slickpoppa
09-27-2004, 01:28 AM
~6.80% or 13.7:1? Can someone verify my math:
Assuming 10 players and anyone with a pocket pair seeing the flop (a pretty good assumption)
Prob. of 1 pocket pair = (3/51) *(48/51)^9*10 = .3408
"""""""" 2 """"""""""" = (3/51)^2*(48/51)^8*45 = .0959
"""""""" 3 """"""""""" = (3/51)^3*(48/51)^7*120 = .0159
"""""""" 4 """"""""""" = (3/51)^4*(48/51)^6*210 = .0017
ignoring 5 or more PP's ...
Prob of 1 pp hitting a set = (2/50)(48/49)(47/48)*3 = .1151
at least 1/2 """"""""""""" = 1-(44/48)(43/47)(42/46) = .2343
"""""""" 1/3 """""""""""""" = 1-(40/46)(39/45)(38/44) = .3491
"""""""" 1/4 """""""""""""" = 1-(36/44)(35/43)(34/42) = .4609
So adding these up: (.3408)*(.1151) + (.0959)*(.2343) + (.0159)*(.3491) + (.0017)(.4609) = .0680
Assuming 10 players and anyone with a pocket pair seeing the flop (a pretty good assumption)
Prob. of 1 pocket pair = (3/51) *(48/51)^9*10 = .3408
"""""""" 2 """"""""""" = (3/51)^2*(48/51)^8*45 = .0959
"""""""" 3 """"""""""" = (3/51)^3*(48/51)^7*120 = .0159
"""""""" 4 """"""""""" = (3/51)^4*(48/51)^6*210 = .0017
ignoring 5 or more PP's ...
Prob of 1 pp hitting a set = (2/50)(48/49)(47/48)*3 = .1151
at least 1/2 """"""""""""" = 1-(44/48)(43/47)(42/46) = .2343
"""""""" 1/3 """""""""""""" = 1-(40/46)(39/45)(38/44) = .3491
"""""""" 1/4 """""""""""""" = 1-(36/44)(35/43)(34/42) = .4609
So adding these up: (.3408)*(.1151) + (.0959)*(.2343) + (.0159)*(.3491) + (.0017)(.4609) = .0680