9 flop is a great draw because you get the nuts with a K while others get fooled.

The odds:
Unique Flops = 50*49*48 cards / 3*2*1 orders = 19,600
Flops that fit the bill = 4*4*4 (i.e. 4 (J) * 4 (T) * 4 (9)) = 64

64 / 19,600 = 1 / 306.25 (~0.33%)

right flop for 1-gappers 42 and KJ:
More likely to flop open ended here as it only takes two specifc cards on the flop, but these are still limited on one end...
We still have the 3 board from before (i.e. 89T for KJ)..
but we also have QTx and 53x for each respective hand, so we add more combos into our total flops that qualify as open ended:
4 * 4 * 40 (40 others that do not make A FLOPPED STRAIGHT) = 640 additional ways
flopped straights:
4*4*8 = 128 (i.e. QT9 and AQT)

So bringing our open ended tally to 64 (three cards) + 640 (two cards) = 704

704 / 19,600 = 1 in 27.8 (~3.6%)
(note odds of flopping straight are 1 / ~153)

right flop for 1 gappers 53,64 and QT,J9:
Now we have two different ways to flop 2 cards to make open ended (i.e. J9x and KJx for QT). All the previous mentioned ways still exist too (including 987 run, etc) - let's just add the new way (like KJ for QT hole cards):
4*4*40 = 640

So 704 (from before) + 640 new ways =
1344 / 19,600 = 1 in 14.6 (~6.9%)
(note 64 more combos of flopped straights too, up to 192 / 19600 flopped straight, ~1%)

right flop for 1 gappers 75 through T8:
..still going.. these can get 3 run boards on both sides (T8 could get 567 or JQK)... not that you really want one end of this for combos < Tx:
but it is still 4*4*4 = 64 more flops to make an open ender by definition:

1408 / 19,600 = 1 / 13.9 (~7.2%)

Note, some of these include paired boards, monotone flops.. and a lot of 2 to a flush, so results not guaranteed even if OESD comes in!

9 handed game you are UTG with pocket 5s... odds that another player has a higher pair?

I will make an effort to get to all.. just going in no particular order here..

This one is not too hard to calculate very close to perfect. The exact answer is much more complicated and maybe I will answer it in the future.

But anyways, a good estimate:

Starting hand combos:
50 * 49 cards / 2 orders = 1225

There are 6 combod of each PP [66-AA] = 6 * nine ranks = 54 combos.

You have to fade (1225-54) / 1225 for eight players:

= (1171 / 1225) ^ 8 = 69.7% chance your 55 is the highest PP

..or alternatviely, 30.3% of the time at least 1 person has you dominated with a higher PP.


(to explain why the real answer is much more complicated.. the hands are not independent events.. i.e. if V1 has XX then those cards cannot be in V2s combos.. however my shortcut models it as independent)

Very nice... I read somewhere a few years ago (I'm thinking Phil Gordon) an easy approximation for this.

Percentage chance of higher PP = (Number of higher PPs)*(Number of players left to act)/2

So in this case 9*8/2=36% which is a little off from your number. It's kind of close but a little off.

Found this on Wikipedia too...
http://en.wikipedia.org/wiki/Poker_p...)#Pocket_pairs

Odds of 3 players flopping a flush in a full-ring game with only 3 players seeing the flop?

Saw this the other day and was shocked.

Equity Question:
This is definitely for poker stove or a similar program as the combinations for a full board run out, then figuring win/loss/tie go well beyond paper and pencil math.

I tried it in stove and it seems poker stove has a bug... same results as you.

edit:
it works if you switch to monte carlo (random runouts rather than waiting for all to enumerate):
Text results appended to pokerstove.txt

8,606,824 games 14.461 secs 595,174 games/sec

Board:
Dead:

equity win tie pots won pots tied
Hand 0: 31.600% 30.67% 00.93% 2639868 79886.33 { random }
Hand 1: 31.635% 30.71% 00.93% 2642841 80004.83 { random }
Hand 2: 36.765% 36.36% 00.41% 3129231 35210.67 { 44 }

.. and thus likely not a bug, just probably the order that it goes through all V2 hands while V1 hand sits at 22 or something.. monte carlo FTW

Strategy Question:
My advice guaranteed squat.. I am not much of a tournament short handed expert
I do not shove with 44 there because any thing that calls you either (a) dominates your hand -or- (b) is a coin flip with your hand. And because the payout is radically different from 3rd, 2nd, to 1st, I do not risk my tournament life in such a thin spot where villains will be likely to call. So basically, you are hoping for folds, and your might as well have been 72 if they fold. You have position, pots are easy to steal postflop in tournament play.. I would not shove, I would actually 2.5x raise... or even fold? Some expert may LOL at that advice.. but oh well. Also, BIG difference between 44 and 88 in this spot. I would raise and call any shove with 88 there. But again, in tournament play, even down to 10 bb, you are looking for the lowest variance gains that are practical - not slim edges in big pots. (I know 10 bb keeps you from being too picky.. but I wait on 44). However, 3 handed a lot of these decisions will depend on payout structure too.

bip! This thread is awesome! I'm a math/numbers nerd, got an 800 on the SAT, have always been quick with mental math, and still would be confused on how to start with a few of these. However, would you mind if I took a shot at some of the easier ones? I love problems like these! If you'd rather this be exclusively your thread just let me know. I'm gonna help out a fellow bosox fan either way though

For this one I think we need to assume everyone's suited in the same suit. Otherwise the odds of everyone flopping a flush is 0.

So if everyone is suited, let's say in hearts, there are 6 hearts being used. That leaves 7 left, and we'd need to see 3 fall on the flop to have everyone flop a flush. The odds of the first one coming would then be 7 (for 7 hearts) in 46 (for 46 unknown cards, since 6 are already being used). The second card would then be 6 in 45 since it's the same as the first but one heart and one card is now known. The third would be 5 in 44. We multiply these fractions together to get our answer.

7/46 * 6/45 * 5/44 = 210/91080 = .0023 = 0.23%

Now that's all given everyone going to the flop is suited. If bip! wants to go into the odds without knowing anyone's whole cards that'd be sick because I have no idea where to even start there.

Ignoring the bias players have towards seeing flops with suited hands (which I cannot quantify) - I will treat this as 3 random opponents hands:

We will work from flop backwards:

There are 52*51*50 / 3*2*1 orders unique flops =
22,100

Monotone combinations = 4 suits * 13*12*11 / 3*2*1 orders = 1,144 monotone flops:

(1,144 / 22,100) that the flop cooperates. Hold this number

Now, we need specific villain hands:
49*48/2 * 47*46/2 * 45*44/2 = 1,258,543,440 V1 V2 V3 potential lineups (*hold off on the reduction because of order)

They all need the specific suit:
10*9 /2 * 8*7/2 * 6*5/2 = 18,900

These can happen in 3*2*1 orders = 6

Odds = 6 * 18,900 * 1,144 / (1,258,543,440 * 22,100) = 1 in 214,400

(NOTE: The increase in likelihood for 10 players if all saw flop is 10*9*8 seats / 6= 120X.. the real odds are between the answer and 120X the answer to account for folding)

I'll have to double check this later.. something seems off with the 10 hand example..

Jump right in, no problem by me! It would be nice to have some cross-checking on the math. I did not get a chance to review this one yet - but will sometime tomorrow.. hitting the hay.