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May 10th, 2016, 04:58 AM  #1 
Newbie Joined: May 2016 From: UK Posts: 1 Thanks: 0  Card removal effects on distribution of remaining cards
Hey, apologies if this isn't the appropiate subforum for this.. In a game of two card poker, the first four players fold their hands implying they are much more likely to have folded smaller cards than higher cards. Given this I'd like to find out the probabilities of each card in the last remaining hand. The probabilities of each card being folded by the 1st, 2nd, 3rd and then 4th player will be on the left below, followed by the relative probabilities of each card in the remaining deck. There is some rounding errors along the way but does my methodology seem like it's correct? I imagine there's a less brute/more simplified method and would love to hear it if anyone can think of one! Player 1 folding frequencies of each card.. A 8.9% K 12.7% Q 12.7% J 12.7% T 14.5% 9 15.3% 8 16.4% 7 16.4% 6 17.1% 5 17.5% 4 18.6% 3 18.6% 2 18.6% If Player 1 folded any two cards, remaining distribution of each card would be equivalent to.. 50/52 * 4 = 3.846 of each card. With the card removal the deck now effectively has has 4 (8.9/15.385 * 43.846) = 3.91 aces in the deck. A  3.911 = 1.017 times as likely as average K  3.873 = 1.007 Q  3.873 = 1.007 J  3.873 = 1.007 T  3.855 = 1.002 9  3.847 = 1.000 8  3.836 = 0.997 7  3.836 = 0.997 6  3.829 = 0.996 5  3.825 = 0.995 4  3.814 = 0.992 3  3.814 = 0.992 2  3.814 = 0.992 As the remaining deck is now slightly heavier in higher cards we must multiply the probability of Player 2 folding each card given by the numbers above given there is a higher chance of recieving those cards. Player 2 folding frequencies.. New distribution.. Ave of each card is.. ..48/13 = 3.692 A 8.4% * 1.017 = 8.5% 3.911 0.085 = 3.826 1.036 times as likely as average K 12.3% * x = 12.4%  3.749  1.015 Q 12.3% = 12.4%  3.749  1.015 J 12.3% = 12.4%  3.749  1.015 T 11.9% = 11.9%  3.736  1.012 9 15.1% = 15.1%  3.696  1.001 8 17.1% = 17.1%  3.665  0.993 7 17.5% = 17.5%  3.661  0.992 6 17.5% = 17.5%  3.654  0.990 5 18.3% = 18.2%  3.643  0.987 4 18.7% = 18.6%  3.628  0.983 3 18.7% = 18.6%  3.628  0.983 2 19.9% = 19.7%  3.617  0.980 Player 3 folding range ... New distribution... Ave of each card is.. ..46/13 = 3.5385 A 2.8% * 1.036 = 2.9 3.826  0.029 = 3.797 1.073 times as likely as average K 9.7% = 9.8  = 3.651  1.032 Q 12.4% = 12.6  = 3.623  1.024 J 13.8% = 14.0  = 3.609  1.020 T 13.4% = 13.5  = 3.601  1.018 9 14.7% = 14.8  = 3.548  1.002 8 17.1% = 16.9  = 3.496  0.988 7 17.5% = 17.4  = 3.487  0.985 6 18.4% = 18.2  = 3.472  0.981 5 18.4% = 18.2  = 3.461  0.978 4 19.4% = 19.0  = 3.438  0.972 3 21.2% = 20.8  = 3.42  0.967 2 21.2% = 20.8  = 3.409  0.963 Player 4 folding range... New Distribution... Ave of each card is 44/13.. .. = 3.3846 A 0% = 0  3.797  1.122 times as likely K 7.1% * 1.032 = 7.3 3.651  0.073 = 3.578  1.057 Q 10.6% * 1.024 = 10.8  = 3.515  1.038 J 11.8% = 12  = 3.489  1.031 T 12.9 = 13.2  = 3.469  1.025 9 12.9 = 13.0  = 3.418  1.010 8 12.4 = 12.2  = 3.374  0.997 7 19.4 = 19.1  = 3.296  0.974 6 18.8 = 18.5  = 3.287  0.971 5 21.8 = 21.3  = 3.248  0.959 4 22.9 = 22.3  = 3.215  0.950 3 24.7 = 23.9  = 3.181  0.940 2 24.7 = 23.8  = 3.171  0.937 Now I would also like to find out the relative probability of the remaining player recieving a certain 2 card hand. For a nonpaired hand can I simply multiply the probabilities together? For example could I say the relative probability of getting AK compared to 32 is the 1.122*1.057 = 1.186 divided by 0.940*0.937 = 0.881 > 1.186/0.881 = 1.346 => AK is 1.34 times as likely to be recieved as 32?? Thanks a lot for reading, if any further explanation is wished for please dont hesitate to ask! 

Tags 
card, cards, distribution, effects, remaining, removal 
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