(Betting) How many different full time score-combinations can it possibly be out of a 3 game-combination in football?

I'm betting on sports from time to time, mainly football (soccer) and I usually bet very small amounts of money on the "correct score". I usually make a combination of 3 games so I get the odds up. Of course since it's extremely hard to get 3 scores correct in the one and same combination I loose almost every time but it's fine since it's such a small amount of money. I bet like 0.10€ per combination.

But I have always wondered one thing and since my math skills are extremely limited I need help with this.

So here's the question:

How many different full time score-combinations can it possibly be out of a 3 game-combination in football? If you follow some of these rules...

Rules:

1. In this calculation you will not count any away team wins. Meaning the only scores possible are home team wins and ties.

2. There can only be a maximum of 1 tie score per combination.

3. The only scores that should be a part of this calculation are: 0-0, 1-0, 1-1, 2-0, 2-1, 2-2, 3-0, 3-1, 3-2, 4-0, 4-1.

Even with these rules I obviously understand that it will be a lot of different combinations, probably over 200 but I'd still like to know if anyone would like to help.

Example:

We have 3 games. Of course it's the same 3 games that we bet on in all these correct score - combinations.

(G1 = Game 1, G2 = Game 2, G3 = Game 3).

Combo 1. G1: 1-0 G2: 1-0 G3: 1-0

Combo 2. G1: 1-0 G2: 1-0 G3: 0-0

Combo 3. G1: 1-0 G2: 0-0 G3: 1-0

Combo 4. G1: 0-0 G2: 1-0 G3: 1-0

Combo 5. G1: 1-0 G2: 1-0 G3: 1-1

Combo 6. G1: 1-0 G2: 1-1 G3: 1-0

Combo 7. G1: 1-1 G2: 1-0 G3: 1-0

Combo 8. G1: 2-0 G2: 2-0 G3: 2-0

Combo 9. G1: 2-0 G2: 2-0 G3: 1-0

Combo10. G1: 2-0 G2: 1-0 G3: 1-0

Combo11. G1: 2-0 G2: 1-0 G3: 2-0

Combo12. G1: 1-0 G2: 1-0 G3: 2-0

Combo13. G1: 1-0 G2: 2-0 G3: 2-0

Combo14. G1: 2-0 G2: 2-0 G3: 0-0

Combo15. G1: 2-0 G2: 0-0 G3: 2-0

Combo16. G1: 0-0 G2: 2-0 G3: 2-0

Combo17. G1: 2-0 G2: 0-0 G3: 1-0

Combo18. G1: 2-0 G2: 1-0 G3: 0-0

Combo19. G1: 0-0 G2: 1-0 G3: 2-0

And so on...

As you can see I don't really have a good counting system for solving this and it gets pretty messy and not very easy to follow. But I hope some of you guys understand the question and what I'm looking for.

I want to know ALL different scores possible, in any order, for a 3-game combination following the rules and only using the scores mentioned in the rules.

I don't want to give away what I'm supposed to do with this calculation if someone can provide it but I promise, it's not what you probably think.

I would be very satisfied if someone just could give the exact number of possible score-combinations for 3 games, it's not necessary to demonstrate ALL the possible scores. Wouldn't mind if someone did that too though hehe... :)

Thanks for reading all of this and I hope I can get an answer on this! Take care

OK, given that there are $3$ possible tie outcomes, and $8$ non-tie outcomes, you have:

Number of games without any ties: $8*8*8= 512$

Number of games with $1$ tie: $3*8*8*3$ (two games no tie, and one game a tie, and that tie game can be the first, second, or third game) = $576$

Total: $1088$

And you're right, I'm not going to show all $1088$! :)

But if you were to do this, you should do it a bit more systematically than the way you do. And, doing it more systematically, should also reveal why the number is what it is.

First, let's do the no ties:

1. 1-0 1-0 1-0

2. 1-0 1-0 2-0

3. 1-0 1-0 2-1 ...

4. 1-0 1-0 4-1

Now change the second game to 2-0, and vary the third one:

1. 1-0 2-0 1-0

2. 1-0 2-0 2-0

3. 1-0 2-0 2-1

...

1. 1-0 2-0 4-1

change second one again, and do the next 8:

1. 1-0 2-1 1-0

2. 1-0 2-1 2-0

...

1. 1-0 2-1 4-1

... I think you get the pattern now, so after we have gone through all 8 games for the second game, we eventually end up with

...

1. 1-0 4-1 4-1 (64 is 8*8)

OK, so now change the first game, and start again with second and third:

1. 2-0 1-0 1-0

2. 2-0 1-0 2-0

...

so we get another 8*8:

1. 2-0 4-1 4-1

2. 2-1 1-0 1-0

Etc.

... until ...

1. 4-1 4-1 4-1

Now let's do the combination with 1 tie:

1. 0-0 1-0 1-0

...

1. 0-0 1-0 4-1

...

1. 0-0 4-1 4-1

2. 1-1 1-0 1-0

...

1. 1-1 4-1 4-1

2. 2-2 1-0 1-0

...

1. 2-2 4-1 4-1

OK, but second game could be a tie instead of the first, so:

1. 1-0 0-0 1-0

...

1. 4-1 0-0 4-1

2. 1-0 1-1 1-0

...

1. 4-1 2-2 4-1

and now the third game a tie ...

1. 1-0 1-0 0-0

...

1. 4-1 4-1 2-2