[aquaman]

Sorry if this in the wrong place, but couldn't find a better one.

Let's say I have pocked deuces and I'm up against pocket aces. The flop is 57J rainbow. My only 2 outs are the 2 remaining deuces, but I have 2 chances at those 2 outs. What is the proper way to state what my chances are? Do I have 4 outs (2 chances at 2 outs)?<--------I don't think this is right, but what do I know?

[ACE'S FULL]

You would have 2 outs with 2 cards to come or 2 outs twice. If these are the only 2 players left 45 cards total left in the deck you have 2 of those to hit your hand so that leaves you with a 4% chance to hit your hand (2 outs / 45 remaining cards). That is if I did the math right.

[tomb1]

What I've heard most often is "xxx outs twice" meaning you have two more cards to come.

[jdunford]

Originally Posted by ACE'S FULL

You would have 2 outs with 2 cards to come or 2 outs twice. If these are the only 2 players left 45 cards total left in the deck you have 2 of those to hit your hand so that leaves you with a 4% chance to hit your hand (2 outs / 45 remaining cards) on the turn...

Argument is correct; math is not (FYP).

Wording: You have 2 draws at 2 outs (aka "2 outs twice"). You do not have 4 outs, as there are not 4 cards that can get you "out of this mess".

Math: The probability of hitting 1 of those outs on the turn is 2/45 (only if you know your opponent doesn't have a 2; if you don't know your opponents cards, then it's 2/47). If you do not catch an out on the turn (43/45), then the (dependent) probability of catching 1 of those outs on the river is (43/45)*(2/44) (or (45/47)*(2/46) if you don't know your opponent's cards). The probability that you catch (at least) 1 of your 2 outs by showdown is the sum of these, which is about 8.8% (or 8.4% if you don't know our opponent's cards).

If you want your true probability of winning the hand, you should also subtract the probability that you catch one of your 2 outs AND that your opponent also catches an A... but that will be a pretty small correction.

*EDIT* I keep a spreadsheet open where I've worked out all sorts of probabilities, including a table of "outs" (even includes some common pre-flop situations, like flopping a 4-flush with suited cards, or flopping a set from a PP, or making a flush or set by the river, etc.). It took me about an hour to make. There's a simple version of this available here:
http://www.homepokertourney.com/docs...let-charts.pdf
(see page 2)

Note: I don't like how it says that you have 1 out for 3-of-a-kind to improve to 4-of-a-kind (bottom of the 2nd page)... as most often you're up against a straight or flush and just need a boat to win. In that case, you have 7 outs on the turn and 10 outs on the river, or a total probability of 33.4% to make your hand by showdown (i.e. when considering an all-in bet on the flop against an unknown hand that may or may not contain some of your outs, but who you expect has a made straight or flush). You have a 1-outer only if you're up against a bigger set (set-over-set situation)... and I've lost a few of those in my days...

[jojobinks]

it's the same in the end, but i like 1-(chance of missing it twice).

so 1-(43/45*42/44) = X

[shanghai_sparky]

Originally Posted by jojobinks

it's the same in the end, but i like 1-(chance of missing it twice).

so 1-(43/45*42/44) = X

I like this formula better, thanks Jojo.

Side question to go with the original one:
What happens to the numbers if there are two others dealt into the hand with unknown cards?

[jdunford]

Originally Posted by shanghai_sparky

I like this formula better, thanks Jojo.

Side question to go with the original one:
What happens to the numbers if there are two others dealt into the hand with unknown cards?

Unknown cards are unknown cards: they don't change the probabilities of making your hand. Doesn't matter if those unknown cards are in someone else's hand or in the stub or in the muck. Doesn't matter if it's a full table, 6-max, heads-up... But if you have additional information (like, "my opponent would only bet that way if he had an open-ended straight draw or a 4-flush, and I'll put a 50% probability on him having one of those hands"), then you can include that in your calculation.

[tomb1]

Outs x 2% x number of cards to come is pretty close most of the time.

So:
Turn 2 outs x 2% each = about 4% chance of hitting (plus a tad)
River 2 outs x 2% each = about 4% chance of hitting

Total about 8%-9% to get one of your two outs on one of the next two cards, or about 10-11:1 against you.

[shanghai_sparky]

Originally Posted by jdunford

Unknown cards are unknown cards: they don't change the probabilities of making your hand. Doesn't matter if those unknown cards are in someone else's hand or in the stub or in the muck. Doesn't matter if it's a full table, 6-max, heads-up... But if you have additional information (like, "my opponent would only bet that way if he had an open-ended straight draw or a 4-flush, and I'll put a 50% probability on him having one of those hands"), then you can include that in your calculation.

Thanks. That's what I suspected, but needed verification.
I never claimed to be a math whiz.

[aquaman]

Thanks guys. That was quick!

So it looks like the math for 2 outs twice is ballparkishly close to having 4 outs with only the river to come. Sound about right?