I was a losing player five years ago who had dropped $200 in buy-ins. Joined FTR, turned it around with a $100 buy-in and Poker Tracker + a HUD. That was back when you bought the HUD separately. I cannot thank FTR enough for the five years of advice and encouragement. I'm a hobby player and a mathematics professor with a teaching load that consists of Game Theory and 4 different probability and statistics courses. Lifetime in poker, I've made back all the money I donked off to start with plus all the money I spent on software and books (nearly $500 worth, as I use some of them for background/examples for my classes), plus an additional $2,500.
I'm a math geek and an underachieving poker dumbass. Ask a mod. They won't disagree. I'm happy with my poker life, though I get accused of "mathturbation" regularly. I get to play a game I love and, every now and again, get to withdraw some money. Just back after a year-long, post-Black-Friday layoff.
For the special 3k occasion, let's play "stump the math geek." I'm going to post some probabilities that I think about at the tables. You don't have to know these factoids to be successful at poker, but some of them help a good bit. Then I'll take questions. Ask me the probability of anything you like, and I'll take a stab at it. I'll do better with gambling related probabilities, but I will attempt any problem you post.
I'll use spoilers to hide the maths, so the post is less scary. The first spoiler is just some notation we'll use in later spoiler-math sections.
Spoiler:
We'll need factorials like 4! = 4 * 3 * 2 * 1 = 24. Note that 0! = 1. We'll need choose notation for the binomial coefficients, so let's go with 5c2 = 5! / ( 2! * 3! ) = 10. This is all the ways to choose 2 objects from a collection of 5 objects where order does not matter, e.g. AK = KA. I'll leave the maths in spoilers so the reading's not so daunting. Check my maths if you're of a mind to.
Example 1: When playing HU, I love a 2 pair hand - bottom two, top two, top and bottom - love 'em all. You can often get all the chips in with 97 on a A97 against any Ax hand. And it's usually the nuts. Hero flops exactly two pair (when starting with unpaired cards) about 2.02% of the time, or just slightly more than once every 50 hands. This should also be an argument AGAINST playing hands like 73o in any game, since the likelihood of a big flop with it is about 1 in 50.
Spoiler:
Let 2PrF be the outcome of Hero starting with two unpaired cards and flopping exactly two pair, e.g. one of each card Hero holds. (We're ignoring times when we flopone pair and the board pairs). Then:
Explanation: suppose Hero has 97, the "3c1" expressions choose exactly one 9 (and one 7) from the three remaining 9's (and 7's), and the "44c1" chooses any one of the remaining 44 non-9's and non-7's.
Example 2: I love paired boards. I view them as "inflection points" when analyzing a villain. How you play paired boards tells me how good your poker awareness is. Instead of hitting the flop 1/3 of the time, a Villain playing a hand like AJ+ will only hit a paired board 1/5 of the time. And when he hits, he has either a big hand or a one pair hand knowing there's almost no chance we've hit something. You can see why paired boards are gold mines for reads on postflop play. When do they barrel? Do they recognize my donk into them with an underpair for what it is, just the mathematical knowledge they can't have hit the flop very often? Donk/fold is easy, as is bet/foldin position. When they hit, they hit hard and know they're way ahead. So they usually let us know, too. This gives a great deal of insight into how they play toppair type hands on other board textures. Same is true for overpairs they hold - they know they're likely way ahead and usually can value bet with impunity. We get a chance to recognize it for exactly what it is and note that for later.
2a) In general, we like playing hands like AK because they flop a pair or better about 1/3 of the time. The exact probability is 32.43%, with 2.02% two pair, 1.35% trips, .09% full houses, and .01% quads. That leaves 28.96% one pair hands that are TPTK. Villains like playing them to, and even the brain dead villains know to cbet. When I see cbet > 75%, I'm looking for a way to extract value from all those times he's barreling with nothing but overs.
Spoiler:
Assuming we are using Hero's hole cards to make pairs, etc, there are 6 cards that improve our hand. The overall probability of hitting at least one of them:
Explanation: the "6c1" picks one card to pair up with one of our while the "44c2" picks any two other cards - the "6c2" picks any two cards, giving us our two pair hands and trips - the "6c3" picks all three flop cards to match one of ours, so these are boats and quads.
