A player can choose his own set of a maximum of 10 numbers and, depending on the number of selected and guessed numbers, his bet is multiplied by a different factor.For example, the player chose 3 numbers.Then his chances are:

- guess 0 numbers: 41,093%
- guess the 1st number: 44.028%
- guess 2 numbers: 13,664%
- guess 3 numbers: 1.215%

For each of the outcomes, we consider the coefficient using the formula

`100/probability`

.It turns out:- if you guessed 0 numbers: 0
- if you guessed 1 number: 2.271
- if you guessed 2 numbers: 7,319
- if you guessed 3 numbers: 82,305

But with such ratios, the"casino" turns out to be a big loser.For example, 100 players bet $1.Of them, according to probability, 1 number will win 44.028 and be picked up, at a rate of 2.271

`2.271 * 44.028=$99.9`

- i.e., the whole prize fund is
already.And 2 and 3 more numbers will guess - there, too, each
option spends $100.Of this, payments are 3 times higher than the
bets made.**How to calculate the odds so that “casino"Remained in the black, and did the rates reflect the real probability of winning?**

Also tell me, please, the literature about this kind of games and their mathematics: both probabilistic and financial.

You can not charge for these factors. – Nice64 Jul 25 '18 at 17:07

As I wrote in the question, if you select 3 numbers, then the chance that all these 3 numbers will be 1.215% in the final “ten”.The coefficient in this case will be x82.305.

This coefficient only works if there are only 2 outcomes of the game - I guessed 3 numbers and did not guess.

Do you understand? – Nice64 Jul 25 '18 at 17:22

That is, for guessing 2 numbers, the coefficients will be different when betting on 3 numbers and, say, on 5.Most of all, of course, if you bet only 2 numbers, and both won. – Vivacious Vendace Jul 25 '18 at 18:11

I understand that.For example, I made the player choose 3 numbers.

But what will they be? How to count them? – Nice64 Jul 25 '18 at 18:17

The total number of cases

`C(10.40) * C(3.40)`

- we divide into it the number of favorable options.This is a choice of 10 winning C(10.40) out of 10, you must choose 2 hits C(2 , 10) out of the remaining 30, you must choose 1 Overshot C(1.30) and multiply.Total You can certainly shorten something then.Delivered this comment in response. – Vivacious Vendace Jul 25 '18 at 18:51

I need on the basis of these chances to count the coefficients. – Nice64 Jul 25 '18 at 19:36

do not intersect: either guesses 0, or 1, or..., or N.Each option has its own probability=>its coefficient.The probabilities of all outcomes add up to 1.Tasks are discrete.Each specific outcome can be isolated and put in a separate box.It remains to recount these boxes.1 red box “guessed 3 out of 3”, 2 yellow “2 out of 3”, 3 green “1 out 3", 4 blacks" did not guess a single one."Everything is all possible outcomes at all, there are no others(numbers are fictitious).In the sum of 10 possible outcomes.The probability of red is 1/10, and the bet on it is taken with a coefficient.1/(1/10)=10: 1, on yellow, whose probability is higher, coefficient 5: 1.On green 3.33.On black it would be 2.5, but this is a winning institution. – Vivacious Vendace Jul 25 '18 at 21:04