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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 ...


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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

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 ...


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 ...


0

The short answer is that the other cards don't matter? Why, because the cards are just as likely to have been folded, as they are to still be in the stub of cards to be dealt out. You can't differentiate the cards in that way, so you don't need to account for them in any special way.



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