2 divided by 0 equals pi

On September 28 2008 16:30 Loco wrote: you don't need maths in poker to be successful just like you don't need position, leave it to the lesser players

is that a giant level or am I missing something

On September 28 2008 16:14 Shakedown88 wrote: Question: How do you take that range that you assigned to your opponent's hand, and very roughly, convert it in your head to some kind of usable odds/percentage that you can apply to make a more profitable decision?

how does the brain do any of the other amazing crap it does?

On September 28 2008 16:30 Loco wrote: you don't need maths in poker to be successful just like you don't need position, leave it to the lesser players

you're obviously taking for granted all of the math your brain subconsciously does while playing

ok lets do a rough combinatorial approach right now

the basic ideas is that you want to give him a range, and then group parts in the range together as "basically the same thing" and then do the rough math on that. If you give his range "AT-AK, KQ, 22+" then i would divide it into "big cards, big pairs, small pairs" and you can do some math to figure stuff out specifically. for example:

PREFLOP STUFFS:
each pocket pair is 6 combinations. Each unpaired hand (AK, KQ etc) is 16 combinations. So, each unpaired hand is a little over twice as likely as a pocket pair.

So, suppose you assigned his range to be AK, AQ, 66,55,44,33,22 and you are holding 77. You can group his hands into "big cards" and "underpairs." In this case, about half his range is big cards and the other half is pocket pairs (16*2 = 32 big card combos and 6*5 = 30 pocket pair combos). Against AK/AQ you are 50%. Against the other pocket pairs you are 80%.

Now we just do a weighted average (50% * 50% = 25% and 50% * 80% = 40%) so your total chance of winning vs that range is 25+40=65%.

POSTFLOP STUFFS:
You can do the same thing, but numbers change slightly, if you think he has a set, there are only 3 combinations possible. if you think he has 1 pair, then there are 12 combinations. So, on a board of K23, if an opponent is a nit and starts raising you alot, he probably has 22, 33, KQ, or AK. There are 6 combinations of PP and, and 24 combinations of hands that have a K in it. proceed accordingly.

---

obviously stuff changes if you have some part of the hand. in the K23 example, suppose you have AA. Now, there are only 6 combinations of AK (2*3), but still 24 combinations of KQ and 6 combinations of sets. So, you have 30 combinations of hands that you beat, and 6 that beat you. AKA you have 5x more combinations of hands you beat. I'd go broke yo.

hope that was helpful!

to continue what travis is saying:

i think its useful to do the math a handful of times for some basic situations because oftentimes the result is really striking. once you've seen some basic stuff, you'll have a good feel and be able to play w/o doing any math directly

what day said. and even once you have a good feel you can always improve your feel by augumenting it with doing ever more involved ev calcs from your 'tough' hands and noting the results. the main reason i almost never do is laziness. dont think for a second you need to be able to deal with actual numbers and big calculations in your head in the heat of the moment when you're actually playing and making the decision right there though, at that time you rely on your 'feel' - ie what travis said, your brain's natural ability to qualitatively approximate weighted estimates based on multiple variables. math is for off the table to improve your feel.

edit: OP if you havent already, google and download a program called "PokerStove" and learn how to use it. Then play and experiment with it A LOT putting hands against ranges, ranges against ranges, etc. just do that whenever youre bored or looking over previous hands you played. it takes into account everything day said for you with combinations, so if you wanna see a hand against a range you just put in that range and your hand and it'll do it all and give you the % equities.

On September 28 2008 16:48 roflcopter wrote: Show nested quote + On September 28 2008 16:30 Loco wrote: you don't need maths in poker to be successful just like you don't need position, leave it to the lesser players is that a giant level or am I missing something

On September 28 2008 16:49 asdf2000 wrote: Show nested quote + On September 28 2008 16:14 Shakedown88 wrote: Question: How do you take that range that you assigned to your opponent's hand, and very roughly, convert it in your head to some kind of usable odds/percentage that you can apply to make a more profitable decision? how does the brain do any of the other amazing crap it does?   On September 28 2008 16:30 Loco wrote: you don't need maths in poker to be successful just like you don't need position, leave it to the lesser players you're obviously taking for granted all of the math your brain subconsciously does while playing

http://www.liquidpoker.net/poker-forum/570308/Floating!!_By_a_pro!!!_Rofl.html

lol travis

On September 28 2008 16:49 Day[9] wrote: ok lets do a rough combinatorial approach right now

Adding to what Day said (which was pretty much perfect) you can also take shortcuts when doing the math. For example, say you have JJ and you put your opponent's range on AA, KK, QQ, TT, 99, and 88. As Day said, each pocket pair has 6 combos, so each hand in his range has an equal amount of combinations. Now you can see how many pairs he has that are above and below yours, and match up one overpair for one underpair, and their equities "cancel out" to 50%. Put another way, you can match up "opposite" hands in a villains range that can cancel to 50% against each other, like bottom set and top set when you have middle set, or higher flushes and lower flushes when you have a flush, and etc.

