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Where is the flaw in this logic?
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? |
Re: Where is the flaw in this logic?
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. |
Re: Where is the flaw in this logic?
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 |
Re: Where is the flaw in this logic?
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. |
Re: Where is the flaw in this logic?
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. |
Re: Where is the flaw in this logic?
[ 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. |
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