1. ## Permutations

Ok i need someone to help me figure this out since math isn't my strong point

I do weekly sports bets on football and thought of a perm idea

Now i want to pick 12 games and in each game there are 3 outcomes: win, lose and draw

Obviously you cannot select more than one outcome for each match per bet such as say Chelsea to win and Chelsea to draw in same bet

How many possible combinations are their with 12 teams (3 outcomes) without duplicating?

2. ## Re: Permutations

Originally Posted by fruityfreddy
Ok i need someone to help me figure this out since math isn't my strong point
Now i want to pick 12 games and in each game there are 3 outcomes: win, lose and draw How many possible combinations are their with 12 teams (3 outcomes) without duplicating?
That phrase is very confusing. Because it is impossible.

Twelve games, two teams per game.
If we make a table with twelve rows, one row for each game, and two columns,
Any row, host team first column and guest team second column, can have one of three entries: $\displaystyle \boxed{W|L},~\boxed{L|W},\text{ or }\boxed{D|D}.$

That means that there are $\displaystyle 3^{12}$ possible tables.

However, it is impossible for any table not to contain duplicate rows.

3. ## Re: Permutations

Ok so we are betting on football(soccer) and we are selecting 12 random games where the odds are even money or better. If we bank a 12 timer at minimum odds we get 4096/1 minimum with the potential to reach around 1,000,000/1+. The games are tricky because there are no clear favorites which the odds reflect. However since we are having no emotional connection to the teams since we are just assigning them a number and selecting from random combinations it makes betting easier. Now as mentioned both teams are similar in strength with no clear favorite to win the game there is a higher probability of draws(ties). So with that in mind we want to cover every combination that includes 4 draws and 5 draws mixing the rest with homes and aways.

so for example 1=home 2= away 3 = draw we want to have something like this

3 3 3 3 1 1 1 1 2 2 2 2
3 3 1 2 1 2 1 2 2 1 3 3
2 1 3 1 3 2 1 3 2 2 1 3
1 3 2 1 2 1 2 2 2 3 3 3

etc etc

how many potential bets are we facing at $1 unit stake? we can bet upto 4096 times before we no longer break even or profit fro the minimum win but odds can extend to 1,000,000/1+. so depending on the combinations we can reduce our stake to include more bets. 4. ## Re: Permutations Originally Posted by Plato That phrase is very confusing. Because it is impossible. Twelve games, two teams per game. If we make a table with twelve rows, one row for each game, and two columns, Any row, host team first column and guest team second column, can have one of three entries:$\displaystyle \boxed{W|L},~\boxed{L|W},\text{ or }\boxed{D|D}.$That means that there are$\displaystyle 3^{12}$possible tables. However, it is impossible for any table not to contain duplicate rows. what i meant when i said duplicate you can't have same match twice in 1 bet i mentioned this because if you were to to number 12 teams and 3 outcomes you would arrive with 36 options but this wouldn't work since you can't back say chelsea to beat liverpool and chelsea to draw in the same bet so we have 12 matches with 3 possible outcomes to each match win lose or draw 5. ## Re: Permutations Originally Posted by fruityfreddy what i meant when i said duplicate you can't have same match twice in 1 bet Say there are only two games: Liverpool and Chelsea Manchester United and Arsenal. You bet on each match only once each. There are$\displaystyle 3^2=9$different you can place the two bets. Originally Posted by fruityfreddy i mentioned this because if you were to to number 12 teams and 3 outcomes you would arrive with 36 options but this wouldn't work since you can't back say chelsea to beat liverpool and chelsea to draw in the same bet so we have 12 matches with 3 possible outcomes to each match win lose or draw That wrong. Placing exactly twelve bets, one on each game, there are$\displaystyle 3^{12}$possible ways to do that. 6. ## Re: Permutations Ok we all confused now Lets try a different line Ok the combinations are gonna be crazy with 12 selections so lets reduce them somehow I pick 3 set results say home wins so now we are only looking at perming 9 games and we figure 4 of the 9 will be draws 1) home 2) home 3) home 4) draw 5) draw 6) draw 7) draw (8-12 random picks) so based on this if we were to write out the combinations on paper we would have something like this bets 1-3 are constant H= Home A= Away D= Draw X= Random(non draw) HHHDDDDXXXXX HHHDDDXDXXXX HHHDDDXXDXXX HHHDDDXXXDXX HHHDDDXXXXDX HHHDDDXXXXXD HHHDDXDDXXXX HHHDDXDXDXXX So on how many combos we looking at now? 7. ## Re: Permutations Originally Posted by fruityfreddy Ok we all confused now Lets try a different line Suppose there is just one game played. Liverpool and Chelsea in Liverpool. You can place exactly one bet. That can be done in three ways. That is what I thought you meant by the question. If not, please take the one game example and expand. 8. ## Re: Permutations Originally Posted by fruityfreddy Ok the combinations are gonna be crazy with 12 selections so lets reduce them somehow how many combos we looking at now? Just putting 4 D's and 5 X's in 9 picks, we have 9!/(4! * 5!). That's (9 * 8 * 7 * 6) / (4 * 3 * 2 * 1) which works out to 126. But each of those X's need to be replaced by an H or A. For each of those 126 combinations, there are 32 ways to pick the H's and A's. So the total number of combinations is 4032. That's way down from the 531441 we started with. 9. ## Re: Permutations Originally Posted by Plato Say there are only two games: Liverpool and Chelsea Manchester United and Arsenal. You bet on each match only once each. There are$\displaystyle 3^2=9\$ different you can place the two bets.
Bet#1: L wins or C wins: that's 2 ways
But you can't count L loses and C loses as 2 more ways; example:
L loses is already taken care of with C wins .... right?

I'm probably missing something...

10. ## Re: Permutations

Originally Posted by Wilmer
Bet#1: L wins or C wins: that's 2 ways
But you can't count L loses and C loses as 2 more ways; example:
L loses is already taken care of with C wins .... right?

I'm probably missing something...
"I'm probably missing something" What about the possible a DRAW?

11. ## Re: Permutations

Originally Posted by Plato
"I'm probably missing something" What about the possible a DRAW?
Sure, there is a possible draw...I was only commenting on L wins automatically means C loses,
so only one bet covers both.
On the draw, ONLY one bet also: L will draw with C; no need to bet C will draw with L !

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