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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

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 ...


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, ...


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 ...


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 ...


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 ...


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

You are calculating pot odds a very unusual way. Your formula is mostly correct (but only works some of the time), and I'll get back to that in a moment, but typically you would just use two variables: costToCall and sizeOfPot. Pot odds don't depend on the number of players to have called the bet. One player putting 400 in is the same as four players each ...


1

Tie is neither win nor loss. Therefore, you should have a counter for ties the same way you have a counter for wins. In any case a round with a tie should be removed from your simulation.


1

As far as the math for determining this, it gets very involved and complicated. The basic formula is the same though, the cards you can hit to win / the total remaining cards unseen. Now I'm sure the problem your having in the situation you posted is that you don't know the other players hole cards. That throws a wrench in the works. Since you don't ...


1

It's a reasonable improvement for the overcards when they are suited. In the example of AK versus pocket tens, you improve from about 42.5% chance of winning to about 45.9% chance. AK does a little better against 99 than against tens, because the 10 blocks the possible straight. Here are some examples from pokerstove. First, unsuited AK against pocket ...


1

Let C(n,m) = n!/(n - m)!. That is, the number of ways to choose m items from n. The calculation you did for the first value is correct: 1 - C(52-4, 30)/C(52, 30) = 97.3% You can arrive at that calculation by first calculating the probability that no eights are drawn. That is the number of ways to choose 30 cards from the deck excluding the 4 eights, ...


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 ...



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