I got a math problem for you...
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I got a math problem for you...
So I'm no Mike Caro, but I was thinking about simplifying a way to figure outs on the turn and river combined. I did a little 9th grade algebra and came up with the problem below. It seems to work, but once again, I'm no math genius. Any comments?
x=# of outs after the flop
y=probability of hitting one of those outs on the turn and/or river
y = 1 - ( 1 - x/47 )( 1 - x/46 ) [the standard out-figuring equation]
y = 1 - ( 1 - x/46 - x/47 + x2/2162 ) [that's x squared, remember FOILing?!]
y = -x/46 - x/47 + x2/2162
y = x2/2162 - x/46 -x/47 [just rearranged it to look nicer]
y = x ( x/2162 - 1/46 - 1/47 )
y = x ( x/2162 - .043) [changed it to a decimal for simplicity]
So you take you outs, divide it by 2162, subtract .043, and multiply that by the number of outs again. The absolute value of that should roughly be your chance of hitting at least one out on the turn or river combined.
Right?
x=# of outs after the flop
y=probability of hitting one of those outs on the turn and/or river
y = 1 - ( 1 - x/47 )( 1 - x/46 ) [the standard out-figuring equation]
y = 1 - ( 1 - x/46 - x/47 + x2/2162 ) [that's x squared, remember FOILing?!]
y = -x/46 - x/47 + x2/2162
y = x2/2162 - x/46 -x/47 [just rearranged it to look nicer]
y = x ( x/2162 - 1/46 - 1/47 )
y = x ( x/2162 - .043) [changed it to a decimal for simplicity]
So you take you outs, divide it by 2162, subtract .043, and multiply that by the number of outs again. The absolute value of that should roughly be your chance of hitting at least one out on the turn or river combined.
Right?
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Adamm - Admin
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