How many outs do we REALLY have?

[Panamajoe]

I'm going try to simplify this to it's most basic form to help me understand.

20 cards, 5 of each suit on the table face down, 4 people "playing".

I win if I grab the most diamonds.

Initial odds that I will grab a diamond are 5:20 or 1:4.

We each take 5 cards from the pile.

Odds are that each of us grabbed one of each suit EXCEPT for the extra one. My odds of having grabbed the "extra" diamond are.... 1:4.

OK, I am beginning to see.... I think. My chances for and against change equally... hmmm.

So back to outs, the remaining diamonds are equally likely to be gone as the remaining non-diamonds.

By George I think he's got it ... maybe

 

[NineLions]

So back to outs, the remaining diamonds are equally likely to be gone as the remaining non-diamonds.

By George I think he's got it ... maybe

Yup.

Exception is if you're broadcasting this on television, and the television audience actually sees the cards of the other players. Then the audience/announcers can eliminate those outs, but none of the players can do so.

A lot of those diamonds may be gone and the audience may know it, or maybe none are gone. So there is a difference between the "reality" if you are one of the players, or the "reality" if you are one of the announcers, or the "reality" if you're the dealer and you fixed the deck.

 

[SavagePenguin]

Ok I see the point now.

the 4 and 2 rule does not use max outs to figure pot odds but the average number of outs in that situation.

I thought that this calculated best case scenario but now I see that it doesnt.

Yes!

The x4 on flop, x2 on turn trick calculates your average chances of catching your outs, assuming that an average number of your outs has been given to other people or are used as burn cards.

The best case scenario would be that none of the 8 other players had any of your outs, and none of your outs would be used in the burn cards so you'd have your "whatever number of outs" out of 28 other cards in the deck (rather than out of the 47 that the trick calculates against).

But we can't see their cards and don't know what the burn cards are, so the rule assumes that an average number of your outs will be dished out to other players or used in the burn.

 

[Sean Pilgrim]

Alright lets go balls out with this one, literally.

In the Powerball you are supposed to pick 5 white balls and 1 red ball

White one's numbered: 1 - 59
Red one's numbered: 1 - 39

Here are your odds that we all can find online:

5 white balls + Powerball Jackpot 1 in 195,249,054

5 white balls $200,000 1 in 5,138,133

4 white balls + Powerball $10,000 1 in 723,145

4 white balls $100 1 in 19,030

3 white balls + Powerball $100 1 in 13,644

3 white balls $7 1 in 359

2 white balls + Powerball $7 1 in 787
1 white ball + Powerball $4 1 in 123
Powerball $3 1 in 62
K now... does anyone know of a company that does research and development as to how the weight of the ink printed on the ball will effect the possible outcome of the ball due to .001mg difference in weight with the double digit balls opposed to the single digit balls?

Just Kidding Just being a jerkface.

 

[sharkyo01]

Right going to pass this question on to the father hes very wizzy with mathematics and statistics... I'm pretty sure he come up something clever... I'll return with hes answer in a few days....

Also if we going to start talking about the lottery you have 14.5 million to 1 to hit the jackpot

 

[Sean Pilgrim]

Also if we going to start talking about the lottery you have 14.5 million to 1 to hit the jackpot

1 in 195,249,054

 

[Emperor IX]

I'd say you have about a 100% chance of your two books being horrible based on this thread, but who knows maybe you still have outs.

I think I folded them.

 

[Emperor IX]

The only possible way that any of this drivel makes sense is when you are considering calling with AK/AQ or something similar and think that you may not have all 6 outs vs. an underpair due to another players table talk/fold. Otherwise, you treat all unknown cards as unknown.

