Yes, you can and you should. The concept you're describing is called implied odds (the estimated profit you'll make if you make your hand).
Notice is a much less concrete value as it is an estimation of whether your opponent will call when the draw comes and the amount he'll be willing to pay. There's also the concept of reverse implied odds which are the ...
Your question can be re-written like this:
What are the odds of having 4 cards with the same suit, from the 5 available community cards?
The answer can be found by using basic probabilities and computing probabilities of composed events.
You can have 5 boards in which 4 cards are of the same suit:
Here, X means the suit ...
As you can imagine, your equity in a heads up hand with no rake, where you bet preflop and deal out all community cards without betting, will be 50%. Other variations of this, such as the dealer winning ties or the introduction of a rake, will lower your equity (and since this is a casino game, I'm willing to bet that they have something in their favor). ...
There are thumb rules for the preflop equity (against a single random opponent) of pocket pairs and suited-connected combos.
For the equity of a pocket pair, you calculate how many cards away from 2 your cards are (for example, Queens are 10 cards away from 2), multiply by 3 and add 50%.
So QQ's preflop equity is approximately:
(10 * 3) + 50 = 80%
I'm not entirely sure what you are trying to say with the math that you have in your question, but I think you are trying to show how you get the odds of hitting a flush or a straight on the turn or river when you have 4-to-a-flush/straight-draw on the flop. The same basic strategy of calculating odds can be done to see what your chances are to hit a set on ...
In order for two people flopping a straight flush we need:
a connected three straight flop, all cards having the same suit. No cards can be lower than a three or higher than a queen.
player 1 holding two connected cards of that same suit making the higher or the lower straight.
player 2 holding two connected cards of that same suit making the last ...
You need to calculate the odds of getting the exact flop that you need.
Since the order doesn't matter, the first card dealt would have three possibilities, and then if you got one of those you would have two possibilities on the second card, etc.
It would look like this:
3/50 * 2/49 * 1/48 = 1/19,600 = 0.005%
Updated based on your comment/updated ...
It all depends on pot odds. If you have better than 50% pot odds and have 50% equity versus your opponent's range, and you have the bankroll to handle the variance, then you should be looking to play for stacks. This will always produce a long term winning strategy, because you're getting >50% return on a 50% bet. Do you see why?
The only situation where ...
Assuming you know nothing about the cards dealt, they don't matter, so the 66% holds up. In most calculations we would just ignore the folded cards since we don't have any definite information about them. If you want to factor them in, you can no longer calculate your exact pot equity, since you don't know how often your opponents are folding hands like Ax, ...
Before anyone speaks, no matter how many players there are the distribution is still totally random. Each card has, for example, exactly the same probability to be in anyone's hand.
However as soon as someone speaks then things change...
What does this do to the overall win/lose odds of a given hand?
Simply put: as soon as a person folds the probability ...
Based on your calculations... If you hold 4 cards (A to 4), you will ALWAYS make a pair?
Estimate with 3/50 * 5 instead of the annoyingly similar fractions:
(3/50 * 5) * 4 = 60/50 !?
The reason you won't get the right answer this way even though I can see the logic in your math is because of "double counting". You think the chance of hitting the ...
Note 1 in the article on Hold'em Odds elaborates on this a bit further:
[Note 1] By removing reflection and applying aggressive search tree pruning, it is possible to reduce the number of unique head-to-head hand combinations from 207,025 to 47,008. Reflection eliminates redundant calculations by observing that given hands h_1 and h_2, if w_1 is the ...
When facing an "at least one of" problem, you can't just add probabilities. You have to calculate the probability of missing everything, then subtract from one.
Assuming 47 unknown cards, 6 of which are outs, the exact probability of missing both the turn and river is 41/47 * 41/46, or 1681/2162, so the probability of hitting either or both the turn and ...
22 vs AK is a "coinflip" (or just a "flip") because there is approximately an equal chance of either hand winning, just like the equal chance of either side of a coin coming up when flipped. In reality, 22 has a slight edge against AK when all in preflop, being a 52% favourite (assuming nothing is known about the suits of the hands).
To deal with your ...
Here is another way to look at it:
Gained if you call and win: 30+50 = 80
Lost if you call and lose: 50
Your Equity = 0.36
EV = Equity(Gained when win) + (1-Equity)(Lost when lose)
EV = 0.36(80) + (1-.36)(50)
EV = 28.8 + -32
EV = -3.2
Plug in 0.38 for your equity and you will see that its indeed near a break-even call.
How did they get that formula?
The article is correct in the way it uses 4:1 and 5:1. Under their assumptions (actual value given their example is more like 4.2:1), you are "4 to 1" to make it while you are getting "5 to 1" on your money. I'd say that this is precisely because both are written / pronounced / thought of this way that it's convenient.
If you check the Wikipedia article on ...
In tournaments, there are often factors that trump pot odds or implied odds when making decisions. Your stack size (and the comparative positions you'll be left in if you call-and-lose vs call-and-win vs fold) is often first and foremost in that list. In no particular order, other factors include table dynamics (e.g. are you going to have opportunities to ...
Without taking into account the fact that the very act of seeing the flop with one or several other player(s) influence the distribution of the flop (*), here's one way how you could compute these odds:
you have C(50,3) possible flops: that is 19 600 flops
out of these there are 48 cases where you'll improve directly to quads, so the probability to flop ...
It turns out that each shuffled deck is in the order that may have never before existed in the history of the universe! :)
The odds of you getting two 52 card decks arranged in the exact same order are 52! ~= 8 x 10^67, which is waaay more than the number of atoms on Earth (~ 10^50).
For a detailed explanation, please check out a great video answer on TED....
Sometimes during play you want a quicker method that isn't 100% but gives you very close percentages.
The rule of 2 and 4.
If you have 2 cards left to reveal (ie the turn & river) then you multiply your outs by 4.
If you have 1 cards left to reveal (ie just the river) then you multiply your outs by 2.
So in your example, 6 outs with 2 chances left = ...
These are the chances (assuming you have no ace):
These are valid only preflop assuming that there are 50 cards left in the deck (you are holding 2) and you are one of the players (2 players = 1 opponent with 2 cards; 3 players = 2 opponents with 4 cards and so on)
From a probability perspective, you can think of this as two events: the first being each of the five player getting dealt a pocket pair, and the second dealing five cards that match those pocket pairs exactly.
First let's begin with the probability that there are 5 different-suited pocket pairs dealt in one hand (note that 78 / 72 / 66 / 60 / 54 are the ...
At first, I thought, OK, lets quickly answer this, returning the favor. But it turned out to be a rather tricky one! One online calculator gives 48.18%, the other 48.28%, off by 0.1%, already very strange. My first result was 49.09%, off by exactly the runner-runner 0.909%. You also confused me with your cases and some calculations.
The tricky part is - of ...
If you have an android phone you can download "Poker calculator", logo is a clover as you can see topleft.
It is a great app. It supports hold 'em but also 4 and 5 card omaha high and omaha high/low. You can add multiple players and can input the cards by voice. I recommend it to anyone.
Nice question and I think the factors you are considering are by far the most important ones to remember. However I'd also say remember, as you also mentioned, the position of players who raised before, as often that can be used to some degree to infer strength.
Other factors to consider could also be (please note I know you said cash game, but I'm also ...
First off, the statement in your question is not the norm:
They have an obvious advantage of knowing about more cards that are in the play than they should, and they can strategically bet against one another to increase the pot (encouraging other players to play in hopes of winning).
Almost always when players collude, they are typically protecting each ...