Single/full system betting and winnings calculating.

System	3	4	5	6	7	8	9	10	11	12	13
System 2 out of ...	3	6	10	15	21	28	36	45	55	66	78
System 3 out of ...		4	10	20	35	56	84	120	165	220	286
System 4 out of ...			5	15	35	70	126	210	330	495	715
System 5 out of ...				6	21	56	126	252	462	792	1287
System 6 out of ...					7	28	84	210	462	924	1716
System 7 out of ...						8	36	120	330	792	1716
System 8 out of ...							9	45	165	495	1287
System 9 out of ...								10	55	220	715
System 10 out of ...									11	66	286
System 11 out of ...										12	78
System 12 out of ...											13

Let`s assume, that I wanted to play a system 2 out of 4 (2/4) with the stake €2. In this case I will have to pay €12 for the whole ticket. Look on the table: for the system 2/4 there are 6 combinations, this is also a price of the betting slip where the stake for a single bet is €1. Because I decided to play for €2, I have to multiply this value by 2; 2x6=12. I am choosing four matches, from which at least two have to be typed correctly. Let`s assume that the matches I`ve chosen are:
A with an odd 1.50
B with an odd 2.00
C with an odd 1.85
C with an odd 2.15
I am betting on each of the matches for the host`s victory, so I put 1 on the ticket. I have given quite low odds examples, but if somebody will decide to play using any of the betting systems, then it`s the best to choose draws or higher odds taking the higher margin of error, so for e.g. the possibility of making three mistakes. While playing system betting the highest possible winning is lower than the winning we could make while betting all of the games on the single tickets, but with the system betting I can make two mistakes and even, if I won`t be on plus, then at least I will get some part of the money from my stake back
This system and the above odds will give the following results - the highest winning will be €21, because:
A x B = 1.50 x 2.00 = 3
A x C = 1.50 x 1.85 = 2.80
A x D = 1.50 x 2.15 = 3.20
B x C = 2.00 x 1.85 = 3.70
B x D = 2.00 x 2.15 = 4.30
C x D = 1.85 x 2.15 = 4.00
We add all of the values together getting a result of €21. This is the winning`s amount if I will bet on all of the events correctly. But if I will make a mistake and lose on the A match, then I have to cross out all of the combinations containing the A match and adding together all the values I was left with. In this case the winning would be €12. So I wouldn`t be successful, but also I wouldn`t be a loser.

If I would decide to play for e.g. the 3/5 system, then I would choose 5 matches and at least 3 of them would need to by typed correctly. For the ticket, with the stake`s amount €1, I would pay €10, with the stake €2 - €20, with the stake €3 - €30 and so on. To count the winning I would need to write out 10 combinations with 3 events each. If I would like to count the winning for let`s say system 8 out of 11, then I would need to write out 165 combinations, each having 8 events. Although with such big system I wouldn`t like to count manually my winning.
We can also connect few systems on one ticket, for e.g. 2/4 + 3/5 + 3/4, then with the stake €1 we would have to pay for this ticket €240. We multiplay by each other the number of combinations from each of the systems, so in this case it will be 6 x 10 x 4 = 240. If we would like to bet a stake of €2, then the ticket would cost €480 and so on.
The rules of the game are the same, but instead of betting on three different tickets and each of the systems separately, we are doing it on only one ticket. All together on the ticket, there need to be 13 events. From the A system (2/4) we need to bet correctly at least two matches, from the B system (3/5) at least three matches and from the C system (3/4) at least three matches. Even for e.g., if we will bet correctly full set of the events in systems A and B, but in system C we will make three mistakes, then the whole ticket is treated as a loss.