Thanks for the answer to the original question and also the one I didn't think to ask "what the hell is the Monte Carlo option in pokerstove?".

Props for the effort. I hope you find tutoring LLSNL helps strengthen your fundamental game as much as it gives us.

respect.

not easy at all and you proved yourself. I will be back with moar

Its play money, shove ATC.

The odds of the last 20 times I have had 99, my opponent has had QQ dealt to them.
(It doesn't matter who wins or loses with these hands)

Odds don't change

I never asked if odds change, I asked what the odds were.
Thx.

Hey bip what are the odds that none vs. at least one of a certain number of cards flop (obviously the former is 1-the latter and vice versa). So if I had a PP I could know exactly how likely it is that an overcard flops or if my opponent's range was heavily weighted to AK/AQ I could know how often we get a flop with no A/K/Q etc. Thanks buddy.

Odds of a KK preflop thread popping up before April 1st?

Part 1 of this is very similar to the "dual flopped straight flushes" question earlier in the thread. Here is that math made more specific that one hand is a Royal flush:


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

Number of flops that include 3 coordinated royal flush + straight flush cards:
4 suits * [TJQ] = 4 * 1 = 4

(4 / 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 4 / 22,100 * 90 * 4 / (49*48*47*46) =1,440 / 112,379,030,400

~= 1 in 78,040,993

Odds for a royal over a straight flush by the turn with both opponents using two hole cards (note this includes flopped RF and SF from above.. this is just anything "by the turn"):

(52*51*50*49) cards / (4*3*2*1) order = 270,725 boards

Boards that satisfy a royal + another SF using both hole cards:

(4) AJT9 with KQ and 87
(4) AJT8 with KQ and 97
(4) AJT7 with KQ and 98
(4) KJT9 with AQ and 87
(4) KJT8 with AQ and 97
(4) KJT7 with AQ and 89

(4 * 45xs) QJTx with AK and 98

204 qualifying boards / 270,725 (save for later)

Still the same math for two exact villain hands:
1/(48*47 / 2) * 1/(46*45 / 2) = 1 / 1,167,480

..and 90 ways those hands can be given to the villains in a 10 handed game (10*9)..


90 * 1 / 1,167,480 * 204 / 270,725 = 1 in 17,214,925


.. Any guests want to try the river?... I will get you started:


(52*51*50*49*48) cards / (5*4*3*2*1) order = 2,598,960


?
?

(4 * 44xs) AJT9x with KQ and 87
(4 * 44xs) AJT8x with KQ and 97
(4 * 44xs) AJT7x with KQ and 98
(4 * 44xs) KJT9x with AQ and 87
(4 * 44xs) KJT8x with AQ and 97
(4 * 44xs) KJT7x with AQ and 89

(4 * 45xs * 44ys / 2 orders of xy) QJTxy with AK and 98


Still the same math for two exact villain hands:
1/(47*46 / 2) * 1/(45*44 / 2) = 1 / 1,070,190

..and 90 ways those hands can be given to the villains in a 10 handed game (10*9)..

So, basically, figure out any more unique boards that make a royal flush and a straight flush.. and add that value Z into formula below:

(5016 + Z) / 2,598,960 * 90 * 1 / 1,070,190 = ?

Guaranteed. And 99% chance APD locks it.

I will assume this is generic quads (i.e. quads on the board, trips on the board + 1 in hero's hand, and paired board + pair in villain's hand)..

Odds of making quads with 7 cards:

(first combos of 7 cards) = 52*51*50*49*48*47*46 / 7! = 133,784,560

Number of these combos that include quads:
13 ranks * 48*47*46 junk cards/ 3*2*1 orders of junk cards = 224,848

Odds of quads on hand X if you stay in to river = 224,848 / 133,784,560 = 1 / 595


So.. chances of getting 2 or more of those in 30 hands = 1/595 * 1/595 * 30 hands * 29 hands / 2 hand orders =
1 in ~813.85

.. wow - not that bad of odds. Lesson to be learned... never fold

math to get it exactly twice... = (594/595)^28 * (1/595)^2 * 30*29 / 2 orders = ~1 in 853... don't want to include those times you get 3 or more quads in an hour

This is awesome BIP. Thanks, your the best. I was searching for this at the table for hours. Probably missed a few important things that could have made me money instead.

