# We've been calculating our outs wrong!

• Bronze
Joined: 08.06.2007
Ok, this one is going to be huge if it’s correct. If it’s not, I’m just making an ass of myself, but I have yet to find the flaw in this. Please, feel free to search for it. In fact, I WANT you to find a flaw in this reasoning. I want to be able to play normal poker again as I was taught, because if this is true, it’s going to change the way you and I play.
Yesterday night, I was lying in bed and I was thinking about some hand where I got it all in with a straight draw, a flush draw and 2 overcards on the flop. So, I assumed I had 21 outs, because I put my opponent on top pair. He indeed had, but he had a also had a better (nut) flush draw, together with one of my straight outs. So my REAL number of outs was actually 12. This got me thinking: we always assume that, if we count our outs, they’re still in the deck, so they’re live. Maybe here and there, we discount one given jow the hand was played. But that’s pretty rare. If we have a flush draw, we always assume to have 9 outs. For this to be true, all 9 outs have to be still in the deck. Is this the case? Let’s find out.
When you’re playing full ring, the number of opponents at the table is 9. That means, if you’re playing a hand that saw the flop, there are 5 known cards: 2 hole cards + 3 on the flop, and there are 47 unknown cards. If you’re drawing to a flush, the previous way to calculate your outs was 9/47=19.1% to complete on the turn, and 9/46=19.6% on the river. But this is assuming that all 9 flushcards are still in the deck. If we have 9 opponents, that means that 18 hole cards are out of the deck. What’s the chance that none of those 18 hole cards was a flush card? It’s pretty small, that’s for sure. The way to calculate it is: 38 of the 47 unknown cards are non-outs. So the chance of 1 card being a non-out is 38/47. If one non-out is being dealt, the chance of the next one being a non-out is 37/46. The chance of the next one 36/45. And so forth. The chance of none of the 18 unknown cards being an out, is therefore the result if you repeat this process 18 times and multiply the chances of each possibility with each other. Calculations are found below.
# non-outs # unknown cards p
38 47 0.808511
37 46 0.804348
36 45 0.8
35 44 0.795455
34 43 0.790698
33 42 0.785714
32 41 0.780488
31 40 0.775
30 39 0.769231
29 38 0.763158
28 37 0.756757
27 36 0.75
26 35 0.742857
25 34 0.735294
24 33 0.727273
23 32 0.71875
22 31 0.709677
21 30 0.7
If you multiply these chances with each other, you’ll find that there’s only a 0.736% chance of all your outs being live, being that all of your 9 flushdraw cards are still in the deck. So next time you’re assuming that you have 9 outs full ring, better think about this. Shocking? It gets worse.
Now, you might ask, “if my number of outs is not 9, then what is it?” The answer is: we don’t really know. It’s still 9 a good 0.7% of the time, but that’s just the probability for that specific case. If we can calculate this probability though, then we’re also able to calculate the expected value of your number of outs as well, based on the number of opponents (and therefore, dead cards) at your table. Let’s do that for 9 opponents.
The probability of a card being dead is: #dead cards / #unknown cards = 18/47 = 38.298%
So, if you have 9 presumable outs, there are probably 9*(18/47)=3.45 outs dead. That means, on average there will be 5.55 outs still left in the deck. That’s a lot less than 9! This is getting worrisome. We can extrapolate this to any number of outs for any number of players if we want to. Basically, the higher number of opponents, the more dead cards and the less live outs we have. I can’t believe no one ever noticed this before!
All this time PokerStrategy, TV-shows, Twoplustwo, etcetera, have been spewing fog in terms of how many outs we have with a flush draw. Why? Do they want to us to keep drawing with worse odds to make more money? Are they conspiring with the online pokersites to produce more rake?

Luckily, no. The answer is simple. If you have 9 outs out of 47 unknown cards, that’s’ basically the same as 5.55 outs out of 29 live cards. So our equity doesn't change one tiny bit. Phew, scared you for a minute now, didn’t I

But next time you're saying: “I have 9 outs” when you're drawing to that flush, realize that you’re actually making an assumption that most likely isn’t true. A better way to say it is. “I can expect to have 5.55 outs”.

Yes, I was bored last night and didn’t feel like sleeping
• 24 replies
• Bronze
Joined: 23.02.2007
damn you must've been _very_ bored
Actually we consider that the mucked cards are still in the deck, since they're unknown. It makes sense that as long as they're unknown, there'll be the same proportion of outs:other cards (both in the deck and the mucked cards). but thanks for the big theory
• Black
Joined: 10.09.2007
tl;dr

HAHAHA actually while typing this I read it and I just wanted to say the exact same thing as you did in your "conclusion", so much for me being a genius
• Bronze
Joined: 14.04.2006
WTF???

