2 Questions on Hand Probabilities

[J_Lebowski]

Hi, I had a few hands the other day that got me to thinking about certain probabilities and was hoping someone could help me out.

First Hand. Playing in a relatively loose NL .5-1 game. Get 6-6 in the BB. One off the Dealer raises to $3, Dealer calls, SB calls, I decide to call the extra $2. Flop comes 3-6-Q with two hearts. Small blind checks, I raise the size of the pot. Dealer comes over the top. Feeling pot committed and with a set, I call. Instead of turning over a flush draw or top pair, which I had been hoping for, Dealer turns over Q-Q and I am done for.

What I have not been able to figure out is what are the probabilities of two people flopping sets?

Second hand:
Big Blind Special when I check J-6o to two limpers. Flop is 2-6-J with 2 spades. SB bets small, I reraise and Dealer comes over the top. SB folds, I call. Showdown is my J-6 against his 6-6. I was thinking about it in the way of the probability of him having either 6-6 or J-J was miniscule, and was more concerned about 2-2. Nonetheless, I think I made the right call from a numbers perspective and am hoping that someone can help me by confirming or poking holes in my logic.

Thank you in advance for any assistance.

[Louie Landale]

[1] Hard to calculate set-over-set; but its easier of you presume that both players start with a pair.

[2] Nit-pick: there is 1 way he can have a set of Js and one way he can have a set of 6s and 3 ways he can have a set of 2s; so if you know he has a set its 3:2 its 2s. But that doesn't help you at all. The real issue here is what's the chances he raises with a hand WORSE than yours? Might he play J2 and 62 on the button AND put himself all in? Well there are 6 ways he can have each of those hands and if you are sure that he WOULD play them AND raise all in with them, then you are a 6+6:5 or 12:5 or 2.4:1 favorite to have the best hand. If he MAY put himself all in with a lessor hand (like the A-flush draw or AJ) then you have an easy call. Gotta know your player.

Noting that the chances he has you beat is "miniscule" isn't enough; you'll need to compare that to the chances he does NOT have you beat; for some folks that's zero so you should fold.

"miniscule-to-infintesimal" are pretty bad odds.

- Louie

[1 bag]

The odds of flopping trips is %11.76 or 7.51:1 against when holding a pair.

[flopmeister_1]

pre flop you are both looking at 25:1 to flop a set. 50 unseen cards divided by 2 cards that will make your trips.

I am not so sure this will be different for two sets flopped w/ hole pairs by both players.

Eric [img]/images/graemlins/confused.gif[/img]

[Louie Landale]

25:1 is completely wrong.

[1] If you are drawing 1 card it would be 48 bad cards to your 2 good cards or 24:1 against. Odds are "BadOutcomes-to-GoodOutcomes" not "TotalOutcomes-to-GoodOutcomes".

[2] You get to a flop of 3 cards, not one. Odds against you flopping a set if you hold a pair is 7.5:1 against.

Calculating 2 players holding different pairs, both flopping a set would be pretty tough. If its AA vrs KK then the flop must contain [a] One A, one K, and some non-A/K card. That's 2*2*44 [b] Two As and one K. That's 1*2 [c] One A and two Ks. That's 2*1. There are 48 cards in the deck and the total number of such possible flops are 48*47*46/3*2 = 17296 flops of which 2*2*44+2+2=180 feature an A and a K. 17296-180=17116 bad flops and 180 good ones.

Odds are 17116:180=95:1 against. If I've got Ks I hate it when that happens.

- Louie

[J_Lebowski]

Louie,

Thanks for your replies. I have one further question: why in calculating in the possible flops would it be 48*47*46/3*2? I am confused in particular by the /3*2, understanding that the 48-46 are just the remaining cards.

Thank you again for your assistance.

[SumZero]

It is 48 choose 3. Where x choose y is x!/(y!*(x-y)!).

And n! = n * ((n-1)!) where 0! = 1.

You use combinations like this (as opposed to permutations) where order doesn't matter. So if you don't care if the flop comes A[img]/images/graemlins/spade.gif[/img], 9[img]/images/graemlins/heart.gif[/img], 8[img]/images/graemlins/diamond.gif[/img] as being different than 9[img]/images/graemlins/heart.gif[/img], A[img]/images/graemlins/spade.gif[/img], 8[img]/images/graemlins/diamond.gif[/img] then you want combinations. If the order didn't matter it would be 48*47*46. But since there are 6 different ways of arranging the flop (3*2) then you divide by that.

BTW do you play as Jeff_Lebowski on UB? The Dude?