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.