What are the Odds of this???

jimbling

It just doesn't get much better than this.......


sry about the full hand history.. the hand converter is not working properly.

***** Hand History for Game 2482177701 *****
NL Texas Hold'em $50 Buy-in + $5 Entry Fee Trny:14566943 Level:2 Blinds(15/30) - Friday, August 05, 13:31:07 EDT 2005
Table Table 11240 (Real Money)
Seat 1 is the button
Total number of players : 9
Seat 1: AO_RO_PWAL_W ( $740 )
Seat 2: Nikaj ( $2287 )
Seat 3: PERTHCHAMP ( $1020 )
Seat 4: laxmonkey ( $735 )
Seat 6: TyroneW777 ( $1610 )
Seat 7: whitepolobee ( $898 )
Seat 8: callDARESyou ( $735 )
Seat 9: jimbling ( $1375 )
Seat 10: RUPPERTJONES ( $600 )
Trny:14566943 Level:2
Blinds(15/30)
** Dealing down cards **
Dealt to jimbling [ 7h 7d ]
laxmonkey folds.
TyroneW777 folds.
whitepolobee folds.
callDARESyou folds.
jimbling calls [30].
RUPPERTJONES folds.
Nikaj: ty
AO_RO_PWAL_W folds.
Nikaj calls [15].
PERTHCHAMP checks.
** Dealing Flop ** [ 4s, 5d, 7s ]
Nikaj checks.
PERTHCHAMP checks.
jimbling bets [50].
Nikaj calls [50].
PERTHCHAMP raises [125].
jimbling calls [75].
Nikaj raises [550].
PERTHCHAMP is all-In.
jimbling is all-In.
Nikaj calls [745].
** Dealing Turn ** [ 3d ]
** Dealing River ** [ 9h ]
Nikaj shows [ 4c, 4d ] three of a kind, fours.
PERTHCHAMP shows [ 5c, 5s ] three of a kind, fives. :rolleyes:
jimbling shows [ 7h, 7d ] three of a kind, sevens. :eek:
jimbling wins 710 chips from side pot #1 with three of a kind, sevens.
jimbling wins 3060 chips from the main pot with three of a kind, sevens.
PERTHCHAMP finished in ninth place.
unreal
wow
PERTHCHAMP has left the table.



I have heard of this happening once before (that is I have heard of it once.. im sure its happened more than that) but what are the odds of this. Especially since there were only 3 of us in the hand.... unreal.

DeVore

in the short time I have been dealing cards, I've dealt such a thing twice.

DeV.

jimbling

DeVore wrote:

in the short time I have been dealing cards, I've dealt such a thing twice.

DeV.

lol... seriously? I wouldn't have believed it. Now Im only starting out, but I've played about 15000 cash game hands, and about 20000 tournament hands, and have never seen this. In fact I have only seen two people hit the set on a few occasions....

jimbling

jimbling wrote:

lol... seriously? I wouldn't have believed it.

and what I meant by this is that I wouldn't have believed it if it wasn't coming from such a knowledgeable and trustworthy fellow as yourself dev

Crumbs

(6/46)*(4/45)*(2/44) = 0.000527

1896.5 : 1

bmc

May I be so bold as to suggest that it would be stranger if this did not ever happen.

Between all of those who contribute to this forum there must be tens of millions of hands played during a year. The probability of something like this (or any particular bad beat) not happening to someone is miniscule compared to the probability of it happening to someone.

It seems though that most people don't see it as interesting enough to bother posting it.

jimbling

bmc wrote:

May I be so bold as to suggest that it would be stranger if this did not ever happen.

Between all of those who contribute to this forum there must be tens of millions of hands played during a year. The probability of something like this (or any particular bad beat) not happening to someone is miniscule compared to the probability of it happening to someone.

okay fair point... and your probably right, and everyone has come across it. But I thought it was an interesting situation that I have not ever seen, and every other person at the table also admitted they had never seen it happen, so I wanted to know the odds of it.

bmc wrote:

It seems though that most people don't see it as interesting enough to bother posting it.

no need to be so ****ty about it buddy.... And in fact I don't aggree. So your saying that people have more interest in posting a bad beat of his AA getting cracked by 8 3o???? cause I can't count how many of those types of posts I have come accross.... and yet i have only seen this hand once before.