2b) How does this change when the board pairs? If we assume Villain has unpaired cards to begin, he has only a 20.43% chance of hitting the flop. We can get a read on his barreling since we know that only 1/5 of the time he has a hand, and the rest of the time he's got air. Also remember that more than 2/3's of the time the board pairs (depending upon his and our hole cards) it pairs with T's or lower, another problem for toppair hands. They're pretty sure we don't have anything, but they KNOW they don't have anything and are unlikely to improve.
Spoiler:
We proceed in two parts and assume Villain is holding two unpaired cards. First, we count all paired board flops that MISS him. Second, we count all the possible paired boards flops that HIT him.
Part I: To count all the flops that DO NOT hit either of Villain's two hole cards: Combos = 11c1 * 4c2 * 40c1 = 11 * 6 * 10 * 4 = 2640
Explanation: choose one of the 11 card values NOT in Villain's hand with "11c1", then "4c2" to get a pair; choose any one of the 40 remaining unused cards with "40c1"
Part II: To count all the flops that HIT at least one of Villain's hole cards: Trips = 2c1 * 3c2 * 44c1 = 2 * 3 * 44 = 264
1 Pair + Board pair = 2c1 * 3c1 * 11c1 * 4c2 = 2 * 3 * 11 * 6 = 396 Full House = 2c1 * 3c2 * 3c1 = 2 * 3 * 3 = 18
Explanation (Trips): pick one of Hero's two hole cards with "2c1", then count the trips with "3c2", then pick any of the 44 remaining unused cards with "44c1"
Explanation (1 Pair + Board Pair): pick one of Hero's hole cards to pair with "2c1", then make the pair with "3c1"; choose any of the remaining 11 card values for the board pair with "11c1", then make the pair with "4c2"
Explanation (Full House): pick one of Hero's hole cards to be the trips with "2c1, then make the trips with with "3c2"; we've already selected the card to pair by NOT selecting it to be trips, so just make the pair with "3c1"
Note: there is an easy double-counting / half-counting mistake. If interested, I will explain in the thread below. The difference is easy to see when dealing 5 cards and counting all the possible two pair hands compared with counting all the possible Full Houses.
Summarizing, there are 2640 + 264 + 396 + 18 = 3318 paired boards, and Villain hits 264 + 396 + 18 = 678 of them, or ~20%.
Example 3: You know how you hate those flushdraw boards? Get used to 'em. If we're observing a table, and know none of the hole cards, we would see the flop come with 2-of-a-suit about 55% of the time. Thinking about playing 98s? How often can you expect a flushdraw in your suit? About 11% of the time. Additionally, about 1% of the time we flop the flush immediately.
Explanation: The "11c3" picks three cards from Hero's suit, with nothing else needed
Note that we use 52c3 = 22100 when counting total possible flops with no hole cards known, and 50c3 = 19600 when the hole cards are known
Now, it's your turn. Ask me anything. What poker probabilities have you wondered about but never felt like calculating? Or any probability problem that interests. Ask me, and I'll do my best.
Start with 2 identical decks and take all the diamonds out of one of them. Turn the diamonds into suns by adding 4 points with a red pen or marker. Shuffle them into a standard deck. They can always be separated into 2 standard decks later (if you can handle that the diamonds are a little different on one deck). Oh yeah, ask a question:
If you had a 5-suited deck (65 cards), what would the hand rankings be (including 5-of-a-kind)?
After your 3-year-old f**ks up your only deck of cards two minutes before your home game starts, you can play 39-card poker by switching the flushes < straights, and no quads obv. Here's the breakdown.
We have 39c5 = 575757 hands - interesting number - who knew?
Found a typo in the 3-suit frequencies: the formula is correct, but the result should be 0.67%
Also: Is the numerator for the straight flush eq's based on:
10c1 * [n]c1
where [n] is the number of suits, or are there missing terms (all of which are 1, so moot)?
Found a typo in the 3-suit frequencies: the formula is correct, but the result should be 0.67%
Also: Is the numerator for the straight flush eq's based on:
10c1 * [n]c1
where [n] is the number of suits, or are there missing terms (all of which are 1, so moot)?
You are exactly right. Another way to do it is simply to count them since they're so few. There are 10 possible straights, 5-high through A-high, and there is one of each variety in each suit, so 10 * [n] is the correct number.
Hand rankings are based on taking 5 random cards out of a 52 card deck. We assign certain combinations of numbers and suits to be scoring hands. We then determine the frequency of drawing those scoring hands, which determines the rank of the hands.
How do the probabilities change when we're looking at the best 5-card hand from any 7 random cards?
