Calculating the EV of folding & questions about Calling on Draws

So, I have a math related quandary for you guys. How do you calculate your EV for folding? Obviously its negative(?), but I'd like to know how negative.

The reason I ask is this. I was manually going through some pot-odds calculations on the following scenario, and I ended up with some funny results. Here's the scenario:

Chance for Our Draw to Hit: 10%
Our Equity in the Pot before calling: 10%
Potsize: $100
No implied odds at all & raising is not an option
Vary how much Villain bets & calculate EV's for folding & calling.

The way I calculated the EV for folding was to assume that since we have 10% equity in the pot, we're folding away $10 each time we fold. So the EV for folding is -$10 constant.

Then if I calculate the expected value for calling, it becomes zero at the expected 9:1 pot odds that you need to hit a 1:9 draw. However, it doesn't become less than -$10 until pot odds of 5:1! By this rationale, shouldn't we all be chasing draws a lot more, since folding will cost us more money? Am I calculating the EV of folding wrong? Should we just ignore any equity in the pot we have, and say that folding costs us nothing? Obviously I'm not discovering some great new truth about poker, so would someone enlighten me why the EV of folding must be 0 for the rules about pot odds & draws to work out?

Potshots, insults, and helpful discussions are welcome, but I always prefer the latter.

I also made an excel sheet to do this stuff, cuz I'm bored & a nerd. If you wanna take a gander at it, you can find it here. But *caution*, Rapidshare sometimes has some banner ads that aren't the most work-friendly.

I could be wrong... but isn't your EV for folding $0, not -$10? The $10 you put in the pot earlier isn't yours any more.

EV of folding is zero.

In regards to missing out on that %10 equity, it just means that we don't make that $10 of the pot because we don't call. We don't lose $10 more by folding.

Ayup.

My inderstanding is that money already contributed to the pot should be considered as dead money.

When calculating the EV, you are deciding whether or not there is value in continuing to play the hand. As such, betting / calling raising will cost money, but folding is free. The EV of folding is therefore nil.

That said, there may be justification for making a 'crying call' in some circumstances.

Boltneck

This raises another question. When, if ever, are you "pot committed"?

When the chance of having the best hand, or chance of drawing to the best hand, is greater than or equal to the pot odds you'd receive if you had to go all-in at this point.

Obviously folding when you have a large amount of equity in the pot costs us something. Folding KK when there's an K on the flop will usually be a value losing play. You think there'd be some way to quantify how much value we're losing by folding. Maybe I just can't get used to the idea of folding always being a 0 EV play...

Obviously you 'lose' value in your general example because you are making an EV-neutral play instead of a +EV play. You 'lose' the difference by folding.

Folding is 0 ev, just the same as calling with exact odds. I'm assuming you already agree that calling with exact odds is 0 ev. For example:

you're a 2:1 dog, pot is 100, opponent bets 100. Before the flop your equity was 33.3333. After that bet if you call your equity is 100. Follow so far?

OK, so on the flop decision, you paid exactly 100 chips to get exactly 100 chips out at the end (long run here). You broke even. Now let's say you fold. You pay exactly 0 chips to get exactly 0 chips in the end. The exact same thing as before. If someone is giving you a free $100, and you refuse it, that's a 0 ev play, not a -ev play. The net is -, because accepting would have been +$100, but declining it was 0.

Therefore folding is a 0 ev play.