Hypothetical Freeroll

The 2.14:1 figure is based on losing 360/990 times (~36.3%) and winning 630/990 times (~63.6%).

Well to get the figure, we divide 15/7 = ~2.14, but we'll leave it as 15/7 for the math. Now assuming it calls us $10 to call into a pot of size 15/7 * 10 = 150/7, there will then be 150/7 + 10, or 150/7 + 70/7, or $220/7 in the pot. We will win half of this 630 times over 990 hands. So we win $110/7 * 630 over 990 hands. The average win per hand will therefore be ($110/7 * 630) / 990. This is calculated easily.

110/990 * 630/7
= 1/9 * 90
= $10 average win.

break even.

[ QUOTE ]
The 2.14:1 figure is based on losing 360/990 times (~36.3%) and winning 630/990 times (~63.6%).

Well to get the figure, we divide 15/7 = ~2.14, but we'll leave it as 15/7 for the math. Now assuming it calls us $10 to call into a pot of size 15/7 * 10 = 150/7, there will then be 150/7 + 10, or 150/7 + 70/7, or $220/7 in the pot. We will win half of this 630 times over 990 hands. So we win $110/7 * 630 over 990 hands. The average win per hand will therefore be ($110/7 * 630) / 990. This is calculated easily.

110/990 * 630/7
= 1/9 * 90
= $10 average win.

break even.

[/ QUOTE ]
I agree. Anything over 2.14to1 odds will be +EV or about 2.5to1 odds to keep it simple.