Can Someone with skills help me...

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?

If anyone wants me to explain something that is not clear above...I can answer any questions.

Oh...and almost forgot.  I was just figuring to add 8 additional parlays for each additional game/team parlay.  So I came out with a 6 team parlay (covering all possible combinations) to be 36 different parlays.  Can someone prove this wrong?  Thanks for any help!

not possible.....wow

Dude.

If you have 3 games with 2 possible outcomes u just multiply the number of outcomes to the power of the number of games.

So

2 outcomes 3 games = 2x2x2 = 8 possible 3 team parlays

So the odds on a 3 team parlay IF they were all even money would be 7-1

With juice a book would pay nearer 6-1

Hope this helps.

my brain started hurting right after the second sentence.

Don't bother to try and figure out a way to beat vegas they are a little bit smarter than U and I

UMM. What?

just send your money to me, i will put it to good use

Unless you only took underdogs, and all those underdogs hit, there's practically no way to turn a profit over the long haul (unless you have really, really deep pockets).

It's like Powerball or any other lottery. There's a limited number of  potential combinations, but how on earth would you ever bankroll that?

Ok so here is how I was looking at it...

If you bet on a 6 or 7 team parlay...and covered every possible combination...and bet a small amount on each...would that not turn a bit to profit? 

But in order to see if that would be realistic I need to know how many combinations would have to be used.  For example:  would one be able to cover every combination of a 6 team parlay...like I said above...with 36 different parlay combinations.  If so, a dollar bet could be made on all 36 combinations...$36...and the payoff for a 6 team parlay would be greater than $36.  So can it be done...that's the question?  How many different combinations would have to be used?  If it takes 63 or whatever, then that would mean a certain loss.  But does it only take 36? 

I know it's confusing and I will have to sit down and work with it on paper...but I was just trying to see if anyone here had any ideas about it or knew if it would/would not work.

We have the sports lottery here in Canada where you can bet on 3-6 games per ticket and, trust me, there's only one way to turn a profit (and a really small profit at that): all the dogs (or at least 4 really big dogs) have to cover. Otherwise you're just breaking even or working at a loss, and what's the point of that?

Trust me, Vegas already has these scenarios covered.

The problem is that none of you have skills.


Skills

"I know it's confusing and I will have to sit down and work with it on paper"


Just wipe your ass with it instead... at least something will get done.

Just kidding, but to be real for a second can you do me a favor and scan a copy of the paper when you're done, and post it online?

 Very nice comments from some of you...thanks so much.

For those of you who are saying NO IT WON'T WORK...would you care to explain mathematically why it won't.  I'm not saying I am certain it is possible to cover every combination of a parlay...but if it's not...then prove me wrong with some numbers I can understand...don't just say it's not possible.

Do half of you even understand what exactly it is I was actaully saying? 

I have thought about it and it seems like there would eventually be too many different combinations to cover them all without taking a sure loss.  But I have not actually 100% determined that it's not possible...

You're very welcome.  How did it feel wiping your ass with printing paper?

There's a tiny little idea called a house edge.  That is your problem.  You can play every combination and guarantee yourself one (and only one) winning parlay, but it won't cover the losses of the other tickets.

Are you being serious?  Come on dude, you obviously have the ability to think and be creative, more than a lot of people.  Apply it to something better than this.

1 X (house edge) = less than 1.

I put it into an equation for you

If you doodled any math on paper you really should post it.

I understand that the house has the edge.  But I still would like to know how many different combinations it would take???

If one bet a dollar ($1) on a 6 team parlay...it should pay out about $44-46 if it is correct.  So are you saying that there are more than let's say 44 different combinations in order to cover every possible scenerio of a 6-team parlay?  Well how many does it take...if you have any ideas...

If you buy points on the games your edge goes up even more - think about it.  You buy points on every one of those games and you have a chance to hit more than 1 parlay - hell you could win every one!  I think you are on to something.

I understand what your saying. There are 64 different possibilities for a 6 team parlay.  Like everyone else said, you're not gonna outsmart vegas.

vvryan, how did you come up with 64?  Just curious because I should know this but I tried 2 different methods and came up with 28 (which I knew was wrong because it is less than the 46 you could win) and 72.

And no there are no other x number of team parlays that can cover themselves. Here's an article that shows actual odds of several x number of team parlays, even has house edge.  https://www.hotstreaks.com/articles/35.asp Hope this helps. 

Ok so all sites are different...one site said a $40 payout for a $1 bet on a 6 team parlay. 

And yes, I would like to know how you are coming up with the numbers.  I don't really think it will work after thinking about it, but I want to know why it won't. lol

An even easier way to look at it is do figure on the over/unders.  I started with 2 teams and figured out that by doing 4 different parlays you could cover all possoble outcomes and win one parlay.  But that would result in sure loss.  So to get the payout higher, I just added more teams.  For each additional team I added to the 6-team parlay I counted 8 more possible parlay combinations...not sure that it is right though.  I simply added a let's say under bet, and an over bet for each additional team.  Which is to say, 4 unders and 4 overs for each team added to the first 2 teams I started with.

I know something has to be wrong with that way of doing it...but then what is it?  And how many would it take to cover them all?  How do you reach your figures?