This was not a bad beat post, and if your only input is snide remarks like that then please refrain from giving your opinion on the matter.

jimbling

Crumbs wrote:

(6/46)*(4/45)*(2/44) = 0.000527

1896.5 : 1

Thanks for the reply crumbs....

bmc

Alright, I got a bit ratty at the end there. Sorry. Feeling a bit that way today.

I did though, want to point out that it would be stranger if these things weren't happening (and more so for the bad beats).

As for the actual odds, (I think) they're a lot lower than crumbs' calculation.

P(three people being dealt different PPs) X (9/46) X (6/45) X (3/44) should give you the answer as far as I can see.

Three people getting different PPs I think should be...

(48/51) X (44/50) X (3/49) X (3/48) X (3/47).

So the final figure is...


(48/51) X (44/50) X (3/49) X (3/48) X (3/47) X (9/46) X (6/45) X (3/44)

Of course I may be wrong, but at least I've "contributed" now.

DeVore

Crumbs reply presumes that the pairs have been dealt already. There is that to consider, which makes it considerably less likely.

DeV.

DeVore

bmc. You're not right there, as each player with a pair has only *2* cards left in the deck rather then 3 (as you seem to be using).

DeV.

bmc

Damn it.

Yes.

The answer should be...

(48/51) X (44/50) X (3/49) X (3/48) X (3/47) X (6/46) X (4/45) X (2/44)

I must remember that 4 - 2 = 2...

Otherwise I think we're there.

DeVore

The calculation of the odds depends on how many players were at the table. (I presume 9).

So thats 9 chances at a 3 PP. That requires a logarithm to calculate (its not 9 * 1/17 as thats like saying rolling 6 dice is CERTAIN to bring you a six).

I dont have a log book to hand and I cant be arsed working it out but I'm sure someone will

DeV.

bmc

Sure?

There may have been nine pocket pairs but six folded. All we need is the probability that the three specific people who saw the flop had different pocket pairs to each other.

DeVore

No not sure. Will think about it.

DeV

DeVore

bmc wrote:

Damn it.

Yes.

The answer should be...

(48/51) X (44/50) X (3/49) X (3/48) X (3/47) X (6/46) X (4/45) X (2/44)

I must remember that 4 - 2 = 2...

Otherwise I think we're there.

Can you explain your reasoning behind these numbers again? I understand the last 3 (for hitting trips) but the first ones confuse me.... can you give me your rationale?

DeV.

NickyOD

DeVore wrote:

The calculation of the odds depends on how many players were at the table. (I presume 9).

So thats 9 chances at a 3 PP. That requires a logarithm to calculate (its not 9 * 1/17 as thats like saying rolling 6 dice is CERTAIN to bring you a six).

I dont have a log book to hand and I cant be arsed working it out but I'm sure someone will

DeV.

I'll have a go. Very roughly......

You need to work out the the odds of 3 player being dealt a pair first and that really depends how many players are at the table. If its 3 handed its aver 4000-1. or (16*16*16)

9 handed its more like. 135-1 "I think"

For all 3 players to flop a set.

(6/46)*(4/45)*(2/44) = .000525 or 1904:1 Times the odds of the players getting the pairs 1904*135=257,040.

And that's on a full table......


....hmmm think I forgot to carry a 1 somewhere.

bmc

I've got to go, so last post for this evening...


Of the three players, the first card dealt is irrelevant... P(1)

The second and third cards dealt must be different to those previous ones...

(48/51) X (44/50)...

The players second cards must match their first cards...

(3/49) X (3/48) X (3/47)...



I'll read back in tomorrow to see what ye think...

Haven't had time to read yours Nicky...

