Wondering if this is correct or am I missing something?

I'm using pin vig free closing line as the real probability of an event to calculate ev winnings for some past records of sportsbets w soft lines. I'm subtracting vig free prob from the soft prob to get the % (i.e. pin vf says 52% prob soft says 48% prob = 4%) then multiplying that by the sum bet to get ev (i.e. 1k x 0.04 = 40$ ev).

If pinny's closing line is vig-free then yes, that's correct at least in theory.

I didn't know their closing line was vig-free, I thought it was still 5 cents.

No it's not, some are higher/lower juice depending on the league too. Just have the formula set up to convert it to vig free when I plug lines for record keeping into excel. Just going over some of the bets I have tracked to confirm my suspicions I'm down a lot in ev (since pin closing should be pretty close to real after all, or at least the closest we can have).

Is the closing moneyline ever such that the favorite is laying exactly as much as the underdog is getting? For instance -120 favorite vs +120 dog?

Or a closing spread where the vigs are -105 and +105 (instead of both -105), or both +100?

If it's not like that, then it's not vig-free, and you have to factor the vig into your calculation.

I guess when people say vig-free line they mean that if they see the closing spread with vigs -110 and +100, the true vig-free line is that spread with -105 and +105. Or if they're both -105 then the true vig-free line would be that spread and both +100.

So actually I think this benefits you, and if anything you may have been underestimating your EV.

Yes of course, i.e. -110 and +100 is at -104.7619048//104.7619048, mostly doing it for some higher $ wagers recently and up to the beginning of the year where I lost a bunch while crushing the closing. Numbers look bleak but w.e. I guess it's mostly for peace of mind ala looking at allin ev when running bad, although this should be a much more conclusive number imo since variables like card distribution, bluffs caught/called, etc. don't play a part.

I'm confused on this point. Let's say we have a real probability, i.e. pin vig free closing, of 14.66 on a parlay, and I bet that same parlay at the line of 18.9.

This means 14.66 = 6.82%; 18.9 = 5.29% meaning a 0.0153 difference in prob.

Now doing it like I previously i.e. sum bet, let's say we have a $500 bet, it would mean a +$7.65 ev per bet ($500 * 0.0135). But this is wrong no? It should be sum bet + possible winnings ($9450 + $500 = 9950), so the actual ev is 9950 * 0.0135 = 134$ per bet.

Sorry, in my first post I ignored payout odds. Your edge isn't given by the probability difference unless it's a 1:1 payout. Calculate the EV for a $1 bet and that's your decimal edge.

For this example, if vig-free is 14.66 then that means your probability is 1/15.66 (if the 14.66 didn't include getting your wager back).

(1/15.66)18.9 - (14/15.66) = 245/783 or ~.3129

EV of a $500 bet is (245/783)500 = +156.45

First bracket is chance of hitting * your edge derived from the soft line no? But in the second I don’t get where the 14 is from, I thought it would be chance of not hitting but the two don’t add up.

I don’t see why doing the previous -($9450 + $500 = 9950), so the actual ev is 9950 * 0.0135 = 134$ per bet – is wrong (giving a different ev) could you explain why exactly?

EV is always P(win)*(amount won) - P(lose)*(amount lost)

EV is also edge*wager, but if we don't know edge directly, we can know EV directly so we can say edge = EV of a $1 bet

Ah sorry, 15.66-1 is 14.66 not 14 lol.

(1/15.66)*18.9 is chance of winning * net payout
(14.66/15.66) is chance of losing * amount you'd lose (which is 1)

So edge is 212/783 or .27075
and EV of $500 bet = 135.376756

I don't see the reasoning behind what you did, so I can't pinpoint what's wrong with it except to say that whatever it is, it's not an EV. An EV is just a weighted average; the dollar outcomes are weighted by the probabilities.

Yeah that's what I thought. Also what I thought I was doing was the same thing as EV is always P(win)*(amount won) - P(lose)*(amount lost) but done differently? But I guess it's wrong somwhere, just don't see where/why?

Also it was supposed to be 0.0153 instead of 0.0135.

I'm doing real probability of hitting (14.66 = 6.82%) - probability of the soft line (18.9 = 5.29%) which gives us 1.53%.

Now I take that 1.53% implied edge and multiply it by the total amount (sum bet + possible winnings) which is $9950 ($500+$9450) and this gives us 152.26.

6.82% is right if 14.66 includes getting your wager back. But if 14.66 would be your net profit, the probability is 1/15.66 = 6.3857%
Because (1/15.66)(14.66) - (14.66/15.66)(1) = 0
I'm saying that's not an implied edge unless you were talking about a 1:1 bet like ATS.

No because algebraically they're not the same.

You have: [P(win) - fair P] * (amount won + amount risked)
= P(win)*(amount won) + P(win)*(risk) - (fair P)(amt won) - (fair P)(risk)

Much different than P(win)*(amt won) - P(lose)*(risk)

Yes, 14.66 is the decimal line, so it has your wager back in it as well, it’s an actual line from my log.

So only in events that are presumed 50:50 + juice?

What about in the instance that the real prob is 1.29 soft 1.33, or 1.87/2.06? Usings P(win)*(amt won) - P(lose)*(risk) is giving me off numbers.

Actually I shouldn't have said "like ATS" since ATS isn't 1:1 payout. If you bet red/black in Roulette you get 1:1 payout and the house edge is equal to the chance of landing on green. In sportsbetting I guess there are no examples except maybe some teasers where you should be laying some odds but are instead getting even odds.

(1/1.29)*.33 - (.29/1.29) = +0.031
Or just (.33-.29) / 1.29
For the next one: (1.06-.87) / 1.87 = +0.1016

So in general: (actual won - fair win) * (actual probability)

Since the odds are decimal, remember to subtract 1 from "amount won" since getting your wager back doesn't count as money won.

Equivalently: (actual win - fair win) / (fair decimal odds)

Aha right, that's what I was doing wrong then since I was using 1.33 instead of .33

Also if you do (actual win - fair win) / (fair decimal odds) why is the answer different for the previous example where we got 135$ev?

(18.9 x 500) – (15.66 x 500) = 1620
1620 / 15.66 = 103.45

500 * (17.9-13.66) / 14.66 = 144.6

500 * [(1/14.66)17.9 - (13.66/14.66)] = 144.6

Got it thanks, did the same error again by not subtracting the 1.