I.e. There is ~51% chance of having a 'high card' hand with 5 random cards, what is the chance of having less than a pair with 7 cards?
Last edited by MadMojoMonkey; 05-13-2012 at 10:24 PM.
How do the probabilities change when we're looking at the best 5-card hand from any 7 random cards?
This is harder, as "weird" hands exist: 3 pair, double trips, 6-flushes, 7-flushes, straight + pairs (trips), flushes + pairs (trips), quads + trips. Counting them isn't any different, it just takes a while to think of all the oddities and figuring out which "normal" categories they fit into.
For one pair hands, we'd like to do something simple like:
P = 13c1 * 4c2 * 12c5 * (4c1)^5 / 52c7 ~ 47.28%
This gives a pair and 5 other card values. The problem is that several straights and flushes are in that figure, and it takes a while to count them and subtract them out.
Also, there are 133 million 7-card poker hands, almost a 2 order of magnitude increase in the size (and difficulty) of the problem.
I'll give it a shot in the morning, but it's easier to just freakin' google it.
OK, I haven't sorted out how to partition out straights/flushes from paired/trips hands, yet, but here's the first batch of probabilities for the best 5-card poker hand.
1 pair (includes some straights/flushes) = 63258624 /133784560 ~ 47.28%
2 pair (includes some straights/flushes) = 29652480 /133784560 ~ 22.16%
3 pair (no full houses) = 2471040 / 133784560 ~ 1.85%
Trips (includes some straights/flushes) = 6589440 / 133784560 ~ 4.93%
As you can see, the big hands are much more likely, while naked 1 pair hands should occur with about the same frequency, after the straights/flushes are subtracted out. Everything else is much more likely.
The hand raking is still the same, though flushes and full houses are nearly identical in frequency. That seems strange, but I'll double-check the maths and think about it some more.
I have some ideas on how to count flushes + pairs/trips, but straights are going to be a pain in my backside. The calculations for straights are unique, quite different from the standard combinatorics we use for pairs, trips, flushes, etc. I'll keep thinking about it.
I am smiling like the Cheshire Cat right now. These things have been gnawing at the back of my brain for months. Thank you so much for this. I am a huge geek, I know, but it makes me so happy to see these results.
Last edited by MadMojoMonkey; 05-14-2012 at 11:25 AM.
Something I've always wondered with suited connectors is as the gap between your cards increases by how much exactly does the possibility of you hitting a piece (gutshot +) of the board decrease. For example 98s vs 96s.
Something I've always wondered with suited connectors is as the gap between your cards increases by how much exactly does the possibility of you hitting a piece (gutshot +) of the board decrease. For example 98s vs 96s.
The only difference in the "flopability" of 98s vs 95s is the straight draws and pairs + straight draws. Pairs, 2 pair, trips, full houses, quads, flushes and flush draws occur in the exact same frequencies for any two suited cards.
Counting straights and straights draws is straightforward (pun intended). It comes down to figuring out all the possible patterns. There are 19600 flops given that we know Hero's hole cards, so I'll just count the flops that are straights and SD's.
Straights
For sc's like 98s, we have the following possible straight patterns:
xxx98
xx98x
x98xx
98xxx
Each can occur in 64 ways since we have 4 of each card value in the pattern: (4c1)^3 = 64. So there are 256 flopped straights for 98s.
For 1-gappers:
xx9x7
x9x7x
9x7xx
Three patterns, so 64 * 3 = 192 flopped straights.
For 2-gappers:
x9xx6
9xx6x
Two patterns, so 64 * 2 = 128 flopped straights.
For 3-gappers:
9xxx5
One pattern, so 64 flopped straights.
To keep things easier to read, I'll do this in multiple posts.
For sc's like 98s, we have the following possible OESD patterns:
xx98
x98x
98xx
Each can occur in 544 ways since we have 4 of each straight card and 28 other cards to choose from that are NOT a 9, an 8 or one of the "ends" of the draw that would be straights and were counted above. We also have 6 cards that would pair the board. So we have 4 * 4 * 34 = 544 ways for each of the patterns to occur.
Three patterns, so there are 3 * 544 = 1632 flopped OESD's for 98s.
How many are there? We have 4 * 4 = 16 ways to draw the SC cards, plus 6 cards that would pair one of Hero's hole cards, so 6 * 16 = 96 flops for each pattern. There are 3 * 96 = 288.