Appreciate the response. I am no math person, but I have to say that 1 in 400 seems much too small?

Well I'm assuming the three players taking the flop are all suited in the same suit. In that case we just calculate the odds of flopping three of the same suit when there's 7 of them left and 46 unknowns. 1 in 400 seems about right for that.

I actually factored an extra "6" in the flush problem.,, Here is one easy way to look at it:

If just 3 handed (note: this is not as witnessed where 3/10 players saw flop),.. but if just 3 handed, what are the odds all 3 flop a flush:

Well, simply the dealer must deal a card and then not stop dealing that same suit for a total of 6 hole cards and 3 flop cards:

52/52 * 12/51 * 11/50 * 10/49 * 9/48 * 8/47 * 7/46 * 6/45 * 5/44 = 19958400 / 25,674,286,176,000

= 1 in 1,286,390 .. again the longer math was ok procedure too, I just goofed by giving 6x orders for villains hands

Now if only required that 3 out of 10 players made a flush on the flop:
10*9*8 / 3*2*1 ways for these hands to be distributed = 120X factor.

So the odds are between 1/1,286,390 and 1/10,720.. Likely closer to to the 1/10,720 number because how people like to see flops with suited cards.

Odds of losing set over set in back to back hands? If in the future I only play live and lets say I play an average of 10 hours a month, should this happen again to me? (I am 27 and in this example make it until 80)

Thx

Okay. Odds for quad tens losing in hold'em using both cards?

my question would be to easy for you [IMG]http://s8.************/vtj6clagl/are_you_wizard.jpg[/IMG]

Does villain have to use both cards too? (like a bad-beat-jackpot)

Flopped set over set? Or any set over set by river?

Flopped set over set in consecutive hands(losing both).

In omaha, assuming a limpy or stationy table pre where most hands see a flop.

what are the chances of flopped :

set over set?
flush over flush?

by the river, flush over flush?

Odds of a perfect bracket? Always wanted to know this.

You have to correctly guess the outcome of 63 games (32 first round, 16 second round, 8 Sweet 16 matchups, 4 Elite 8 match ups, 2 Final 4s and the championship). If you were to close your eyes and guess, making each match up a 1/2 shot of being correct, then it'd simply be 1/2 * 1/2 * 1/2 * ... 63 times, or (.5)^63. This comes out to 1.08 x 10^-19 on the calculator, or 1.08 / 10000000000000000000, or .000000000000000000108%

That's if you closed your eyes and guessed. I assume since you do have a little bit of logic to be used (a 1 seed has never lost in the first round so you might want to give yourself 4 locks there) then that number should be slightly higher, but when a team like Florida Gulf Coast makes the Sweet 16 it feels like you might as well throw out a random guess on every game. That's why some websites and organizations don't have to be too worried about offering a huge prize, such as $1,000,000, to anyone who completes the perfect bracket.

Pretty much what redsox said above. There are 2^63 possible outcomes to a 64 team single elimination tournament. However, not every outcome is equally likely. Some matchups (especially early) one team may be as high as 98% to win. Even in later rounds, I would expect a gambling shark to be able to call a winner with >60% accuracy.

Let's just simplify the problem by saying every matchup has a 70% favorite. If you filled out one bracket, you would have the greatest chance of getting it right by identifying and picking the "favorites". Thus the odds of your bracket playing out as predicted would be:
0.7 ^ 63 = 1 in 5.74 billion

By comparison, if all outcomes were equally likely the probability would be:

1 / 2^63 = 1 in 9.2 quintillion

Also, a good shortcut for working with large 2^x problems... every 2^10 (=1024) can be closely estimated by 10^3 (=1000). So without a calculater one could estimate:
2^63 = (2^10)^6 * 2^3 ~= 1000^6 * 8

And
1000 ^ 2 = million
1000 ^ 3 = billion
1000 ^ 4 = trillion
1000 ^ 5 = quadrillion
1000 ^ 6 = quintillion

So without a calculator I could have estimated the odds if all games were 50/50 at 1 in 8 quintillion.

Expanding on the more practical odds of 1 in 5.74 billion for a favorites bracket to be realized in outcomes...

If gambling shark X played one of those brackets for the next 1 million years, the odds he would ever call a tournament right are:
1/[1 - (1 - 1/5.74 billion) ^ 1,000,000] ~= 1 in 5740