Correlated Parlay Math Question

[OCKO]

Can anyone answer this academic question?

Consider this hypothetical NCAA football game:
Spread = 48.5, Total = 48.5

Assume you can bet this game as a correlated parlay paying 2.6.

What is the probability of the following results:
1) Dog and Under
2) Favorite and Over
3) Dog and Over
4) Favorite and Under (0%)

[HedgeHog]

All things being equal:

+U: 33.33%
-O: 33.33%
+O: 33.33%
-U: 0%

PS Bet the first 3 pars equally and you can't lose.

[Louisvillekid1]

Why, this is will never happen?

[HedgeHog]

That's why it's academic--or hypothetical. I've seen lines that are very close to their Totals. And if allowed, it's worth knowing how to parlay them if the opportunity presents itself.

[Louisvillekid1]

Well of course, but can you give me an example of one of these lines that are close to their totals? . . .

[OCKO]

They're not equally likely. I was thinking something like this:
+U: 44%
-O: 43%
+O: 3%

[HedgeHog]

Ocko: It has to add up to 100% and +O is way too low.

[HedgeHog]

Well of course, but can you give me an example of one of these lines that are close to their totals? . . .

USC was a 49 pt fav/ total 58.5 vs Idaho
USC was a 41 pt fav/ total 57.5 vs Stanford

Both times +/U collected, and no I won't tell which books were dumb enough to accept them.

[HedgeHog]

They're not equally likely. I was thinking something like this:
+U: 44%
-O: 43%
+O: 3%

More likely:

+U: 37.5%
-O: 37.5%
+O: 25%

[idontlikerocks]

i have seen in international hockey a team favored by 11 1/2 goals with an over under of 11 1/2 on another site! unfortunately i am not allowed to play one of those sites ......sigh i think it was some kind of womens hockey russia vs japan or something...

[TLD]

I would say either the side or the total has to be a bad line, because basically the lines taken together are saying not just that the underdog probably won’t score much, but that there is zero chance the underdog will score at all. In effect, the team totals for this game would be 48.5 for the favorite, and 0 for the underdog.

But if we make believe these lines can somehow be accurate (for instance, maybe the underdog has announced in advance that as some form of protest they will refuse to score in the game and just see how well they can play defense), and that therefore the side comes down to 50-50 and the total also is 50-50, here’s how I would approach it mathematically:

A1. Favorite = 50%
A2. Favorite = Favorite with Over + Favorite with Under
A3. Therefore, Favorite with Over + Favorite with Under = 50%

B1. Favorite with Over + Favorite with Under = 50%
B2. Favorite with Under is impossible.
B3. Therefore, Favorite with Over = 50%.

C1. Over = 50%
C2. Over = Favorite with Over + Underdog with Over
C3. Therefore, Favorite with Over + Underdog with Over = 50%

D1. Favorite with Over + Underdog with Over = 50%
D2. Favorite with Over = 50% (From B3)
D3. Therefore, Underdog with Over = 0%

So if both the side and total lines are set correctly (which they obviously aren’t), then I would say:

Favorite with Over = 50%
Favorite with Under = 0%
Underdog with Over = 0%
Underdog with Under = 50%

Now if you want to say, “Let’s not assume the lines are accurate. Let’s assume instead that the book screwed up, that the underdog does have some non-zero chance of scoring, and yet you’re still allowed to bet into these lines. What are the probabilities of the various outcomes now?” I don’t think there is enough information for it to be solvable. If you’re not assuming 50-50 for the side and 50-50 for the total, then whatever other numbers you’d plug in instead would presumably be plucked out of the air. About all I could say would be the more confidence you have that the underdog will score, the more Underdog with Over creeps up from 0% and the more Favorite with Over and Underdog with Under correspondingly slide down from 50%.

[Ganchrow]

TLD hits the nail squarely on the head here.

If we assume that both spread and total are separately accurate (i.e., that each of the over, under, fave, and dog has an exactly 50% probability of winning) then the only possibility for the aggregate outcomes would be:

p(dog+under) = 50%
p(fave+over) = 50%
p(dog+over) = 0%
p(fave+under) = 0%

This is because we know the fave and under can't simultaneously win and so the only way for the fave to win would be with the over. If the over wins exactly 50% of the time then the over and the dog can't win simultaneously either (as the 50% over prob is already "used up" with the fave). The only remaining possibility, under+dog must then account for the reaming 50%.

Again, just as TLD has said, this spread/total combo (assuming perfectly efficient and accurate markets -- i.e., an exactly 50% probability of any given outcome occurring) would make sense if and only if the underdog had, for whatever reason, precisely a 0% probability of scoring.

