If X denotes the face of the die then

E(X) = 1*0.3 + 2*0.3 + 3*0.3 + 4*0.033 + 5*0.033 + 6*0.033 = 2.2950 and

E(X^2) = 1^2*0.3 + 2^2*0.3 + 3^2*0.3 + 4^2*0.033 + 5^2*0.033 + 6^2*0.033 = 6.7410

These gives,

V(X) = E(X^2) - (E(X))^2 = 6.7410 - 2.2950^2 = 1.4740

expected value of Die?

Your xi's are 1,2,3,4,5,6 ? if yes then your claim E[xi] = n*pi of is wrong

The formula for E(X) is SUM[i](p(X=xi)*xi).

@Mr. Karmakar : x is face but n is number of times each face occurred. Here Im taking into account formula for random variable (Multinomial Distribution) not for the data.

@Mr. Chakraborty : How ?

@ Mr. Wilhelm: We'll get a single term out for it or 6 terms, 1 for each face of dice ?

I think it is not multinomial distribution...You can refer this document...

http://cseweb.ucsd.edu/~dasgupta/103/3.pdf

Anjani, you gave p(X=xi) for each of the possible outcomes xi. Plugging in these values gives exaclty what Sandipan already wrote. You get one number (2.3 [Sandipan's 2.295 has a rounding error bcause 0.033 should be 0.1/3 = 0.03333...]), and this is the expected value of "eyes" you will get from such a die (per roll!). If you roll the die n times, you expect n times this much.

The variance is VAR(X) = E((X-E(X))²) = sum[i](p(X=xi)*xi²) - E(X)²

You can again plug in your xi's and pi's and you will get what Sandipan wrote. Again this is for a single roll. If you consider the variance in a series of n rolls, the variab´nce needs to be multiplied with n².

See

http://en.wikipedia.org/wiki/Expected_value

http://en.wikipedia.org/wiki/Variance

Check any basic book on probability and restate your problem.

Yes the formula i was following was for the single face of the dice and not for the entire dice. Thanks to all for helping me out.

?? The expected value of a single face is its value...

No no the I was searching for entire dice but was constantly looking at the formula which dealt only with single face of dice. So I got wrong answer every time when compared with these.