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Calculate simple slots • Absender: lola, 16.05.2019 23:15

Calculate simple slots PinUp

Slots for me personally are the only gambling entertainment that never gets boring. This is due to the huge variety of this game, in which it is simply impossible to find something boring. Of course, to maintain the excitement you need to win regularly. In this article, I will tell you by examples how simple mathematical calculations can be used to calculate the possibility of winning in simple slots.

I must say that for the calculations you need to take it simple slots, in which there are no bonus games, wild and other things, as in complex slots calculations are very difficult and often paid; or to build an emulator requires knowledge of programming languages. Everything else, for the correct calculation you need to know the layout of the slot, in some casinos it is freely available. Alternatively, the slot layout must be bounded, as in the example below.

Example # 1: a game slot, Craps, casino Aplay

Calculate simple slots

Simple slot with three regular six-sided dice. The win depends on the amount of points or the combination.


1 to 1 - any pair;
2 to 1 - if the sum of the three dice is seven. It is important to take into account that a pair of units plus five, a pair of triples plus one and a pair of twos plus three are considered as seven points, not as a pair;
4 and 1 - one plus two plus three;
10 to 1 - if all three numbers are the same;
Mini-jackpot. For the convenience of calculations, we take it for a standard payment of 2 to 1 when a four plus five plus six.
The number of possible variants of bone loss on each line is 6*6*6=216. In this example, the method of calculation will take too much. It is necessary to calculate how many pairs, sums and so on will be in these 216 combinations. First, let's count the pairs. For ease of understanding, we introduce symbols: x — a repeating number, Y — a random number from 1 to 6, not giving a set or number 7 in total. There are only 3 variants of falling out of the pair. This is x-x-Y; Y-x-x and x-Y-x. Substitute in the first case the number 1 instead of x. it Turns Out 1-1-Y=2,3,4,6 (5-ku throw, as with it in the sum of the cubes is the number 7). Multiply by 3 possible combinations. Get 12 different pairs with units. We also deal with numbers 2 and 3. It turns out already 36 options. Take the number 4 instead of x, and Y will be equal to 1,2,3,4,5, or 6. Again multiply by 3 and get 15 possible combinations. We do the same with numbers 5 and 6. Now let's sum up all the combinations. 36 to 45 added and the resulting 81 combination of the pairs.

Now let's go through all the options in which the sum of the numbers on the dice is 7. It turns out 3 options 1,1,5; 3 options 2,2,3; 3 options 3,3,1 and 6 variants with numbers 1,2,4. A total of 15 options.

Sequence 1+2+3 can fall 6 ways. The mini-jackpot and the set can also fall only 6 times each.

Sum up. A total of 216 combinations may fall. One of them could be the pair 81, 15 sevens, sets, mini-jackpots and sequence sequence 6 combinations of each.

To calculate the amount, you need to substitute the amount of payments for each loss: 81*1 + 15*2 + 6*4 + 6*10 + 6*2=207. So, of all the possible 216 combinations, 207 will bring returns. The difference between these values is 9. It turns out that we will lose 9 out of 216 bets. Thus, the advantage of the casino is 9/216*100=4,167%. The advantage of the player could be said if the return for the back would be more.

In this example, the number of possible combinations is small, so it was easier to consider it as a brute force. If the combinations of several thousand or even millions, this method is clearly not suitable. The second example will focus on how to deal with such cases.

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