Expected Value at the Casino . I told her I had a 90.25% chance of being up $810 (two pots: $640+$770 subtract $600 buy-in), and a a 0.0025 chance of being stuck $600. Of those combos, 32 include a K. So the probability the CO holds a K is 32/159 = 0.20.We have to ignore a second order effect here which is that the specific combo that CO holds impacts the combos BTN can hold, but to a first order approximation it’s reasonable to also assign BTN a 0.20 probability of holding a K.To find out if at least one of them holds a K, we calculate the probability that neither of them holds a K.Thus the probability at least one of them holds a K is 0.36.A more accurate result would require using the initial ranges and running a Monte Carlo simulation which assigns CO and BTN a specific hand drawn randomly from that range for 10,000 iterations or so.Pot is $640 after two players are all-in on the turn.Pot is $770 after two players are all-in on the turn.These were the only two hands of significance for me during this session. This calculator will tell you the expected value for a binomial random variable, given the number of trials and the probability of success. This calculator is used when determining whether to call a bet, assuming no further action. Expected Value Calculator for a Binomial Random Variable. {"smallUrl":"https:\/\/www.wikihow.com\/images\/thumb\/5\/5a\/Calculate-an-Expected-Value-Step-1-Version-5.jpg\/v4-460px-Calculate-an-Expected-Value-Step-1-Version-5.jpg","bigUrl":"\/images\/thumb\/5\/5a\/Calculate-an-Expected-Value-Step-1-Version-5.jpg\/aid1612308-v4-728px-Calculate-an-Expected-Value-Step-1-Version-5.jpg","smallWidth":460,"smallHeight":345,"bigWidth":"728","bigHeight":"546","licensing":"