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Wasn't sure where to post this one, but here goes.
You hold a pocket pair, say KK. The turn is an innocuous 2 7 T.
What are the odds of hitting a full house or better (i.e. quads) ?
It sounds simple at first but I don't think it is, as when the turn card gives you a set instead of 2 pair, the numbers change.
Naturally, an explanation of the math would be nice, as opposed to a computer generated figure.
Bench.
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it is not so complex bench
the board just needs to pair any of the cards, on each of the two remaining cards to be dealt.
so a K, 2, 7 or J.
there are 11 remaining on the turn, from 40 cards ..let say a 25% chance. similar numbers for the river.
25% x 25% = 0.16
say a 1 in 6 chance
5 to 1 odds
Turn odds are 11/47 (unless you know whether or not your out cards have already been dealt)
Then
If a K hit (2/11) you have 10 outs to a FH or quads - 10/46
If a 2,7,T or hit (9/11) you have fewer outs.
Only 2Ks and 2 of the card you just picked up will improve to FH - 4/46
So overall the odds are (11/47)*(10/46)*2/11 + (11/47)*(4/46)*(9/11)
=2.6% or 1/50 roughly (edit 1/40)
But my stats are a bit rusty so could be wrong
when you say you could be wrong, you are correct. you are wrong.
Yes Phantom, I like the look of that. Was wondering how the odds would be proportioned for the river, but your factors of 2/11 and 9/11 share them out. Nice work.
Aussie I may well be wrong and at least I have the humility to recognise it.
Odds of improving a pair to a full house on the turn and river = 2.5% from a poker odds website so I think I am close.
If your answer is correct then I am running worse than I thought not to get so many full houses
2. You have used one of your outs when you hit it on the turn yet you haven't considered that
3. If you hit say a 2 then the 7 and 10 are no longer outs on the river
0.25 x 0.25 = 0.0625
which is 1 in 16
or 15 to one odds.
accurately, without trying to keep it simple, the chances are
11 cards from 47 multiplied by 10 cards from 46
which is,
0.243 x 0.217 = 0.051
which is slightly better than 1 in 20
odds are 19 to 1
You cant have 10 outs still on the river otherwise you would be counting 27, 2T, 7T, as acceptable run outs which they are not
My head hurts
it is just a question of having pocket pairs and the odds of getting to quads or house, either will do. for that, the turn must pair the visibile cards, and so too the river.
the workings for the turn we agree on 11 from 47. however, the river must be any one of ten cards. that is the 11 required less 1 (whatever came on the turn, it doesn't matter what occurred). that 10 from 46.
your workings were fine for K on the turn scenraios but didnt account for no King. (imo)
you'll be right stewarty. i'm on a defo wrong streak.
Pocket Cards: KK Flop: 2 7 T
To hit our full house or quads we need to hit both the turn and the river, as we need 2 more cards.
For the turn, we need any of those cards (2 kings, three 2's, three 7's or three T's) = 11 cards
That's 11/47. Okay? All agreed so far ?
If the turn card is a K, then we need either K (for quads) , a 2, a 7 or a T = 10 Cards (10/46)
If the turn is a 2, 7 or T, we need either a K or a card to match the turn card = 4 Cards (4/46)
This is where it gets tricky...
both 10/46 and 4/46 can't be correct, so we need to allocate a factor based on how likely each is to occur.
This comes from the turn.
At the turn we have 11 outs. Two are Kings and Nine are the 2, 7 and T's.
So, the likelihood of us hitting the K is 2/11 and the likelihood of hitting a 2,7, or T is 9/11.
These ratios are then applied to the river odds
For the Full house with Kings or Quads it's (2/11)*(10/46)
For the Full house with either 2's, 7's or T's it's (9/11)*(4/46)
Both these calculations need the turn odds adding to them, making Phantom's
(11/47)*(10/46)*(2/11) + (11/47)*(4/46)*(9/11)
Bench.
Do the math?
Did you go to school in America Bench?
Math or maths is the real question here. I'm going with maths 100% of the time
phanto, you know when i said youwere wrong? sorry
it was meant to be funny/witty
my figs were meant to be rough
soz