So if the underdog has 0% probability of scoring, then this implies that the spread = the total.

Similarly, if the spread = the total, this implies that the underdog has 0% probability of scoring.

[TLD]

I have nothing to add here, except to show off my new signature line.

[Data]

While I basically agree with what TLD is saying and Ganchrow is confirming I think they both are not strictly accurate as they both somewhat missing the context. I think that instead of saying "there is zero chance the underdog will score at all" ("precisely a 0% probability of scoring") the more accurate statement would be "there is CLOSE TO zero chance the underdog will score at all but it is not big enough to justify line adjustment".

[Ganchrow]

While I basically agree with what TLD is saying and Ganchrow is confirming I think they both are not strictly accurate as they both somewhat missing the context. I think that instead of saying "there is zero chance the underdog will score at all" ("precisely a 0% probability of scoring") the more accurate statement would be "there is CLOSE TO zero chance the underdog will score at all but it is not big enough to justify line adjustment".

As long as we assume that the spread and total are both unbiased estimators (meaning that in each case there's a 50% probability of either side winning) then it logically follows from first principles that the spread and total are equal if and only if the underdog has an exactly zero probability of scoring. Not just "close to" zero. Exactly zero.

Of course as the absolute bias of either the spread or the total increases (and in the real world we never expect the spread/total to perfectly bisect the outcome distribution) so too would the underdog's probability of scoring.

If the underdog scores in a game where the spread and total and equal then we can say ex post that either the total was too low or the spread was too high.

[Data]

As long as we assume that the spread and total are both unbiased estimators (meaning that in each case there's a 50% probability of either side winning)

That is an incorrect assumption giving the context.

Lets assume that by saying
spread = 48.5
OCKO meant
fav -48.5 -110, dog +48.5 -110

The line has to be ROUNDED, it is not 110.000000000000... but -110. It is rounded to the nearest integer, and therefore the line above does not necessitate assumption of 50% chance. For example, 50.01% for dog covering seems to be just fine.

[Ganchrow]

That is an incorrect assumption giving the context.

Lets assume that by saying
spread = 48.5
OCKO meant
fav -48.5 -110, dog +48.5 -110

The line has to be ROUNDED, it is not 110.000000000000... but -110. It is rounded to the nearest integer, and therefore the line above does not necessitate assumption of 50% chance. For example, 50.01% for dog covering seems to be just fine.

When I opened my first post in this thread by writing (emphasis added) "If we assume that both spread and total are separately accurate (i.e., that each of the over, under, fave, and dog has an exactly 50% probability of winning) then ... " I was making an exogenous operating assumption in order to illustrate what I considered the most salient point.

I'm not claiming the assumption always holds (nor am I even claiming it ever holds). But if (and only if) it does hold then the conclusion logically and directly must follow (exactly).

If you're making the technical point that due to pricing granularity in the sports betting markets a balanced, efficient market need not imply an exactly 50% probability per outcome then it's certainly well-taken. It's just not particularly germane to the above thought experiment, the solution to which is well-defined precisely because of that simplifying 50/50 assumption.

Nevertheless, if it really makes you happy I'll present B) below (after restating A from above):

1. If we assume that each of the over, under, fave, and dog has an exactly 50% probability of winning then:
   p(dog+under) = 50%
   p(fave+over) = 50%
   p(dog+over) = 0%
   p(fave+under) = 0%
2. If we assume that each of the over, under, fave, and dog has an approximately 50% probability of winning then:
   p(dog+under) ≈ 50%
   p(fave+over) ≈ 50%
   p(dog+over) ≈ 0%
   p(fave+under) = 0%

Happy?

[the_fredrik]

Let's make it more interesting. What about the line +40 total 50 - what are the four probabilities then?

[Ganchrow]

Let's make it more interesting. What about the line +40 total 50 - what are the four probabilities then?

Unlike in the limiting case above, that can't be answered solely from first principles.

[Data]

1. If we assume that each of the over, under, fave, and dog has an exactly 50% probability of winning then:
   p(dog+under) = 50%
   p(fave+over) = 50%
   p(dog+over) = 0%
   p(fave+under) = 0%
2. If we assume that each of the over, under, fave, and dog has an approximately 50% probability of winning then:
   p(dog+under) ≈ 50%
   p(fave+over) ≈ 50%
   p(dog+over) ≈ 0%
   p(fave+under) = 0%

Happy?

Well, almost. A and B are not equal that one may think while looking at the way you list those. While B is the solution/answer, A is just a single example covered by B and as such has significantly less value than B if at all.