Note: Not at the old Poker1 site. A version of this entry was originally published (2008) in Casino Player.
Knowing all the esoteric odds and being a wizard at poker math won’t provide you with a big advantage over your opponents. Sometimes you’ll hear players say that you need a solid understanding of poker mathematics to win. That’s a big lie.
Yes, I calculate poker odds and publish them. It’s something I like to do.
These odds are important, because they give you an understanding of how often you can expect things to happen. Without that understanding, you’re likely to be frustrated when cards don’t fall the way you’re hoping. Also, you’re apt to be amazed by outcomes that aren’t really that extraordinary.
Mental energy
When you don’t know basic hold ’em odds, you might use up a lot of the mental energy you need to make quality decisions, while pondering over what you perceive to be weird runs of cards.
I recently heard a player complain, “I haven’t started with a pair of aces or kings in three hours! The best I had was one pair of queens.” That’s a ludicrous statement. You’re not supposed to have a pair of aces or kings in three hours — and you’re slightly lucky when it happens. The math isn’t too complicated.
If you’re curious about how it’s done, follow along.
Some math
There are 52 cards in the deck. Eight of these are aces or kings, four each. So, you have eight chances of looking at your first card and seeing one. When that happens, there are only three matching aces or kings left among the remaining 51 unknown cards to give you that commanding pair.
When you put this into a formula, it looks like this: (8 qualifying aces or kings / 52 cards in the deck) x (3 remaining cards that pair the first one / 51 total remaining cards). That’s (8/52) x (3/51) = 0.00905. That figure is less than 1 percent, and the exact odds against being dealt a pair of aces or kings are 109.5-to-1.
Fine. Now let’s assume that your game averages 35 hands an hour. On average, you’ll be dealt a starting pair of kings or aces only once in 3 hours, 9 minutes. And you might easily go twice that long, and sometimes a lot longer, without being dealt one of those premium pairs.
Not a huge advantage
We’re dealing with the true value of knowing poker odds. Except for counting the money in pots and estimating pot odds, knowing poker math won’t give you a huge advantage over opponents who possess a pretty good feel for the game. But these odds will tell you what to expect and keep you from being either frustrated or amazed by the cards.
Let me do the calculations. Just familiarize yourself with a few basic odds. (I’ve made all my calculations available here at Poker1.com, by the way.)
Knowing the most important ones gives you a good understanding of what will happen. But, despite what you may have heard, you don’t need to do elaborate math at the poker table in order to win.— MC
Thanks for the info and breakdown!
What I would like someone to explain to me is why the outs to a draw that are in the dead cards, are counted as AVAILABLE odds.
10 handed game means that there are 18 cards in other hands, 2 in your hand, 1 burn card, and 3 on the flop, after the flop, accounting for 24 cards, leaving 28 available. So if you have a flush draw, there are 9 cards in your suit out there.
However, you most likely do not have 9 cards available to catch. Considering that 24 cards have been dealt, how can you possibly expect that there are still 9 left in your suit available?
No one has ever addressed this and I don’t know why.
Of course there are 9 outs to make your hand, but if you use 9 to make your decision, you are using what is most likely the wrong number.
I would think it is closer to 5, on average.
The operative word here, when counting outs, is AVAILABLE OUTS.
Of course the true number is not known, but it’s highly unlikely to be 9.
Hi, Bill —
It doesn’t matter whether a card is actually available to be dealt or simply unknown. We can only subtract from possible remaining cards those that are known (those on the board and those in your hand, as examples). The rest might be still available or they might not, but it doesn’t matter, because the likelihood that they are still available remains constant no matter how many cards have been folded already. (Of course, you can make rational assumptions or “guesses” about what cards might have been more likely to have been folded, and doing so will change the probabilities.)
Unless you make assumptions, you should think of both folded cards and remaining cards as being unidentified. As such, the folded and available cards are simply a random sample from the group of unknown cards. The likelihood that you will see them in play doesn’t change. That eight of clubs you need to make a straight flush is either available or it isn’t, but the chance that it is available is part of the calculation. That calculation already INCLUDES the possibility that it might have been previously folded by an opponent.
To simplify: Suppose you are interested in knowing the likelihood of the ace of spades being on top of a freshly shuffled deck. You could say, hey, everyone says it’s 51-to-1 against (one in 52), but what if it’s in the bottom 51 cards? Then it can never be dealt first. That reasoning fails, because the fact there is one card at the top and 51 cards below doesn’t change the probability of the ace of spades being dealt first.
Straight Flushes,
Mike Caro
Straight Flushes,
Mike Caro
Mike, thank you for replying, but I still can’t get comfortable with that, because of the dead cards.
Let me give an extreme example. Let’s add 10 more players to the game. That would account for 20 more dead cards, making a total of 44,.