DeVore

Nicky, you have to give a rationale behind the numbers otherwise you are kinda just coming up with the answer "42" for the life universe and everything


BMC, I'm now certain that the number of players must enter the calculations somewhere. Using a kind of Reductio Ad Absurdum (ooo, isnt that a big phrase )... if we played the hand with 20 players rather then 9, the event would be more likely. And if we played it only with 3 then it is less likely. Therefore the event's probability is sensitive to the number of players and our calculations must incorporate that somehow....

DeV.

NickyOD

DeVore wrote:

Nicky, you have to give a rationale behind the numbers otherwise you are kinda just coming up with the answer "42" for the life universe and everything

It's 43! :rolleyes:

Amaru

I played a Party SNG not so long ago, where 7 pocket pairs were dealt in the first hand(including 10s to myself)! 5 went all in, and pocket Qs held up after making a set. Needless to say that person won the tournament fairly easily after that.

Hectorjelly

what are you trying to work out exactly?

BrendanB

Ah probability.

Alright, once the pairs are out, as has been agreed, I make odds of three sets as (6/46)*(4/45)*(2/44) = 0.000527.

The odds on the pairs being out in the first place are a bit trickier, as Dev has pointed out, NickyOD's taking of P(one player being dealt a pair on a given hand) to the power of three is wrong.

If the events were independent - i.e. the odds of each player getting a pair not changing as other players got / didn't get a pair, we could just use a binomial distribution. Even though they're not independent, you wouldn't expect a significant bias from the binomial - I resorted to simulation over 10 million deals to get a distribution of

0 58.0%
1 32.4%
2 8.2%
3 1.2%
4 0.1%

Which matches the binomial distribution to within 0.1% for each of them.

So for any given deal, the odds of exactly three players receiving a pair and each of them flopping a set, given they all see the flop is:

0.000527 * 0.012213 = a really small decimal = 155363 : 1

So if we take Tom's dealing (he's a bit slow), maybe 5000 hands, odds of him doing it twice are 2000 : 1 against. So, as we've always known, Tom is a bit special.

Crumbs

My way...

There are 1326 two-card combinations. 78 of these are pocket pairs. Therefore, there are 1248 non-pair hands.
If we deal one pair, that leaves 72 other pairs. Deal a second pair and it leaves 66.

In a 9-seated game, we want three people to get a pocket pair and six not to get one. There are 9c3 = 84 ways to do this.

So, the probability of dealing 3, and only 3, pocket pairs in a 9-seater game is

(78/1326) * (72/1325) * (66/1324) * (1248/1323) * (1247/1322) * (1246/1321) * (1245/1320) * (1244/1319) * (1243/1318) * 84 = 0.009424

ie. 0.94%, which is close to the 1.2% from Brendan's binomial distro.

Then multiply it by the set-over-set-over-set probability which gives 0.00049665%

ie. 201,347 : 1

MadsL

Tom is a bit special.

Nah, I've done it at a final table. QA2 flop, 3 players. Ouchies.

bmc

DeVore wrote:

if we played the hand with 20 players rather then 9, the event would be more likely. And if we played it only with 3 then it is less likely. Therefore the event's probability is sensitive to the number of players and our calculations must incorporate that somehow....

DeV.

Hmmm...

Agreed.

P(At least 3 distinct PPs at a nine handed table).

BrendanB wrote:

I resorted to simulation over 10 million deals to get a distribution of

That'll do me.


Jimbling? Any further elaboration required?

jimbling

lol... wow, a lot has happened since I last read the post... wasn't at my computer last night.

I think I'm pretty satisfied now.... thanks for the effort guys. I guess I should buy a probability book at some point

DubTony

Now I know why Im not winning anything. I skipped to many maths classes.

Wallko

yeah jesus i even did a bit of probability/statistics in college last year and still havent a clue whats going on,
mind boggling stuff
Jiberish its all jiberish i tell you

DeVore

Brendan writes: "I resorted to simulation over 10 million deals to get a distribution of"


Oooh, you so cheaty!!

