Odds on this occurance?

Goethe

Playing cash, in the small blind with A,3o - limped in along with four out of the other five players.

Three 3s hit the flop - so big kiss for the poker fairy. But what are the odds of this happening (quading a random card on the flop)?

I made it 19,599-1;    (3/50) x (2/49) x (1/48) = 6/117,600 = 1/19,600

My boys book of knowledge says that the odds of flopping a quad with two unpaird cards is 9,802.92-1, which is just around half of that and allows for the fact that either of the two cards could quad on the flop.

I think this is my best ever outcome from the flop, and easily beats the three times I've hit quads on the flop from a pocket pair (at 407-1). But . . . I picked up just 63p (16 BBs) as a result.

sikas

i make it about 50/50, 

...either u flop'em or u dont

BLACK_MASS

doesn't it start 3/40 * 2/39 etc which makes it less but it's too early for me to be bothered to finish the sum,
nice attitude sorry.

Goethe

In Response to Re: Odds on this occurance?:

doesn't it start 3/40 * 2/39 etc which makes it less but it's too early for me to be bothered to finish the sum, nice attitude sorry.
Posted by BLACK_MASS

No, the calculation is done on the basis of unseen cards not just those that haven't been dealt. By omitting the cards dealt to other players you'd be assuming that the outs you're hoping for won't have been dealt out - which of course they could be. It's always amusing when you've folded a hand and you can see someone throwing money into the pot on the offchance they will hit one or both of the cards you've just mucked.

The small difference between my calc and the boys book of knowledge answer (it's actually a very good coaching book by Ken Warren) is purely down to rounding the decimals on the probability before it was converted to a fraction.

50/50 ? Hmmm . . . . will have to think about that.

BLACK_MASS

In Response to Re: Odds on this occurance?:

i make it about 50/50,  ...either u flop'em or u dont
Posted by sikas

LOL. Not 50/50 tho. It's either inevitable or impossible pre flop. So it can be 100/100 or 0/100 but nothing else

BLACK_MASS

In Response to Re: Odds on this occurance?:

In Response to Re: Odds on this occurance? : No, the calculation is done on the basis of unseen cards not just those that haven't been dealt. By omitting the cards dealt to other players you'd be assuming that the outs you're hoping for won't have been dealt out - which of course they could be. It's always amusing when you've folded a hand and you can see someone throwing money into the pot on the offchance they will hit one or both of the cards you've just mucked. The small difference between my calc and the boys book of knowledge answer (it's actually a very good coaching book by Ken Warren) is purely down to rounding the decimals on the probability before it was converted to a fraction. 50/50 ? Hmmm . . . . will have to think about that.
Posted by Goethe

Ok man but in this specific case the reality dictates it has to start 40 etc because it'd be impossible to hit that flop if any of the outs are unavailable or else the question simply wouldn't occur. I think we talk about same thing pre and post flop. Maybe lol.
I like it more when people are vigorously representing the cards you've just mucked. You can tell who is the sickest puppy and therefore a good online tell.

Goethe

In Response to Re: Odds on this occurance?:

In Response to Re: Odds on this occurance? : Ok man but in this specific case the reality dictates it has to start 40 etc because it'd be impossible to hit that flop if any of the outs are unavailable or else the question simply wouldn't occur. I think we talk about same thing pre and post flop. Maybe lol. I like it more when people are vigorously representing the cards you've just mucked. You can tell who is the sickest puppy and therefore a good online tell.
Posted by BLACK_MASS

Sorry . . but the correct basis is to treat all unseen cards as the same - there's an equal chance of any of the unseen cards being one of the outs you seek, not just the ones left in the deck.

Look at it another way . . . let's suppose you play a game with 24 players, where 48 cards are dealt out and where you need one of four outs.. If you only divide by the number of cards left to deal out, then the probability of you hitting an out based on the cards in the deck only would be 4 (outs) divided by 4 (cards left), or 100%, ie a dead certainty. There's a strong probability however that those four cards will have been dealt out to other players - in reality the odds of pulling one of the four outs will be 4/50, 1/12.5 or 11.5-1.