## Poll

 I've always wondered this as well No votes (0%) Only the Wizard will know! 1 vote (25%) How do I even bet on horses (I really don't know) 1 vote (25%) ZK will think the restaurant is cheating him 2 votes (50%) Axel has action on my sports bets No votes (0%) Something about a a bigot No votes (0%)

4 members have voted

Joined: Jun 13, 2014
• Posts: 2233
June 1st, 2017 at 2:48:17 PM permalink
By the way, since both L.R. Rooster & I have linked to the same web-based converter, as the saying goes: "there's (also) an app for that" at least for Droids...

...and there are probably at least several more, but that just happens to be one I have. Don't know that'll quite get you all the way to what you're after in how you're trying to think of your odds comparisons, but if you're doing this where an easily portable mobile calculation is better, including finding the break-even points (aka "implied odds" or what's listed above as "Win %") then there you are.

I don't know for sure that I'm on the same page with what you're thinking, but I agree with the general way of reasoning about it from Mr. Rooster in the discussion about how to think of comparative value in odds lines. That's how I think of it. "This line means I have to cash this ticket x% of the time, that one means y% is the break even point, I have reasons to think z% is the actual fair-value probability, and the difference, or the estimate of value for me, is therefore #1% - #2%."
Last edited by: DrawingDead on Jun 1, 2017
"I'm against stuff like crack and math" --AxelWolf
lilredrooster
Joined: May 8, 2015
• Posts: 2638
June 1st, 2017 at 2:59:47 PM permalink
Quote: Romes

I'm curious because I figured the difference in odds would be worth a lot more, just figured with an educated guess form my readings they were worth a lot more).

you know what. i think you're right and that my first answer was wrong. i'm pretty sure i'm right this time.

so the -220 gives us implied odds of 68.75% lets say 69%.

so, if you bet \$1 one hundred times you would win 69 times. 69*1.2 (the book's mistaken line) which equals \$82.80. which is what you would get paid for your wins.

you would lose 31 times which means you would net \$51.80.

you would have bet a total of \$100 and subtracting the takeout of about \$4.55 you would net about \$47.25.

i'm pretty sure that this is correct. sorry about the wrong info earlier. surely would be good to double check with Mike.
"but I don't care too much for money..........money can't buy me love".............. the Beatles
TomG
Joined: Sep 26, 2010
• Posts: 1965
Thanks for this post from:
June 1st, 2017 at 3:16:29 PM permalink
Quote: TomG

Convert true odds to a percentage, then it becomes easy:

first example: 66% probability, getting even money, 33% advantage
second example: 66% probability, laying three to two, 17% advantage
third example: 69% probability, getting five to four, 55% advantage

I gave you the answers already, how are you all still getting this so wrong?

If the true odds are -220 or 68.75% and the payout is \$1.20 for every \$1 risked, the edge is 51.25%

For a \$100 bet you have a 68.75% chance of winning \$120, for a value of +\$82.50. You also have a 31.25% chance of losing \$100, for a value of -\$31.25. Which equals an overall value of \$51.25.

Someone must have a much more eloquent algebraic equation (there are online calculators for this sort of thing), but this method comes much more easily to me, so it's what I use.

Miscalculating probabilities is a definite question.

The real question, though, (and my reason for the correct answer a second time after everyone ignored it) is how and where is Romes finding these bets with 50% edges? (past-posting is my guess)

Another thing necessary to do is to compare estimated edge to actual results
TomG
Joined: Sep 26, 2010
• Posts: 1965
June 1st, 2017 at 3:27:56 PM permalink
Quote: lilredrooster

you know what. i think you're right and that my first answer was wrong. i'm pretty sure i'm right this time.

so the -220 gives us implied odds of 68.75% lets say 69%.

so, if you bet \$1 one hundred times you would win 69 times. 69*1.2 (the book's mistaken line) which equals \$82.80. which is what you would get paid for your wins.

you would lose 31 times which means you would net \$51.80.

you would have bet a total of \$100 and subtracting the takeout of about \$4.55 you would net about \$47.25.

i'm pretty sure that this is correct. sorry about the wrong info earlier. surely would be good to double check with Mike.

Much better. Only error is that if -220 is in fact not the true odds, but true odds + a 4.55% commission, you need to take out the 4.55% first, then make your calculations:

0.6875 / 1.0455 = 65.75% probability. Then go from there.

The closer to 50%, the less that matters, but as you get bigger and bigger favorites, it could cause some pretty big errors in calculating an edge
RS
Joined: Feb 11, 2014
• Posts: 8623
June 1st, 2017 at 3:48:10 PM permalink
Bruh I already explained this in my first post...
TomG
Joined: Sep 26, 2010
• Posts: 1965
June 1st, 2017 at 3:55:46 PM permalink
Quote: RS

Bruh I already explained this in my first post...

You didn't answer the most important question: where is Romes finding these bets with 50% advantages?
Wizard
Joined: Oct 14, 2009
• Posts: 21208
June 1st, 2017 at 4:05:21 PM permalink
Quote: Romes

I'm familiar with sports betting and I know how to bet/etc, but I'm not sure on the verbiage of what to say the -200 is... is that "points"... I know that -200 means you have to bet \$200 to win \$100, but if you were to assign a word to that -200... -200 "what?" Let's pretend the word is "points." So Team A is the favorite at -200 "points."

First, let me say I oppose the American way of expressing odds and wish we go with the European model of showing what get back for one unit bet. I also oppose our units of measurement and favor going to the metric system. Fat chance of either thing happening in my lifetime.

To address you question, if forced, -200 means you have to bet \$200 to win \$100. If you take out the minus sign, which I interpret to not imply a negative number, but that you're laying odds, then it is how much you need to bet for a net win of \$100.

