2. If X is a discrete random variable, all its possible values are a 1, a2, …, an, …, and the corresponding probabilities are p 1, p2, …, pn, …, then its mathematical expectation E (x) = (A1) (P

Expectation of uniform distribution: The expectation of uniform distribution is the midpoint (a+b)/2 of the value interval [a, b]. ?

Variance of uniform distribution: var(x)=E[X? ]-(E[X])？ .

Extended data:

With the knowledge of probability theory, it is not difficult to know that A is likely to win and B is unlikely to win.

Because the probability of A losing the last two games is only (1/2) × (1/2) =1/4, that is to say, the probability of A winning the last two games or any of them is1-(1/4) = 3.

However, if B expects to win 100 francs, it must beat A in the last two games. The probability of B winning the last two games in a row is (1/2) * (1/2) =1/4, that is, B has a 25% probability of winning 100 francs.

It can be seen that although the game can't be played again, according to the above possibilities, the objective expectations of both parties for the final victory are 75% and 25% respectively, so Party A should get 100 * 75% = 75 (francs) and Party B should get 100×25% = 25 (francs). The word "expectation" appeared in this story, from which mathematical expectation came.

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