My advice: Never bet on more than 1 game!
Why not? Quite simply, in a game there are 3 possibilities, if you place a bet, you have the victory (regardless of the odds, purely mathematically) to 1/3 in the bag, but the risk is still relatively high.
How to reduce the risk with my strategy looks like this: Bet on matches between teams of almost equal strength (see betting odds), then bet on 2 match outcomes – usually tip 1 (home advantage!) and tip X (of course you are also free to bet on tip 2, just as you like).
So you can either play it safe or pay attention to the odds ratio to get a higher profit
So you give 2 tips for 1 game (2/3 chance of winning). e.g. the match Slovenia vs. Norway
you bet on the following tips:
Tip 2: odds 2.7
Tip X: odds 3
Calculate the ratio between the two odds: 3/2.7 = 1.111 (i.e. you have to bet more for the weaker odds) + 1 = 2.111 (the ratio between the two picks should be as close as possible to achieve the highest possible return, in this case it is excellent).
For example, you bet a total of 100 euros So you divide as follows: 100/2,111= 47,37 EUR on tip X For tip 2 you logically have to bet a little more money, as the odds are lower.
The following scheme: For tip X you bet like this: 100/2,111(odds ratio)*3(betting odds) = 142,11 -☻ minus the stake you can expect so with result X: 42,11 EUR profit
For tip 2 you have to bet more, as said, the following scheme for the bet calculation: 100/2,111 = 47,37*1,111 = 52,63 EUR on tip 2 -☻ minus the bet you can expect therefore also with result 2 42,11 EUR profit
Total stake = stake 1 + stake 2 = 47.37 + 52.63 = 100 EUR
When calculating the odds ratio proceed as follows: higher odds / smaller odds = stake for the smaller odds in this example: 3/2.7 = 1.111 in total therefore results in 1 + 1.111 = 2.111 (calculation basis for dividing the stake)
2. Why should you never bet on several games (combined bets, accumulator bets)?
Let’s assume you want to use this method on 5 games (on one match day).
In one match you have 2 out of 3 possibilities secured. In 5 matches there are already: 3*3*3*3*3 or in short 3 to the power of 5 possibilities that is a total of 243 possibilities (there are per match: tip 1, tip X and tip 2).
But you have no guarantee of 2/3, because: You have 2 possibilities secured per game, so: 2*2*2*2 or in short 2^5 = 32 possibilities are secured, from this follows, your chance of winning is 32/243, that is only 13.17 percent more!!
Combining 5 games might be a bit much, but also the example for 2 games, to illustrate the deterioration of the chance of winning with more than one game shows:
You have 2 out of 3 possibilities hedged per game as I said with 2 games there are 3*3 or short 3^2 possibilities = 9 possibilities.
Secured: 2*2 or short 2^2 = 4 possibilities are secured
This gives a chance of winning of 4/9 = 44.444 percent (gives a lower probability of winning of: 66.666 (with only 1 game) – 44.444 (with 2 games) = 22.222 percent)
A tip to finish: Basically 2/3 are a good hedge, but bet only with the capital that you can bear in case of loss
Good luck to all bettors, and if you still have questionsfeel free to contact me!
René – U2_4U@gmx.at
Appendix for clarification: (Same facts as above, but may help to increase understanding)
W2: Amount for the highest quota
W1: Amount for the lowest quota
W2 < W1
Q2: highest quota
Q1: lowest quote
V: odds ratio
V = Q2/Q1
V: is the relatively increased amount for W1
B: ratio value for the use of W2
B = V+1; 1 corresponds to B
B: Total input/(V+B)
W2 = total input/B
W1 = (total input/B)*V
V+B (ratio value for the use of W1)
B must be lower than Q2
B ☺ Q2
damits understandable:
Slovenia Norway match, betting is on tip 2 and tip X
Tip2: Odds 3 (Q2)
TipX: odds 2.7 (Q1)
Bet: 100 Euro
V = 3/2,7 = 1,111
B = 1,111 + 1 = 2,111
W2 = 100/2,111 = 47,37
W1 = (100/2,111)*1,111 = 52.63
now B must be lower than Q2
B ☺ Q2; condition fulfilled
This is how the profit is now calculated:
Q2/B*100 – total stake
in this case 42,11 Euro
