Quick math question.

[Deleted]

I don't remember how this is calculated so i thought that someone here could help

In gambling, when you 3 choose a winner from 3 groups, it looks like this:

1x2
1x2
1x2

So 1 is first team, x is tied game and 2 is second team.

If i would bet every possible combination, is the formula 3 to the power of 3 to the power of 3, which would make it 7.6255975e+12 so you couldn't bet them all? Or how is it calculated?

[Obby]

theres 2 equations for stuff like this its either a combination or permiutation

one is right for this situation because 1x2 isnt the same as 12x 21x 2x1 x12 x21 where as the other one says they are the same its been so long since ive done those i dont even remember

i think theres 18 possible outcomes 6 per game 3 games.....but you cant chose (winner to loser) 2x1 so it would rule out the tie and loser so youd only be left with 9 choices 3 games 3 choices per game

[Deleted]

You have to extract the base root square of the radical of x. Multiply that by how many extractions you want to divide then apply pythagorean theorem to the % of the n number power of the peclacle.

Wtf OP, didn't you go to school?

[Dendrek]

The way you worded this question makes it a little confusing (for me at least). Let me see if I understand:

You have 3 games, and you're considering placing a bet on each game for the person you think will be the winner. And you want to know what the probability is that you will win all three bets -- that you'll pick the winning team all three times?

Actually, I know that what I just said is wrong because you mention betting every possible combination. So yeah, I don't understand what you're trying to do here.

[Deleted]

I don't remember how this is calculated so i thought that someone here could help

In gambling, when you 3 choose a winner from 3 groups, it looks like this:

1x2
1x2
1x2

So 1 is first team, x is tied game and 2 is second team.

If i would bet every possible combination, is the formula 3 to the power of 3 to the power of 3, which would make it 7.6255975e+12 so you couldn't bet them all? Or how is it calculated?

Not sure I understand you correctly here, but if you want to know all possible combinations of betting on 3 games, it's fairly easy:
Simply multiply the number of options of each game: 3 possible results in each of the 3 games means 3 * 3 * 3 = 27 possible results:

1-1-1, 1-1-x, 1-1-2, 1-x-1, 1-x-x, 1-x-2, 1-2-1, 1-2-x, 1-2-2,
x-1-1, x-1-x, x-1-2, x-x-1, x-x-x, x-x-2, x-2-1, x-2-x, x-2-2,
2-1-1, 2-1-x, 2-1-2, 2-x-1, 2-x-x, 2-x-2, 2-2-1, 2-2-x, 2-2-2,

(4 games would result in 3*3*3*3 or 3^^4)

[Obby]

Not sure I understand you correctly here, but if you want to know all possible combinations of betting on 3 games, it's fairly easy:
Simply multiply the number of options of each game: 3 possible results in each of the 3 games results in 3 * 3 * 3 = 27 possible results:

1-1-1, 1-1-x, 1-1-2, 1-x-1, 1-x-x, 1-x-2, 1-2-1, 1-2-x, 1-2-2,
x-1-1, x-1-x, x-1-2, x-x-1, x-x-x, x-x-2, x-2-1, x-2-x, x-2-2,
2-1-1, 2-1-x, 2-1-2, 2-x-1, 2-x-x, 2-x-2, 2-2-1, 2-2-x, 2-2-2,

(4 games would result in 3*3*3*3 or 3^^4)

ahhh ok but if its limited to set outcomes like team 1 wins tie or team 2 wins it doesnt matter about the loseing team(s) or the tie and if theres 3 out comes per game team 1 wins team 2 wins or a tie you cant have 1-1-1 or x-x-x or 2-2-2 kinda things since team one and 2 played team one didnt play by its self

[Deleted]

Well... 3 possibilities for the first set, 3 possibilities for the second set and 3 possibilities for the third set gives: 3x3x3=27 possible combinations.

At least that's how I recall it from my statistics class. Another way to find it would be to write out all the combinations. I don't think there can be much more than 27...

Edit: Dang it, beaten to it...

- - - Updated - - -

ahhh ok but if its limited to set outcomes like team 1 wins tie or team 2 wins it doesnt matter about the loseing team(s) or the tie and if theres 3 out comes per game team 1 wins team 2 wins or a tie you cant have 1-1-1 or x-x-x or 2-2-2 kinda things since team one and 2 played team one didnt play by its self

If you can't have 1-1-1 or x-x-x or 2-2-2, simply subtract those 3 results. If any team or tie can only occur once, then the equation becomes 3*2*1 = 6. This is because whatever the outcome of the first match, can't be the same as the second, etc.