While we're complaining about notation, why is there no term for a simple bet against the spread? Please don't say it is a "straight bet." That can be any wager on a single outcome (as opposed to a parlay or teaser).

Quote:

EX
True Line is -200, I got my bet at -100... a difference of 100 "points"... Advantage = ?

True Line is -200, I got my bet at -150... a difference of 50 "points"... Advantage = ?

True Line is -220, I got my bet at +125... a difference of 145 "points"... Advantage = ?

At the end of the day I'm looking for a "points conversion" formula so that ANY bet based off the "points" difference you can calculate the percent advantage of the better bet.

The probability of winning is 2/3.

So, if you're getting even money your advantage is (2/3)*1 + (1/3)*-1 = +1/3.

At -150, you're winning 2/3 of a unit, so the advantage is (2/3)*(2/3) + (1/3)*-1 = +1/9

A +125, the advantage is (2/3)*1.25 + (1/3)*-1 = +1/2.

So, 100 points is 1/3 advantage
50 points is a 1/9 advantage
125 points is a 1/2 advantage

I think I'm going to have to use an "if" in a formula.

Let p be the points.

If the p <=100 then the advantage is ((2/3)*(100)-(1/3)*(200-p))/(200-p)

If p >=100 then the advantage is ((2/3)*p-(1/3)*100)/100

Personally, I don't memorize such formulas but figure it out my scratch.
It's not whether you win or lose; it's whether or not you had a good bet.
Romes
Joined: Jul 22, 2014
• Posts: 5458
June 2nd, 2017 at 7:08:17 AM permalink
Quote: TomG

You didn't answer the most important question: where is Romes finding these bets with 50% advantages?

Perhaps there's a local bookie (not friend, but big enough that he usually fleeces my friends and a bunch of other people) who takes up to moderate action and often has lines wrong on things he's not familiar with that I am (or can check online with a bunch of places to find the fair line)?

Also Tom, I definitely am keeping your answer and reviewing it with different scenarios. Just simply explaining to new people exactly what I'm looking for. I definitely did not brush it off in any means and am grateful for your response.

Quote: Wizard

...The probability of winning is 2/3.

So, if you're getting even money your advantage is (2/3)*1 + (1/3)*-1 = +1/3.

At -150, you're winning 2/3 of a unit, so the advantage is (2/3)*(2/3) + (1/3)*-1 = +1/9

A +125, the advantage is (2/3)*1.25 + (1/3)*-1 = +1/2.

So, 100 points is 1/3 advantage
50 points is a 1/9 advantage
125 points is a 1/2 advantage

I think I'm going to have to use an "if" in a formula.

Let p be the points.

If the p <=100 then the advantage is ((2/3)*(100)-(1/3)*(200-p))/(200-p)

If p >=100 then the advantage is ((2/3)*p-(1/3)*100)/100...

Thanks Mike! Is this a formula for telling what advantage the favorite/dog are as the line stands? In this thread I think I've confused people the most by that... I'm saying, if 99 casinos have the line at -220, and 1 casino is WAY OFF at +120 and I get action at +120 when the fair line is -220 (so a difference from -220 to +120 of 140 "points/odds"), how do I calculate that advantage? Past that, I'm looking for a formula to do this with any "points/odds" difference... So say I have a bookie that doesn't know anything about MMA and he offers me -120 on a fighter I look up and know for a fact is -170... that's a 50 "points/odds" difference, where as in the first example it was a 140 "points/odds" difference between the two favorite lines. I'd like to be able to get X "points/odds" difference and know what the advantage is.

You might have answered my question, but for some reason with sports the verbiage seems to confuse me a bit. I apologize for any ambiguity.
Playing it correctly means you've already won.
Romes
Joined: Jul 22, 2014
• Posts: 5458
June 2nd, 2017 at 7:22:05 AM permalink
Quote: TomG

...If the true odds are -220 or 68.75% and the payout is \$1.20 for every \$1 risked, the edge is 51.25%

For a \$100 bet you have a 68.75% (true odds of -220 to win) chance of winning \$120 (+120 payout on \$100), for a value of +\$82.50. You also have a 31.25% chance of losing \$100, for a value of -\$31.25. Which equals an overall value of \$51.25...

Bold added. I think this was my disconnect before. I saw the 68.75% and was like, no that's the -220 but I'm getting it at +120 but you do account for that with the win. Thanks Tom, now I just need to rip that online calculator apart to try to see what the formula is for calculating the true odds and I'll program my own application to simply plug in the two lines (true -220 and offered +120) and spit back a percentage advantage.

What I was looking for I think doesn't exist (or isn't possible) which is my confusion. I was looking for a formula to take the DIFFERENCE between the odds and get an advantage, but I think you "must" know the actual line. Thus, -300 and -250 is a 50 "point/odds" difference... but the advantage won't be the same as a -150 and EVEN MONEY 50 "points/odds" difference. Let's do the math to double check =P.

-300 gives a 75% chance of winning, but I got the line at -250 for a payback on \$100 of \$40. Thus:
75% chance of winning \$40 for an EV of \$30
25% chance of losing \$100 for an EV of -\$25

Overall value of \$5, on my \$100 bet for a total Player Advantage of 5%.

-150 gives a 66% chance of winning, but I got the line at EVEN MONEY for a payback on \$100 of \$100. Thus:
66% chance of winning \$100 for an EV of \$66
34% chance of losing \$100 for an EV of -\$34

Overall value of \$22, on my \$100 bet for a total Player Advantage of 22%.

So in both cases there was a 50 "points/odds" difference, but the advantage is VASTLY different... and it appears correct that the 50 "points/odds" difference makes more of an impact the closer the line is to even money.
Playing it correctly means you've already won.
Rigondeaux
Joined: Aug 18, 2014