Where is the flaw in this logic?

[Pitbull3k]

OK....

Say you find 2 games on a given night that are both 5 point spreads. And say you do the following:

(we'll use $11 for this exercise)

Put:
$11 on underdogg 1 at +5
$11 on underdogg 2 at +5
$11 on a 2-team parlay with the moneylines on the 2 favorites.

Now (as I've figured out), the odds on a 2-team MoneyLine teaser on 2 favorites that are both at -5 is 2.12.

Based on the outcomes of these 2 games, 6 scenarios are generated.

(For arguments sake, I'm saying I found 2 games at +4.5, odds on the ML teaser may be slightly lower but I'll keep it at 2.12 since I already figured it out at that figure)

Scenario 1: Both favorites win by 5 or more (if this occurs I lose $9.68)

Scenario 2: 1 favorite wins by 5 or more and the other favorite wins by 1, 2, 3, or 4. (if this occurs I win $11.32).

Scenario 3: Both favorites win by 1, 2, 3, or 4 points. (Being the BEST scenario I win $32.32)

Scenario 4: Both underdogs win out right (if this occurs I win $9.00)

Scenario 5: 1 underdog wins outright and 1 favorite wins by 1, 2, 3, or 4. (this is the same as scenario 4 and I win $9.00).

Scenario 6: 1 favorite wins by 5 or more and 1 underdogg wins outright. (if this occurs I lose $12.00)

4 of the 6 scenarios have us winning.

Scenarios 1, 2, and 3 seem the most common and, over time, you should find enough instances in which a game or two happens that the favored team wins by 1, 2, 3, or 4.

And you can do this with spreads up to about 10 or so. I haven't figured out the payouts yet but I probably will.

Did I miss something obvious? I mean of course its still a gamble, but you should be able to incrementally win money with this strategy. no?

[BobJoeJim]

At a glance, I would say that the flaw in your logic is:
[ QUOTE ]
Scenarios 1, 2, and 3 seem the most common and, over time, you should find enough instances in which a game or two happens that the favored team wins by 1, 2, 3, or 4.


[/ QUOTE ]

"Seem the most common," and "you should find," are dangerous phrases, you'll need to calculate actual chances of these things happening before you can calculate whether the play is +EV or -EV.

This is an interesting approach, though, I'd like to see you do a more mathematically detailed analysis on it. It might be good.

[TheStation]

I think some deep math on the likelyhood of each scenario will show a flaw but I cant quite spot it yet. You definitely have me thinking.

[ QUOTE ]
Scenarios 1, 2, and 3 seem the most common

[/ QUOTE ]

Scenario 3 is NOT common, pulling off the squeeze play on both bets will not happen often, its a super small % of the time

That also goes into saying you may be over valueing Scenario 2 & 5 as it also implements the ever glorious squeeze

I think scenario 1 will happen the most and you dont have enough other bets to counteract its negatives

Anyway, let us know what you figure out

[DeucesUp]

Here's a rough estimate:

(assumptions 5pt fav wins 66%, covers 50%, therefore dog wins 34% and fav wins 1-4pts 16%

Favs cover: 25% * -9.68 = $-2.42
Favs win <5: 2.6% * 32.32 = $0.83
Dogs win: 11.6% * 9 = $1.04
1 fav >5, 1 fav<5 16% * 11.32 = $1.81
1 dog win, 1 fav<5 10.9% * 9 = $0.98
1 fav >5, 1 dog win 34% * 12 = -$4.08


EV: -$1.84


Let me know if I screwed up the math somewhere.

[PropPlayer]

I'm confused on this?
"Now (as I've figured out), the odds on a 2-team MoneyLine teaser on 2 favorites that are both at -5 is 2.12."

What is a Money line teaser?

I'm not sure on the math because what ML is used to calculate the parlay payouts? I think this was saying that you’re parlaying the money line of the 2 favorites, or taking a 4.5 pts teaser depending on which one has better odds?

If the math DeucesUp did is correct then I would suspect this slightly more of a -EV. 5pt favorites will win the game about 70% of the time as opposed to 66% but shouldn't matter a ton because if this is right its -EV either way.

[DCopper04]

[ QUOTE ]
Scenario 2: 1 favorite wins by 5 or more and the other favorite wins by 1, 2, 3, or 4. (if this occurs I win $11.32).

[/ QUOTE ]

I could be missing something here, but couldn't there be two possible outcomes here? Fav A wins by 5+, and Fav B wins by 1-4. OR Fav A wins by 1-4, and Fav B wins by 5+.

You may have realized this, I don't know, but it wasn't clear in your post. It seems to me that because there are two possible "sub-scenarios" within Scenario 2, then Scenario 2 should be twice as likely to occur as you have currently evaluated.

Same principle applies to Scenario 5; there are two possible occurences that would make that scenario work. Perhaps, then, it would be more accurate to say that 6 out of 8 scenarios would win you money.