[Deleted]

ahhh ok but if its limited to set outcomes like team 1 wins tie or team 2 wins it doesnt matter about the loseing team(s) or the tie and if theres 3 out comes per game team 1 wins team 2 wins or a tie you cant have 1-1-1 or x-x-x or 2-2-2 kinda things since team one and 2 played team one didnt play by its self

I'm not an expert when it comes to betting but afaik you get a list of games and you bet on the outcome, i.e. team1 wins, team2 wins, tied, e.g.

Man City - Everton 3:1 -> 1
W Bromwich - Arsenal 1:1 -> x
Norwich City - Chelsea 1:3 -> 2

In that case any combination is possible.

Then again the OP might need to clarify what's he's actually looking for.

[Obby]

I'm not an expert when it comes to betting but afaik you get a list of games and you bet on the outcome, i.e. team1 wins, team2 wins, tied, e.g.

Man City - Everton 3:1 -> 1
W Bromwich - Arsenal 1:1 -> x
Norwich City - Chelsea 1:3 -> 2

In that case any combination is possible.

Then again the OP might need to clarify what's he's actually looking for.

yea i knew what i was saying =p but its just not comeing out right

useing names makes this a bunch easier on me =p

man city wins, man city loses, or tie
w bromwich wins, w bromwich loses, or tie
norwich city wins, norwhich city loses, or tie

so theres 9 possible out comes =p

[Deleted]

3 different matches where you have to choose the outcome from every match. For example match 1: team 1 wins, match 2: it's tied, match 3: team 2 wins. The multiplier of your winnings is calculated by given odds of the betting office. For example: Team 1 winning first match, multiplier 2, second match is tied, multiplier 4, 3rd match: team 2 wins, multiplier 3. So if that is the outcome, you win your bet times 24.

If there is 27 different possible outcomes in my OP example, then in the multiplier might be somewhere between 6-300, depending of the multipliers given by the betting office beforehand. My reason for asking this is that in some cases it might be wise to bet all the possibilities so that you might get something back even if you don't get the biggest multiplier and if one of the not so probable ones matches, you win plenty. If there really is only 27 possibilies, then it is something worth trying.

Hopefully this clears my idea

[Deleted]

yea i knew what i was saying =p but its just not comeing out right

useing names makes this a bunch easier on me =p

man city wins, man city loses, or tie
w bromwich wins, w bromwich loses, or tie
norwich city wins, norwhich city loses, or tie

so theres 9 possible out comes =p

Not quite: you're simply listing the possible results for each single game in a list, however you need to *combine* the possible results of each game to get the number of total possible outcomes:

man city wins AND bromwich wins AND norwich wins is one result
man city wins AND bromwich wins AND norwich loses is another one
man city wins AND bromwich wins AND norwich ties is the third
...

Following that pattern you can list 27 possible outcomes.

[stormgust]

If there is 27 different possible outcomes in my OP example, then in the multiplier might be somewhere between 6-300, depending of the multipliers given by the betting office beforehand. My reason for asking this is that in some cases it might be wise to bet all the possibilities so that you might get something back even if you don't get the biggest multiplier and if one of the not so probable ones matches, you win plenty. If there really is only 27 possibilies, then it is something worth trying.

Hopefully this clears my idea

Don't try. All betting offices offer you martingales at best, usually supermartingales, i.e. on average you'll just loose money. Unless of course, you actually get a big win on the first few bets, though then you'd gotten more money out of your bets, if all you did was bet on the not so likely one, instead of all combinations.

[Firebert]

You only win 3 times out of the 27 possibilities, so your equal bets on all outcomes requires your specific result to have a huge ratio (like 100:1) in order to make any money, and in that case you may as well have just put money on those outcomes anyway.

[Dendrek]

You only win 3 times out of the 27 possibilities, so your equal bets on all outcomes requires your specific result to have a huge ratio (like 100:1) in order to make any money, and in that case you may as well have just put money on those outcomes anyway.

This may not even be the case. Each outcome has a specific weight (a likelihood of happening that is different than the other outcomes), while the multipliers the betting offices offers are essentially less than the inverse of these weights. In other words, betting an equal amount on all possible outcomes is very likely to cause a player to lose money essentially every time.