#### Topic: How to count probability

Hello! It is not assured that I can correctly count probability  even in sober mind, and in passion - especially. Statements of the problem such: There are three football matches, in each of which the recognized favourite against other command which unexpectedly not bad showed itself on tournament start plays. The bookmaker accepts rates with coefficients 1:4, 1:4, 1:9. A question: how correctly to count probability at least to remain at the? One scoring of any of commands-nefavoritov at least pays back loss in other matches. Empirically I estimated that the chance is quite good, and the Author delivered: Marty Date: 01.07 22:11, and it is similar, did not lose. But the mathematical deal too is interesting

#### Re: How to count probability

Hello, Marty, you wrote: M> Statements of the problem such: There are three football matches, in each of which the recognized favourite against other command which unexpectedly not bad showed itself on tournament start plays. M> the bookmaker accepts rates with coefficients 1:4, 1:4, 1:9. M> the Question: how correctly to count probability at least to remain at the? One scoring of any of commands-nefavoritov at least pays back loss in other matches. M> empirically I estimated that the chance is quite good, and the Author delivered: Marty Date: 01.07 22:11, and it is similar, did not lose. But the mathematical deal too is interesting Well, if the probability of a scoring at least in one match taking into account that chances of a scoring in a match correspond declared, interests: Let A - a scoring in the first rate, B - a scoring in the second, With - in the third. In total 8 total outcomes - ABC are possible! ABC, A! BC, AB! A C! A! BC, A! B! A C! AB! A C! A! B! A C where ABC - simultaneous  all rates, and! A! B! A C - loss in all. Events of matches independent, therefore their total probability is their product. A = 0.25! A = 0.75 B = 0.25! B = 0.75 Cs = 0.11! A C = 0.89 Taking into account that at us only one variant of loss of all matches-! A! B! A C with ease it is calculated its probability which is equal 0.5 It is possible to check up calculations: we Find probabilities of all total outcomes ABC = 0.0068! ABC = A! BC = 0.02, AB! A C = 0.055! A! BC = 0.061, A! B! A C =! AB! A C = 0.166! A! B! A C = 0.5. The total probability should be to equal unit. We add and is received 0.9948 - similar on truth, taking into account a rounding off. Further, we count  a scoring: we Construct a random variable for what we define value of a scoring in each of total outcomes: for what we add scorings in each private outcome taking into account the rate (we accept it for 1) ABC = 4+4+9 - 3 = 14! ABC = A! BC = 4 +9-3 = 10, AB! A C = 4+4 +0-3 = 5! A! BC = 0+0+9-3 = 6, A! B! A C =! AB! A C = 4+0+0-3 = 1! A! B! A C = 0+0+0-3 =-3. We receive a random variable of 0.0068: 14 0.02: 10 0.02: 10 0.055:5 0.061: 6 0.166: 1 0.166: 1 0.5:-3 Matozhidanie of a scoring it turns out summation of multiplication of probability of a random variable on its value, i.e. M = 0.0068*14 + 0.02*10 + 0.02*10 +0.055*5 +0.061*6 + 0.166*1 + 0.166*1 +0.5 * (-3) = 0.0952+0.4 + 0.275+0.366 + 0.332-1.5 =-0.0318. You  is negative, and, means, the rate minus.

#### Re: How to count probability

Hello, SomeOne_TT, you wrote: SO _> Hello, Marty, you wrote: SO _> You  is negative, and, means, the rate minus. Thanks for calculations. I will sober a little from  effect of a scoring of a command of the Wales then I will check up. If only Italy does not benefit. UPD I thump, it is glad to an outcome. Vologda  itself showed

#### Re: How to count probability

Hello, SomeOne_TT, you wrote: SO _> Hello, Marty, you wrote: M>> Statements of the problem such: There are three football matches, in each of which the recognized favourite against other command which unexpectedly not bad showed itself on tournament start plays. M>> the bookmaker accepts rates with coefficients 1:4, 1:4, 1:9. M>> the Question: how correctly to count probability at least to remain at the? One scoring of any of commands-nefavoritov at least pays back loss in other matches. SO _> you  is negative, and, means, the rate minus. While you consider probability of a scoring on the basis of bookmaker rates,  a scoring for each separate match to equally zero, without dependence from bookmaker coefficients. The matozhidanie totals of several rates, also should be zero, without dependence from structure of rates. Distinct from zero of values - a rounding-off error.

#### Re: How to count probability

Hello, SomeOne_TT, you wrote: SO _> a C = 0.11! A C = 0.89... SO _> the Total probability should be to equal unit. We add SO _> and it is received 0.9948 - similar on truth, taking into account a rounding off.... SO _> Matozhidanie of a scoring turns out summation of multiplication of probability of a random variable on its value, i.e. SO _> M = 0.0068*14 + 0.02*10 + 0.02*10 +0.055*5 +0.061*6 + 0.166*1 + 0.166*1 +0.5 * (-3) = SO _> 0.0952+0.4 + 0.275+0.366 + 0.332-1.5 =-0.0318. SO _> You  is negative, and, means, the rate minus. And here if a C not to approximate, and to write C=1/9 and then fairly to count all also the total probability will be equal 1, and  will be zero.

#### Re: How to count probability

Hello, Eugene Sh, you wrote: ES> And here if a C not to approximate, and to write C=1/9 and then fairly to count all also the total probability will be equal 1, and  will be zero. Thanks, and you and Tchertkov.

#### Re: How to count probability

Hello, Chorkov, a C> Matozhidanie of the total of several rates, also should be zero, without dependence from structure of rates. A C> Distinct from zero of values - a rounding-off error. It in the mathematician In life  the totals of several rates value always negative for the bookmaker office too wants to eat bread and butter.

#### Re: How to count probability

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