Soccer betting combinations for accumulators I would like to bet on soccer games, on every possible combination. For example, I bet on $10$ different games, and each soccer game can go three ways: either a win, draw, or loss. 
How many combinations would I have to use in order to get a guaranteed win by betting $10$ matches with every combination possible?
 A: Each game has three possibilities. So after the first game there are three potential outcomes. After the second, game there are three times three. That is for each of the three outcomes of the previous game, there are three outcomes for the second. By induction, each new game increases the number of potential outcomes by a factor of three. So for ten games, it is $3^{10}$, which is about 59,000.
A: If you bet on every possibility, in proportion to the respective odds, the market is priced so that you can only get back about 70% of your original stake no matter what.
A: I will assume that you only bet to one pair of outcomes for each combination. Let's say you want to risk a quantity x of your capital in a strategy involving every possible double combination within a set of games.
The formula to calculate the number of possible combinations can be derived from geometry, for it is like finding the number of sides and diagonals of a polygon of n vertices (it also holds for the line (n=2)). Let be n the number of games of your set:
Number of different bets = n + (n(n-3))/2
See that for n=3, the number of bets is three, as the sides and diagonals (which has none) of a triangle. For n=4 it is easy to see that the formula yields the number of sides and diagonals of a square. Holds for any natural n>1. It is also important to note that a bet on two games B(G1,G2) equals to B(G2,G1), that's to say that the combination of Game 1 and Game 2 is the same as the combination Game 2 and Game1 (kinda obvious).
So we have 10 games and we would make 45 bets as the formula says. Then the fraction of capital which is risked, x, has to be divided by the number of bets. The capital risked on each bet is x/n, or x/45 in this case.
Now we need to think the return r of a bet, and the probability k of winning a bet. r=1.10 would mean that for every dollar wagered you would win $.10. Let's see how big must be r to break even if we only win one bet and lose 44:
Profit = (x/45)(r-1)-44(x/45) = (x/45)*(r-45)
That means that the mean return r of the bets you combine must be 45 to break even if you only guess one. Note that if you apply this strategy, the way to increase interest is increasing r (which also reduces our risk) and x (that is, the fraction of your capital dedicated to this strategy, which when increased increases our risk as well). Increasing the number of bets will reduce our risk but will also cut potential gains.
Let's say we bet on a league handball of nowhereabouts-in-east-europe. We see that it has a good percentage of draws, say 20%, and we want a lotta money. If the r of a simple bet to a draw was 9, then every double would have r=81. If we recall our strategy, we would bet on 10 games 45 combinations. If we win one, a single one of 45, our return would be
Profit = x*(81-45)/45 = 0.8x
That means we would gain an 80% interest on our invested capital. Had we invested x=100 now we would have 180 = x + profit, HAVING RISKED $100, for only guessing one and the rest wrong. I suggest that instead of having a set of games n=10, it would be wiser to use a lower number, like 6 or 7. By doing this you risk less (because it lets you bet a total smaller quantity), and I personally prefer operating with 20 doubles.
If I only won 1 bet among 20 with same r,
Total capital = x*(81-20)/20 = 3.05x.
Which is much bigger than the foregoing, but also less probable. Bet the minimum amount the house allows, like .25 if you can. 20 bets are $5. Risk 5, try to earn 10.25 of profit. If the probability of winning each combination is 0.05 or greater, you win.
The probability of not winning any combination is equal to the probability of failing to predict the outcome of 6 games (given n=20, the number of games should be 7, but we should have done exactly n=21) because we have to guess at least two. If we assume that we are random guessing, so there's 1/3 of succeeding, said likeliness is (2/3)^6, or 8.8%. Then
E(strategy) >= (.912*3.05*x-.088*x) = 2.69*x.