Anyone who wants to check my counting of the patterns or my arithmetic, feel free. It's 1 AM, and I'm not fully functional on even basic tasks like patterns and multiplication, and even the spreadsheet may not have saved me. I will check for errors in the morning.
This is a good point, DJ. AKs is a great hand, but it doesn't have all the straight possibilities of 98s. Also, you end up against a lot of combo hands and other straight draws when playing naked straight draws with broadwayconnectors, so they can be scary to play. On the plus side, you're much better off when you hit TPTK with it!
Where did the table come from? The SD counts seem a bit high, like maybe they include Gutshots, too? Or maybe I screwed up. But a back of the napkin calculation suggests these aren't all 8-out straight draws.
I didn't include naked Gutshots on purpose, as I don't consider them 8+ out draws, as the 6 cards that make Hero a 1 pair hand may not be enough to put us ahead. I think that's the difference in the two.
just wanted to illustrate that some SCs are more equal than others.
What you're talking about is called "full stretch". Pockets that do not have full stretch should not be considered "connected", IMO. Basically if both of your cards are less than 5 or greater than 10, they are not connected. Clearly, broadway cards have other equity.
I'm basically saying that the 42,43,32 family are a bunch of losers and they smell funny even when you dress them up in suits. I heard they took each other to the prom.
Fun side note: every straight has either a 5 or a 10, not both. They're kind of the "fundamental straight cards". What a terrible name. Someone come up with a good name for the 5 and 10 as pertains to this quality.
I didn't feel dissed. I was just scratching my head looking at your tables compared to mine and hoping I got the right answer! One point of the post was show the calculations for anyone interested. I'd love to have your programming skills.
I'm not a night person, so my midnight maths are always suspect. Defo the computer would count them accurately, and probably a good programmer could write and debug a program faster than I did the analysis above. NH!
The "playability" of the draws fors sc's, 1-gappers, and so forth are very different, as you pointed out. The sc's flop the the "nut flushdraw" way more often, so the playability when you do hit your straightdraw is superior with 98 vs 96. I like thinking about the patterns and understanding how they would play out.
We flop the "nut straightdraw" 5/6 of the time we flop an SD with 98, but only half the time with 96, for example.
The SD patterns with 98 are:
xx98
x98x
98xx
If "S" is the card that completes the straight, the three patterns above lead to 6 possible made straights (discounting our hole cards pairing the board and 6-card straights, etc):
Sxx98
xx98S
Sx98x
x98xS
S98xx
98xxS
Except for pattern #1, the suited-connector flops the nut straightdraw every time it flops an SD.
For the 96, just to use an example of "lessened playability," here is the pattern:
9xx6
Not only do we hit this SD less often, only half the times we hit it do we have the nut straightdraw:
As you probably know, JT has the distinction of - every single time it makes up part of a straight - being the nut straight. The sc's are even more playable (vs. gappers) than the "frequency of draws/combos" table show because of these playability problems with the gappers.
Having the donk end of the SD can stack you in a hurry.
I think you may have forgotten to take into account the reduced prob. of straights at the top and bottom of the deck
I believe this is true: Ax hands that can make straights always play like 3-gappers in terms of straights and straight draws. The playability is different, obviously, but the quantities are the same.
Looking at your table, this isn't true - I think the reason again might be the gutshots. If we only count 8-out SD flops, Hands like A4 and AQ have the same number of straights and SD's (and pairs + SD's) as T6.
If we take into account all of Hero's possible hole cards, pairs, unsuited, and suited, these probabilities are exactly the same. But it's a bunch more complicated to do the math.
There are several ways to answer this question. We hate Ace high flops the most when we hold KK or hands like KQ. These are most easily accomplished by using the set complement, and calculating the probability of it NOT happening, then subtracting from 100%.
Probability of at least one Ace on flop, given we have KK:
P = 1 - 46c3 / 50c3 ~ 31.31%
Probability of at least one Ace or King, given we have QQ:
P = 1 - 42c3 / 50c3 ~ 48.05%
Probability of at least once Ace, King or Queen, given we have JJ:
P = 1 - 38c3 / 50c3 ~ 61.83%
Probability of at least one overcard, given we have TT:
P = 1 - 34c3 / 50c3 ~ 72.92%
Probability of at least one overcard, given we have 99:
P = 1 - 30c3 / 50c3 ~ 81.63%
If you have some other scenarios you'd like the counts on, like number of times Hero holds Ax and hits an Ace-highflop, let me know.
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Originally Posted by MadMojoMonkey