 

[PayMeh]

Wow.. so many opinions here.. lol.. Here's what I do. I just mentioned it in another post as a side fact. If I'm playing on a 9 person table I usually figure one of my outs is gone just to be realistic. If I'm on a flush draw I assume 3 of those cards are gone. That being said it doesn't really affect how I play the hand that much. It just makes me feel better when I draw dead. Outs are meant to be rough estimates you can figure out on the fly. If you want to get more accurate than that then you have to look at it based on the percentage of times you hit your out over many many hands which has already been done and well documented.. =/ Using this info on the fly is harder to do and requires a lot of memorization. That's why most people use outs instead. I'm afraid you're trying to take something that's meant as a roughly estimated value and trying to make it an exact value.

 

[Egon Towst]

Wow.. so many opinions here.. lol..

Actually, there are a lot of silly opinions here and one agreed fact, which is this:

you treat all unknown cards as unknown.

..and surely it is self-evident, when summed up as neatly as that.

 

[Roger Muller]

Hello,
Hmm it totally depends upon Probability and Odds equations. Tough to say beforhand.
All depends on the situation..

 

[sharkyo01]

1 in 195,249,054

Well depends on what lottery you talking about... The English lottery is 14.5 million to 1.

 

[Sean Pilgrim]

Well depends on what lottery you talking about... The English lottery is 14.5 million to 1.

Oh man I like those odds a lot better I'm moving to England.

 

[SavagePenguin]

Well depends on what lottery you talking about... The English lottery is 14.5 million to 1.

Oh man I like those odds a lot better I'm moving to England.

Yeah, but you probably don't win thirty million times your buy-in if you buy an English lottery ticket.

 

[Sean Pilgrim]

Yeah, but you probably don't win thirty million times your buy-in if you buy an English lottery ticket.

I loved MASKI III

 

[Egon Towst]

The English lottery is 14.5 million to 1.

True, except that is the UK lottery, my friend. There are many CC members who hail from Scotland, Wales etc. I feel sure you meant no disrespect to their homes.

For the benefit of foreign members who may think, "Eh ? What`s he on about ?": the United Kingdom is federation of member states in roughly the same style as the United States. England is where the capital is, and is the best known abroad. However, to say "England" when you meant the UK or Britain is like saying (for example) "California" when you meant the USA or America.

 

[katymaty]

only way you can really reduce your true outs is by good reading of your opponent, or if someoone accident revealed one of their cards

Example if you limped in a pot with AD 8D and the flop was 2D 7D JC then the turn was 4S and you bet the pot and the big blind who is as tight as a ducks fitting jacket (Quoted from BeardyIan ©) went all in:--

You can assume that they either flopped a set or hit the set on the turn or they just hit 2 pairs.

Normally you would expect to have 9 outs possibly 12 against loose opponent with a pair if you count the aces, but if they have a set or 2 pairs you will have either 7 or 8 possibly 9 only if they have 2 7:icon_bigs

 

[RoyalFish]

Well said. Here's what I'm doing. I'm running my own test. I'm dealing out 9 hands open, burning a card open and dealing a flop. I'm on hand #140 and the results are staggering. Small sample size so far but well worth the time.

This really isn't anywhere near enough of a sample size to be meaningful, nor are you really going to get there dealing cards by hand. I did some tests to determine preflop equity empirically for all hands, and to get numbers I felt like were solid took billions of hands.

You're, uh, gonna be at this a while...

RF

 

[RoyalFish]

Ok, putting math geekness in service of CardsChat...

You have AXs heads up. You flop a flush draw. 9 cards remain, so P(flush on the turn) = 9/47, or 19.14894%.

Let's do it your way. There are 3 possibilities. Villain has 0, 1, or 2 of your outs. P(0) = 38/47 * 37/46 = 65.0324%. P(2) = 3.33025%. P(1) = 1 - P(0) - P(2) = 1 - 0.650324 - 0.0333025 = 0.316374 (31.6374%)

Case 1: Villain has 0 of your outs

P(flush on the turn) = 9/45 (two fewer because you know he holds two cards which are not your outs, which ones don't matter) = 20%.

Case 2: Villain has 1 of your outs

P(flush on the turn) = 8/45 (two fewer because you know he holds two cards which are not your outs, which ones don't matter) = 17.7778%.