Good luck at the tables!
Puschkin81
• Bronze
Joined: 25.11.2006
You could have spent that time actually playing poker! Thats my shocking revelation.
• Bronze
Joined: 08.03.2007
As long as you don't know that a card is dead, it makes no difference. If there were 21 opponents, say, then yes its true that most of your outs are dead, but its also true that the deck only has 5 cards remaining. If even one of your outs is in the deck, you have a huge chance (64%) of making your hand, so the probability balances on average.

To illustrate even more clearly : If only 2 cards remain in the deck (impossible irregardless of opponents), then those two cards will definitely get dealt. Say you're on a flush draw with 9 outs (and you have no information about where they are). It's clear that if one of those two cards is an out, you win. You've seen 5 cards, so the chance that neither of the two cards remaining is an out is 38/47 * 37/47 (first card not being an out times second card not being an out) =
0.636

i.e. you have a 0.363 chance of making a flush and the probability hasn't changed at all.

• Bronze
Joined: 08.06.2007
Originally posted by whipflip15
You could have spent that time actually playing poker! Thats my shocking revelation.
Or I could have been studying for my Data analysis exam tomorrow (notice the link? ). I wasn't trying to make a point, just saying that the expression "we have 9 outs" is inherently wrong if you think about it. I'm not propagating to use the real expectation of outs obviously, because calculations at the table would be a mess but it's just a thought that crossed my mind, and I tried to give it a funny twist.

Back to my books now...
• Bronze
Joined: 21.09.2007
lol Timor you're funny, made me laugh. Don't think to much about this plz. you'd get a stroke or sumtin . I just assume the deck to be all cards that aren't visible, since which cards are out is often a random number anyway. However, lots of flatcalls on a 2flushboard make the flushdraw worse. However (again), the pot odds get a lot better, and the implied odds A LOT better if someone else has a 4-flush too, so that compensates it a bit.
• Bronze
Joined: 15.06.2007
Originally posted by Timor83

Or I could have been studying for my Data analysis exam tomorrow (notice the link? ).
Hey, at least we got a head start for chapter 7 : "Decision Making under Uncertainty"
Lucky for me I ain't going all cookoo over it as you apparently are.
• Bronze
Joined: 08.06.2007
You're right, you have way more important reasons to go cookoo. Like the fact that your favourite soccer team is yet again losing at this very moment But hey, blame it on variance if you like.
• Bronze
Joined: 03.10.2007
yeah i read the start, realised your flaw and was just thinking... OMG fish timor .

but you realised it as well, good!
• Bronze
Joined: 14.09.2007
Don't put poker before your real life. It's just a game o.O
• Bronze
Joined: 15.06.2007
At least he'll be able to explain why flunking his exam was just a statistical anomaly...
• Bronze
Joined: 08.06.2007
You really don't want me to comment on that one given our past grades.
• Bronze
Joined: 15.06.2007
I never used statistical anomalies as an excuse however. Despite having lower grades, I must have gotten one of the best points/study hours ratios out there

So you getting better grades is kinda like you having a bigger BR. I just spend less time on it, getting lower results despite higher efficiencies.
• Bronze
Joined: 04.12.2007
• Bronze
Joined: 08.06.2007
Person A studies 1 hour and still gets a 1 out of 20 grade. Person B studies 48 hours and gets a 15. The ratio is clearly higher for A, but which one is more retarded?...Exactly.
• Bronze
Joined: 15.06.2007
I still had 64 % last year, don't make it look like I'm some gigantic failure ffs

Just for your information, you must use a logarithmic regression, no linear comparison, to measure effiency. Should be more realistic. I'm an extreme outlier at the left hand side of the graph btw, go figure that one out eh.
• Bronze
Joined: 18.12.2007
I personally will never go all-in on a draw. If I want to pull off a semi-bluff I'd rather work on the ammount in the pot and try to screw up the odds to something like 2:1. Where I come from, there's no way in hell your opponent will fold if he has 3 of a kind or top pair, even if you put him all-in.

In the end you have to play the outs against the pot odds. Donkeys go all-in on a semi-bluff. Donkeys also ignore pot odds. Then again, donkeys will sometimes call your all-in on something like 7-9o and get a straight purely on luck. The idea is to take luck out of the equation, isn't it?

The only way I'll go all in with my money is if I have the nuts or a very high top pair on the flop. Most of the time I would beat the player going for the draw. Also, I will only go all in with a top pair if I put my opponent on a draw or a top pair with a lower kicker than mine. Our stack sizes should be more or less the same.

There's no use in bluffing with an all-in against a player who can call you with 1/5 of his stack. If you lose, you lose everything. If he loses, he can always come back.

The whole reason why we have something like outs is because we use it with the pot odds to make decisions. Thus, use the pot rather than your stack to influence decisions around the table. Trust me, it will be much less painful, and if the pot is big enough your opponents will get the same message if you had gone all-in.

I've made a lot of money off people who go all-in on draw. Trust me. Not a good idea.
• Bronze
Joined: 08.06.2007
Originally posted by SonicXT
I still had 64 % last year, don't make it look like I'm some gigantic failure ffs