Approximate odds with 2 cards to come

[Henke]

Another post here got me thinking about deriving an approximation to the odds (against one) with two cards to come given a specific number of outs. Anyway, the approximation I came up with is 23/o-0.69. This is very exact if the nr of outs equals 5 to 15, with a maximum absolute error of about 0.0058 (this is when o=9). The errors for 1 to 4 outs are (rounded to two "value-digits) 0.19, 0.069, 0.031 and 0.013. Don't know if it's practical though, it isn't really that easy to calculate that 23/5-0.69=3.91 [img]/images/graemlins/grin.gif[/img]

To get an approximation to the odds rounded to the nearest integer, just use round(25/o-1). That gives the same result as the exact odds rounded to the nearest integer for o>2 (well, actually not when o=17,18 or 19).

[Bozeman]

You've noticed that your 25/o-1 (for odds) is the same as the /4 approximation (for probability), right?

[Henke]

Now that you say so, yeah. Guess I wouldn't have to have done the series expansion then [img]/images/graemlins/smirk.gif[/img]

[N4CER1]

A simple way to get really close with out all of the fancy math is just count your outs and multiply by 4...say you have 8 outs....8 x 4 = 32.....you have @ a 32% chance of making your hand....but you have @ a 68% chance of not making the hand....this will get you very close without having to learn all of the math..........