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to be honest i just thought you needed a hand coming up with your z-score, and was trying to be helpful. i didn't really read the post thoroughly

... just thinking out loud here... if you include...

I use the simplified version (only applied to 50/50 propositions) I went through it once a long time ago and have always used this simplified approach. There is a more complete version that you have...

it isn't wins - losses.

It is wins - (total number of games)/2

assuming a 50% chance of winning a game, your expected number of wins is the total number of games divided by 2.

Nah, of course not. Why would something so hokey pokey like math work...

assuming this is a point spread sport where the expected win pct is 50% if you flipped coins...

470-375....


(470 - (470+375)/2) / ( ((470+375)/4)^0.5 ) = 3.26

Thank me later for this one: ...

if you search this forum you'll find the answer. It's been asked a few times and answered pretty well already... no need to rephrase it.