Then, even though there would still be 9 outs, you couldn’t possibly have 9 AVAILABLE outs, since there would only be 8 cards remaining.
I just don’t understand why no one ever accounts for the unknown number of outs in the dead cards. It’s never considered, and that makes no sense to me.
Bill, using your example of four to a flush, common wisdom says you have 9 outs. 9 of the suit remain in the unseen cards, the unseen cards being 52 – 2hole cards – 3 flop cards = 47 unseen cards.
9 out of 47 is 19%. 19% of the remaining unseen deck is made up of the cards you need and that is also the probability of hitting an out on the next deal, I.e the turn. The odds of hitting on the turn are 9/47 = 0.191 or 19%.
Ok so now let’s deal a 10 handed game. That means 18 more cards out on the table. Some of those cards will include the outs you need. You cannot ever know how many of those outs are in those dealt cards unless everyone turns them face up. So you have to estimate. How do you do that?? Well all you can reliably do is work on averages.
We already worked out that 19% the remaining deck was made up of the cards you need. So on average 19% of those 18 cards dealt to the other players will include the cards you need. 18 x 0.19 = 3.42. So on average the most likely number of needed flush cards dealt to other players is 3 cards.
So on average you now only have 6 available flush cards left in the remaining undealt deck and the remaining deck is 52 – 2 hole cards to you – 18 hole cards to opponents – 3 flop = 29 cards.
There are 6 outs on average are left in those 29 undealt cards. The odds of the next card dealt being the one you need is therefore ON AVERAGE simply 6/29 = 0.207 or 20.7%.
For all intents and purposes it’s pretty close to the original 19%. It’s pretty close to the standard outs calculation of 2x total outs for the next street (2×9=18%). And that demonstrates that it is futile to adjust the outs because it makes no practical difference ON AVERAGE.
Over the long run sometimes fewer flush cards will be dealt into opponents cards, and sometimes it will be more, but it all averages out in the long run and the probability of hitting remains the same ON AVERAGE.
Bottom line, there is no point adjusting your outs assessment for discarded cards. Hope this helps.
(Note for simplicity burn cards have not been taken into account).
This is why I used to love 7 card stud. I could calculate those odds. Three of your suit out on third? Don’t go for it. Count as you go. Know the odds to each street. Know if the your opponent can likely hit his flush draw. Know if your full house is bigger than his full house. I raised a player to his last $5 chip because I had that information. Then he still got mad and rude. I thought I was being nice letting him keep that chip.
Hi Mike,
Curious, Is there a place on this site where I can find a list of statistics similar to ones stated above?
Hi, Mark —
A complete set of poker odds will be included soon. They already exist at the previous Poker1 site and will be transferred here to the new Poker1 prior to our official opening.
Thanks for making your first comment and joining our Poker1 family.
Straight Flushes,
Mike Caro
Please tweet when they are posted. thanks!
Andy Bloch is about as “wizardly” as one can get, mathematically speaking and yet I’ve played with him and watched him make some moves that made perfect mathematical sense, but were just wrong for the situation and timing. That said, never doubt that a math “wizard” knows exactly how many outs he has, what the odds are of hitting his hand and exactly what your bet sizing means. My experience is that the only safe place at a table with Andy Bloch is immediately to his left. That tends to negate his mathematical superiority and sort of level the playing field. I've also learned not to listen to Mike or Andy at the table. They will dazzle you with their mathematical acumem and talk you right into doing something that you later recognize was stupid. Just saying.
Thats one thing I seem to get in trouble over and destroys my chip stack. Following math in poker just doesnt work for me at all I love it but it dont work for me. Which there is something else that makes no sense to me what so ever. A straight is less value then a flush. But flushs seem so much easier to hit then a straight. Maybe its the software they use ( I dont play much live). I tryed figuring out this problem I had one day and from what I could tell mathematically it seems right. But heres the problem playing suited connecters more often so results may vary. Then there are different types of games but from what I have seen that doesnt make a difference in hand rankings. So im assuming the hand range I play is giving me more flushs I guess I could be getting the cards in your words clumpy, but it has to be my hand range doing this, but this is just something im working on right now and it absolutely makes no sense to me right now.
Hi, David —
You're right. When using a standard 52-card deck, it's easier to connect drawing one card to a flush than one card to an open-end straight. Nine cards can complete the flush; only eight cards can complete the straight.
However, four-card flush draws are less common than four-card open-end straight draws. And getting a flush on the first five cards is rarer than getting a straight on the first five cards. That latter rarity — defining a pat hand in draw poker, without requiring a draw — is the basis for the rankings.
Straight Flushes,
Mike Caro
Thanks, Mike
So poker math can be learned, but is there a way to learn what you name as "feeling for the game"? Without having to play all my life, of course.
I think math wizard would succeed at the table, where the other players were making decisions based primarily on mathematics behind poker. However, I’ve never seen such a nerdy table…