Probablity expectation What is the expected value when a 1 dollar lottery ticket is bought in which the purchaser wins exactly 10 dollar million if the ticket contains the six winning numbers chosen from the set $\\{1, 2, 3, \dots , 50\\}$ and the purchaser wins nothing otherwise? I am not getting this question .. $$P(X=1)=P(X=2)=P(X=3)=P(X=4)=P(X=5)=P(X=6)=1/50$$ $$E(X)=1/50\cdot (1+2+3+4+5+6)=1/50 \cdot 21=21/50.$$ Is my answer/approach correct ..please correct me !

I think that we win if the ticket contains all the six winning numbers chosen and each ticket has six distinct numbers between 1 and 50.

Then the probability to win is $p=\frac{1}{\binom{50}{6}}$. Then the expected value of the final profit should be $$E=(-1)\cdot (1-p)+(10^7-1)\cdot p.$$

P.S. Notice that the expected value for lotteries is "always" negative, otherwise...