After reading the article, I basically found out the same thing I already knew--a 2 team parlay has only 4 possible combinations--one of them will be right. Then it talked alot about odds and how they always make a profit. But the example used only a 2 or 3 team parlay where the house will most certainly come out ahead.
But, as you add more teams to the parlay...the win amount goes up...and if you add over 6...it really starts to move. So as the payouts go up, you have the ability to bet more and more possible parlay combinations (just like when we had 2 or 3, but were taking a loss), with higher payouts...my question is...can all the different parlay combinations be bet using let's say $1...and there be a point where the player would be able to profit...
You telling me there are 128 different combinations for a 7-team parlay?
On a side note: I checked and...a $1 bet on a 12-team parlay would pay $1800.00...too bad I don't know the point total for every NFL game this week. lol
NVM...I finally got it. lol 
I was confusing myself so much. Guess it's not really ever possible guarantee that you get them all right, and at the same time make a profit.
However, this could be used on parlays where you felt really confident about 2 or 3 picks and you wanted to throw in other picks you were not sure about. Then just pick the same 2 or 3 in the all parlays (and hope you were right) and then use different combinations for the other picks you were not as sure on. Maybe that would work. But guess there could always be upsets that would change everything.
I get it now! 
You telling me there are 128 different combinations for a 7-team parlay?
On a side note: I checked and...a $1 bet on a 12-team parlay would pay $1800.00...too bad I don't know the point total for every NFL game this week. lol
My brain is hurting trying to figure this out...so if you can help and explain that would be cool.
Ok so this is about parlays (betting on multiple picks and they all must win or the parlay loses...I'm sure you all knew that).
Ok so let's say I wanted to bet on multiple parlays so that I could hopefully get one right. Is there a way that anyone can think of in that you could bet on multiple parlays and cover every possible outcome from the different games you picked? I have tried to put this on paper, but my head hurts after trying to think of it...it's very confusing...for those mental ones amoung us...continue on...
Ok so for an example:
Let's start with a 2 team parlay using games. Let's say it was Dallas and Oakland; and Green Bay and Detroit on Thanksgiving. Now let's try to cover every possible outcome by betting parlays on those games. (I will use moneyline to make it easier).
Ok so I figure that, to cover every possible outcome and guarantee a win, one would need to bet on 4 different parlays to cover the 2 team parlay.
Let's say we pick Dallas and Green Bay to win in one parlay. Then the 2nd parlay would have to be Oak and Detroit in case of an upset. Now let's say Dal wins and Detroit wins--parlay 3. Then that leaves parlay 4 with Green Bay and Oak winning. So by betting on 4 different parlays, each possible winning combination has been covered. We will end up losing 3 of them, and winning 1.
Now this is where it gets tricky. Let's say we want to add on more games. Let's say up to a 6 team parlay. Let's add on the other Thanksgiving Day game (NYG vs. Den). Ok so I figured that to make the 2 team parlay that we have correct above, into a 3 team parlay...all we would have to do is add a NYG win, and a Den win to each of the combinations above. This would require that we now use 12 parlays to cover all combinations (if I have it right). Because we have 4 parlays above and now we are making a new parlay for each different possible outcome.
So does anyone know if it is possible to cover all possible outcomes of a 6 team parlay? And how many parlays would it take to have a correct combination in one of them?
@After_I_Win
For 5 game parlay there’s 36 combos 1$ each will have a 29$ loss with a 85$ win witch will leave your profits at around the 50$ area the point is to play 5 games that all have a underdog and atleast one team with a hit underdog that you think will hit put those five teams at 1$ each
So does anyone know if it is possible to cover all possible outcomes of a 6 team parlay? And how many parlays would it take to have a correct combination in one of them?
Here is the math.
For 110 110
2 teams 2 squared 4.
This bet loses. Because of the juice.
So 3
2 cubed 2x2x2 one in 8
4 teams
2x2x2x2 2 to the 4th 1 in 16
2x2x2x2x2 2 to the 5th power 1/32
2x2x2x2x2x2 2 to the 6th 1 /64
Ect.....
If the odds are offset say horses. And a 1/8 wins the first race on a 6 race ticket as example implied odds and actual 50/50 greatly exaggerated to add value in bets....
Established in 1995,
Hooraaaay skills! 
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