What are the odds on this: 3 sets and a str8 on the flop.

[bobman0330]

[ QUOTE ]
've seen a similar hand in a live tourney at the Trop in AC

[/ QUOTE ]

Come on, everyone knows the Trop is rigged. I bet your opponents had cashed out recently.

uh....what?

[ohnonotthat]

If we start with the assumption that all the hands necessary for this to happen were in fact dealt out (and were played) . . .

The odds against AA, KK and QQ all flopping sets while JT flops a straight are calculated in the following manner:

There are 8 ways of flopping A-K-Q. (2 Aces x 2 Kings x 2 Queens).

There are 13,244 (44! / 6!) possible flops.

The odds against this occuring are therefore 13,244 divided by 8 . . .

Or, 1: 1,655.5

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If instead of AA, KK and QQ the three pairs were TT, 99 and 88 it would theoretically be a bit easier since there are two "straight draws" that would get there . . . Q-J and 6-7

However, since the question specified three sets plus the NUT straight, the above answer applies.

If we further assume that this does in fact occur, the chance that the lowest set - QQ in my example - quads up while NEITHER OF THE OTHER PPs QUAD UP is 1: 38.

Thus the chance of these unlikely events all occurring in the same hand is . . .

1: 62,909

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Bottom line is it's not going to happen often.

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Also, remember that this figure assumes that these hands were in fact dealt out.

If we instead wish to calculate the chances that:

1. AA, KK, QQ and JT are all dealt out

2. The flop is A-K-Q

3. The river is a Queen

4. The turn is neither an Ace nor a King

Wow, I am tired and will leave it to one of our human calculators to do that equation but it is well into the > million to one range.

That said, I posted some time back about the average person's tendency to confuse the UNUSUAL with the UNLIKELY.

A royal flush is both but an A-K-9-5-2 flush is only the latter. It is precisely as UNLIKELY as the royal but far less memorable.
In other words, based solely on the fact that it did indeed occur, it's extremely unlikely that "the fix was in".

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FWIW, I was on the lowest end of set/set/set and did quad up; this was in a [live] 6-12 some 7 yrs ago. This was the only time I have been involved in a hand where 3 sets were flopped tho I have been in countless hands where the turn or river produced the 3rd set in a pot where 2 sets were flopped.

I have been on the wrong end of set/set and quaded or flushed several times - and have had it happen to me several times.

"Sh!t happens - usualy to me". [img]/images/graemlins/smile.gif[/img]

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P.S. In a hand where there are three sets giong into the river [which pairs the board] and the showdown features all 3 PLUS a straight or flush . . . well, what can I say.

Win or lose you are in a great game; the river almost definitely was capped and yet the straight (or flush) hung on till the bitter end.

If that straigh or flush was not the nuts on the turn, well . . . superwow.

It doesn't get much better than this.

[ohnonotthat]

for sympathy in the dictionary.

It can be just after sh1t and just before syphilis. :-)

[ohnonotthat]

WHo happens to teach probability ?

YIKES !

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.12 cubed is not the correct answer.

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If "Joe", "Bob" and "Bill" all hold pocket pairs and it is known that both Joe and Bob flop sets, the chances of Bill also flopping a set is not .12

BAYES THEOREM is at work here.

If we know that "A" and "B" occured and want to know whether "C" occured, the chances are not static if we are referring to a series of DEpendent trials.

- The same would hold true if the calculation was to determine the chances of both AA and KK flopping sets in the same hand. Since the chance of "B" occurring is dependent on whether or not "A" occurs they are dependent and therefore do not lend themselves to the same method as . . .

If the chance of one die landing on 6 is 1:6 what are the chances that a set of two dice show 6-6 ?

For anyone who doesn't get it . . .

If AA, KK, QQ and JJ all see the flop, what are the chances that ALL FOUR flop a set ?

Zero, right ?

OK . . .

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NEXT [img]/images/graemlins/smile.gif[/img]

[ QUOTE ]
I've seen a similar hand in a live tourney at the Trop in AC that I eventually won. 12 left in tourney, so we're playing 6-handed.

Pre-flop:
UTG (which is me) has 66 and raises
SB has 22 and calls
BB has 33 and calls

Flop comes 6-3-2 Rainbow.
I bet.
SB Calls.
BB raises all-in.
I call.
SB folds & shows his trip dueces. I was pissed, because I should have taken them both out.

Anyway, here's the math:

The probability you flop a set is about .12 (which is about 7.2-to-1).

3 people flopping sets is (.12)(.12)(.12)=0.001728

You need 8 of the remaining 50 cards. Whichever one you get, you need 4 of the remaining 49 cards.

The Probability of flopping the 2-gut straight is (8/50)(4/49)=.013

The Probability of 3 people flopping sets & a 2-gut straight is (.12)(.12)(.12)(.013)=.00002351

If you didn't notice, I'm a math teacher who just happens to teach Probability & Statistics. I'm pretty sure those numbers are correct (even though they've been rounded slightly).

[/ QUOTE ]

I could be wrong, But I think your missing the probablity of the players all being dealt pairs in the first place. It looks like your calculations already assume the four players have the nessasary hands for this to work.

Dealing out 3 pair and a hand that fits a striaght between them is probably also an unlikely event.

That hand history is nuts. Who makes a hand history that doesnt show suits? How are supposed to know if someone was drawing a flush or was just insane?