In this example, there are 3 underpairs and 3 overpairs, so each underpair matches up with one overpair. Even though you are 80% against the underpairs but only 20% against the overpairs, each "match" cancels to 50% equity, so you are 50% against his total range. Doing it this way can save a lot of time and you might not have to do any math (or at least not as much as taking each individual hand and finding the equities).

So to show how much easier this can make things, let's add 77 to the range above and figure out the equity. There are 6 combos of each pocket pair and 7 pairs in his range, meaning there are (6*7)=42 combos in his range. Matching up "opposite" hands, we can match up 3 underpairs with 3 overpairs, and there is one extra underpair. We already determined that we are 50% against the matched pairs, and we know we are 80% agaisnt an underpair. Now all the math you really have to do for this is figure out how percentage of his range each "group" is and multiply it by the equity against that "group" (aka taking a weighted average), pretty much the same way Day said it. Basically this works best when you think someone has a wide range, or in other words you think someone's range is made up of a lot of hands. This way you can reduce the number of groups that you are taking the weighted average for the way Day said it.

On September 28 2008 19:36 tomson wrote: Show nested quote + On September 28 2008 16:49 Day[9] wrote: ok lets do a rough combinatorial approach right now

true true

after every session take a few hands which you thought were close and pokerstove them assign ranges to your opponents etc.. and see how your hand plays against each range, after a while, you just break it into your subconscience and you just know, some people call that "feel" its more along the lines of experience and hard work

uh, just pop up pokerstove and do it before your timebank runs out = done

This was extremely helpful. Much obliged.

i'm a novice at calculating probability and day's assessment doesn't make sense to me

if there are 52 cards in the deck then there are 52 * 51 = 2652 combinations of two cards

anyway so there are 2652 combinations and if we divide that our 6 combinations of AA by our 2652 possible then 6/2652 = .00226 etc. which means that we only have a .226% chance of being dealt AA and according to pokerstove our chance of being dealt AA can be rounded up to .5%... does assuming 6 combinations of AA neglect to acknowledge AsAd can be dealt with AsAd or AdAs

AdAs AsAd AdAc AcAd AdAh AhAd
AcAs AsAc AsAh AhAs
AhAs As Ah

for a total of 12 combinations rather than 6, and then 12/2652 = .0045 which satisfies pokerstove's suggestion that AA composes .5% or at least .45% of hands we are dealt

oh lolz nevermind 52!/(2!(52-2)!) = 1326 and 4!/(2!(4-2)!) = 6 for 6/1326 = ~.45

oh

lol me

On September 29 2008 11:37 failsafe wrote: i'm a novice at calculating probability and day's assessment doesn't make sense to me if there are 52 cards in the deck then there are 52 * 51 = 2652 combinations of two cards anyway so there are 2652 combinations and if we divide that our 6 combinations of AA by our 2652 possible then 6/2652 = .00226 etc. which means that we only have a .226% chance of being dealt AA and according to pokerstove our chance of being dealt AA can be rounded up to .5%... does assuming 6 combinations of AA neglect to acknowledge AsAd can be dealt with AsAd or AdAs AdAs AsAd AdAc AcAd AdAh AhAd AcAs AsAc AsAh AhAs AhAs As Ah for a total of 12 combinations rather than 6, and then 12/2652 = .0045 which satisfies pokerstove's suggestion that AA composes .5% or at least .45% of hands we are dealt

there are NOT 52*51 combinations of cards. there are 52*51 / 2 = 26*51 combinations of cards. In math terms we would use the binomial coefficient (52 choose 2) = (52!) / (2!)(50!) = 52*51 / 2!. The dividing by two is equivalent to ignoring the order. That is, the number we get won't include AdAc and AcAd as two different things. So, there are 26*51 = 1326 possible starting hands.

then, when i say there are 6 combinations, i am using (4 choose 2) = 4*3 / 2 = 6 (by the same logic above). Everything i did in my calculation ignores double counting. In your calculations, you double count, but coincidentally divide out by an appropriate number to get the same result. In other circumstances you won't get so lucky, you just run good w/ your probability ; ].

with poker probabilities, you'll use the binomial coefficient alot alot alot. i'd recommend wikipedia-ing it. its totally awesome and i used it a bunch in my thesis ^_^

On September 29 2008 11:37 failsafe wrote: oh lolz nevermind 52!/(2!(52-2)!) = 1326 and 4!/(2!(4-2)!) = 6 for 6/1326 = ~.45

err wow i didn't see this

oops haha

On September 29 2008 16:36 Day[9] wrote: Show nested quote + On September 29 2008 11:37 failsafe wrote: oh lolz nevermind 52!/(2!(52-2)!) = 1326 and 4!/(2!(4-2)!) = 6 for 6/1326 = ~.45 err wow i didn't see this oops haha

hahahaha yeah actually in an undergrad adv. math course i'm taking we just TODAY went through the proof for the binomial theorem and of course had the mandatory tracing of a few lines/combinatorials from pascal's triangle. this was an epic failure on my part to not IMMEDIATELY recognize that your first answer was exactly a case of what we'd looked at in my class

but i do have a follow-up question! how do you count the combinatorials for a hand like AK?

maths is overrated i just call when i got the best hand and fold when im beat