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8

The first thing to learn is some basic hand requirements, sorted by position. Say, you've been dealt Q♥J♠ and since it's face cards it should be good to play no matter what. No! You need to memorize (and do it quickly) your relative position to the BTN (dealer) and play these cards according to your current position. Memorize the hand rankings ...


5

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, ...


5

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 ...


4

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 ...


4

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 ...


4

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% EDIT Updated based on your ...


3

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 ...


3

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 ...


3

You've been given a very good answer above by @vlzvl in fact, I upvoted it. So, I'm going to go in a different direction and just give you some "meta" advice that I think you will find helpful. This is just in brain-dump format, choose as you see fit: 1 - You are not special: At core, there is still a lot of math & randomness in poker. You will NOT ...


3

if I hold two different cards, what's the probability I get a pair at river? Probabilities computation can quickly get very complicated and you need to be very precise in what you want to exactly compute. What if the board comes with a pair in it? Do you consider that "getting a pair at river" or do you only want to know the probability of making a ...


2

Assuming random hand distribution and players of equal ability, the probability of winning is 1/x. Of course other factors are involved in real poker, so it's impossible to summarize this in a simple formula. If you are interested in the more specific question of how much cash equity you have in a tournament given the remaining players' stack sizes, you ...


2

Unfortunately, this is the kind of probability question that would take a couple of hours and reams of paper notes to get the exact answer. But simulating it is easy (I have loads of C code for that). I got about 1 in 500. (205423 hands out of 100000000).


2

p is 0.0000046642 Here are two different ways to arrive at that result... If you have XY, there is one way to make "quad X" and C(3,2) ways to make "set Y". If you have XY, there is also one way to make "quad Y" and C(3,2) ways to make "set X". C(3,2) is 3, so starting from XY, there are 6 possibilities to make what you called the "big full house" (very ...


1

Since all the cards except the two known to you, are unknown the odds do not change in respect to how you count you outs of hitting a particular hand or card. Having said that Your odds against winning a particular pot go down the more players there are in a hand. Also particular hands play better AK for example against larger fields, and other hands, pairs ...


1

There are (52 choose 7) = 133784560 total possible hands in 7 card poker. Of these, 12 * 4 * 13 are a "Big Full House." To see why, consider the 12 seven-card hands with four aces and three of a kind, disregarding suit: AAAA222 AAAA666 AAAA101010 AAAA333 AAAA777 AAAAJJJ AAAA444 AAAA888 AAAAQQQ AAAA555 AAAA999 AAAAKKK For each of these 12 hands, ...



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