Case 3: Villain has 2 of your outs

P(flush on the turn) = 7/45 (two fewer because you know he holds two cards which are not your outs, which ones don't matter) = 15.5556%.

The overall probability is the weighted average of those outcomes, weighted by the probability of them happening.

P(Case 1)P(flush given Case1) + P(Case 2)P(flush given Case2) + P(Case3)P(flush given Case 3)

(0.650324 * 0.2) + (0.316374 * 0.316374) + (0.0333025 * 0.155556)

==

(wait for it)

0.1914894. It's exactly the same result whether you ignore the fact that villain may have some of your outs or you consider the cases where he does.

If you need a statistics consultant for your book, you know where to find me.

RF

 

[dd_decker]

Re: the original post: Your premise is wrong. You have as many outs as there are. An out may be in the muck or the in the deck or in someone's hand, it's all the same, dude! The thing is, sometimes we think we have an out but it is not really an out because it might give your opponent a hand better than yours, even if you make your draw. Your "out" might give your opponent a full house or a flush. Also, it's hard to calculate "1/2 outs" like runner runner in your head, but it can be done matimatically. :joyman: You gotta learn the ins and outs... (sry)

 

[NBA2K10ROCKETS]

man this is a tough question this might not be able to be answered. This is not like calculating odds of winning you'll have to find a mathematician for this. I do have one questions relating to this thread do they do burn cards for online poker like you know in live poker before you deal the flop you take one card face down and put it a side and then put down the three cards to the flop? Or do they just deal it? This relates to the thread because it will be a factor in calculating how many of you out can possibly be in someone elses hand.

 

[RoyalFish]

man this is a tough question this might not be able to be answered. This is not like calculating odds of winning you'll have to find a mathematician for this.

Done.

I do have one questions relating to this thread do they do burn cards for online poker like you know in live poker before you deal the flop you take one card face down and put it a side and then put down the three cards to the flop? Or do they just deal it? This relates to the thread because it will be a factor in calculating how many of you out can possibly be in someone elses hand.

See my prior post. It really doesn't matter. You can deal the hole cards then throw away all but 5. No effect on the odds or outs at all. The only purpose of a burn card is to defeat card marking. No burn card means you can mark cards and see that a marked card is next to be dealt because it's on the top of the deck. The burn card covers the next real card to be dealt, preventing that. For the same reason, dealing out the rest of the hand face down is bad (or at least renders a burn card useless).

RF

 

[PayMeh]

FWIW just to answer your question about burn cards online.. no they don't.. basically the way it works is like this.. Picture a huge rotating ball of cards, 10000 decks or so. The software deals from that big mess and checks itself to make sure that card isn't already in play. If it is then it redraws until it gets one. Interestingly enough I believe this is part of the reason why you get such odd hands at FT. At some point if it's truly random you're going to have like cards group in the mega deck. If you find a way to add that to any equation you'll be golden.. =D

 

[SavagePenguin]

I loved MASKI III

I haven't messed with that stuff in years. I found another one in September. (I seem to find one every Sept.) but just flushed it instead of tormenting it.

 

[RoyalFish]

FWIW just to answer your question about burn cards online.. no they don't.. basically the way it works is like this.. Picture a huge rotating ball of cards, 10000 decks or so. The software deals from that big mess and checks itself to make sure that card isn't already in play. If it is then it redraws until it gets one. Interestingly enough I believe this is part of the reason why you get such odd hands at FT. At some point if it's truly random you're going to have like cards group in the mega deck. If you find a way to add that to any equation you'll be golden.. =D

You don't have a reference for that, by chance, do you?

I don't know how they do it, but having written poker simulation software, I'd never do anything like this. There's really no reason to have a deck at all, let alone 10,000 of them. Just a random number generator and a 52-bit bit vector. When you draw the card, see if you've dealt it yet. If not, flip the bit. If so, draw again.

RF