There is something wrong here though. Crumbs calculations for the PP are (from a prob/statistical point of view), pretty far from BrendanB's. 0.94% vs 1.2% is quite a gap statistically... about 20% variance.

I'd like to point out, my mommy also says I'm special.

DeV.

BrendanB

Oooh, you so cheaty!!

Not really, I figured it was going to approximate the binomial, and tested it with a simulation that verified that it did. (And we've heard little theory from you, beyond the profundity that the number of players at the table might affect the outcome, oh great maths graduate ).

I'm not happy with crumbs' method, but I need to figure out in my head why. I have a funny feeling that an exact calculation needs brute force on each draw.

Crumbs

BrendanB wrote:

I'm not happy with crumbs' method, but I need to figure out in my head why. I have a funny feeling that an exact calculation needs brute force on each draw.

Most people aren't happy with my methods for anything in life but anyway, if there's a flaw in my reasoning, I'd welcome the enlightenment.

NickyOD

Crumbs wrote:

Most people aren't happy with my methods for anything in life but anyway, if there's a flaw in my reasoning, I'd welcome the enlightenment.

Hi Crumbs.

I think your method was good.

The only thing is it assumes that 3 and only 3 players will be dealt a PP. A more accurate equation would factor in the times 3 or more pairs would be dealt, so the number would be lower. However working out the probability of 3 players seeing a flop isn't possible, and the probability of set over set over set is probably much much higher that 200K-1 since the action would normally force small pairs to fold preflop. It's at least as rare a situation as a royal flush IMO. I've seen 3 royals this year online and not one set over set over set and I play a lot of limit holdem.

N.

BrendanB

Look at it from the point of view of the second draw - you have this as 72/1325. Firstly, the number of two card hands left in the deck isn't 1325, it's 1225 (50*49/2). In the deck at this point are 12 4-card ranks, and one 2-card rank. In other words, say the first hand drawn is a pair of twos. Now there are 4 of every other rank, and two twos remaining in the deck.

So the probability of drawing a pair here is:
P(One of 4 card rank)*(3/49) + P(One of 2 card rank)*(1/49)
(48/50)*(3/49) + (2/50)*(1/49) = 0.05952

As opposed to the (72/1325) = 0.05434 that's used.

Crumbs

BrendanB wrote:

Look at it from the point of view of the second draw - you have this as 72/1325. Firstly, the number of two card hands left in the deck isn't 1325, it's 1225 (50*49/2).

Ah, of course. *Bangs head* Good stuff, Brendan.

So the probability of drawing a pair here is:
P(One of 4 card rank)*(3/49) + P(One of 2 card rank)*(1/49)
(48/50)*(3/49) + (2/50)*(1/49) = 0.05952

Yes but we don't want the second pair to be the same as the first pair otherwise there's no chance of a set, so it's simply (48/50)*(3/49), right?

NickyOD wrote:

A more accurate equation would factor in the times 3 or more pairs would be dealt, so the number would be lower. However working out the probability of 3 players seeing a flop isn't possible

Exactly, so in this case we have to assume that nobody folded a pair preflop. But who in their right mind would fold a pair preflop anyway?

BrendanB

Yes but we don't want the second pair to be the same as the first pair otherwise there's no chance of a set, so it's simply (48/50)*(3/49), right?

Yeah, you're right, I'd forgotten we were even talking about hitting sets and was just looking at pair distribution. And that affects my previous results as well, I'd neglected to enforce a three different pair condition, I'll have a look back.

jimbling

to be fair... it is highly unlikely that anyone would have folded a pair pre-flop.... it was very early and there was no pre-flop raise, so in this particular case I would think it is safe to assume that there was only 3 pps dealt.

DeVore

I know Bren, I got to the conclusion that Crumbs just posted and realised the problem was very non-trivial. I also, funnily enough, hated probability in college and dropped it as soon as I could, after first year!

I think you might be right about the brute force on each draw too. I dont think the problem can be elegantly modeled with a single binomial.

DeV.