Your post is so verbally spewy. First and foremost a decision can be -ev in comparison to another option. The knowns are:
a) villain has a lesser hand that can outdraw hero profitably (breakeven at the very least) if he has 4:1 odds
b) there is $300 in the pot.
c) a $100 bet gives villain 4:1 odds exactly
d) villain will FOLD if not given the correct odds
So the ev of your decision can be significantly -ev. Just jamming allin obviously denies the villain the correct odds of a call, so the ev of the decision to check is that you win $300 four times and villain wins $300 one time. That means in 5 hands you win $900 an expected return of $180 per hand. The same hand in which your decision to open shove (or for that matter to simply bet $101) would have had an expected return of $300 per hand, so yes the decision to check LOST you money.

Please read the OP before slamming other people that gave thought out answers to a hypothetical question.

-ev doesnt equal less +ev as i said, because less ev is still +ev in a vacuum. -ev would be negative "expected return per hand". 180$ per hand is still +ev even if there are more profitable options.

guess i should add quote to my post as i was answering not to OP, but to this
which is just not true.

Again you are not readin OP. It is a true statement. You are GUARANTEED $300 if you bet any amount $100 or greater. Simply betting $99 you LOSE money. We are not talking about any other hand or any other situational decision, we are talking about THIS decision to bet x or check where we will make $300 if x is $100 or more. The ev of the decision to bet less than $100 is negative.

i still stand on my own, the op asked if it would be a mistake to bet 99, yes it would be a mistake. But it will not be negative ev at any point. You dont lose money, you just win less.

How is it you don't understand that we are talking about the ev of the turn decision, not the ev of the entirety of the hand?

and where i'm not talking about turn decision only?
I just ask to not confuse -ev with less +ev.

Thank you rockin for backing me up. I'm new here even though I've been lurking around a while. I'm no expert and have much to learn and am eager to do so. Def not ******ed though

i've used the word ******ed too literally and apologize for that. w/e

liberally*

So then this is easy to solve. No, you would never give your opponent correct odds to call. Ever.

If doing so offers you no increase in expected value then all you're increasing is your variance, which is a bad thing if you're a winning player.

If your opponent would NEVER call with incorrect odds (and you're aware of their holdings) just bet $500,000 and win. More realistically though, your best bet is to get them to call with most money possible without folding. I.e. Can you bet $150 and get them to call? $140? $101?

If the answer is no and we hold steadfast to the original parameters than this question is strictly about variance.

the most basic issue with original question is that it's not very relevant because you will never know for sure if the villain is on a draw. if you insist on the assumptions of this question, then it's more than adequately been explain in MOP. /thread

I didn't read the whole thread, but OP asks a pretty important question, so here's my certified answer (certified by me, that is):

First of all, in the situation that OP describes: if you bet less than 100$ in this situation then you are losing money, compared to betting 100$. It's better for you if villain folds than if he calls a 99$ bet. But it's better for you if he calls a 130$ bet than if he folds to a 140$ bet, so you need to figure out a reasonable amount over 100$ that will maximize your value (if you know exactly the probability that villain calls each size of bet you can compute it by maximizing profit when price-setting just like in economics).

To see the above is correct, just assume both us and villain are playing with out hands exactly face up.

But in poker something more interesting happens. Like EmptyPromises wrote above, villain actually has a range of hands, some of which have less than 20%, some of which have more, and some of which actually have us beat right now. To keep the discussion close to the original question, let's assume that we know for a fact that villain is on a draw with 20% equity, but we don't know which draw. Suppose for simplicity that we know villain is on exactly one of several draws, all equally likely, but we don't know which of them he is on, and suppose that the outs for these draws are all disjoint.

What happens now? Well, it actually depends on stack sizes. Suppose we are infinitely deep. Then it's not hard to prove that if villain has our hand pegged down and knows for a fact that we have none of these draws, then he has 40% "effective equity", meaning the situation is as profitable for him as if his equity was actually 40%. In this case, we'd actually want to bet 600$ into 300$, to deny him the "effective odds" (more like game-theoretic equity or such) from continuing with his draw(s). If his draws had more than 25% equity, the best move for us would actually be to check/fold! This is not a theoretical point: this comes up all the time. There are two ways to ameliorate it: one is to not look at pot odds but "effective odds" and thus to bet much bigger than we think denies villain his odds. But a much more important way is to make sure our range is not capped on the river, so that sometimes we have one of the draws he'll be repping. This is, for example, why it's important to bet our draws OTF even if villains are calling too much: to put them in our betting range, together with our made hands so as to not be exploitable on later streets.

bet an amount thats incorrect for opponent to call but does not scare him off