Probability - how come $76$ has better chances than $77$ on a "Luck Machine" round? a Luck machine is a $3$  $0-9$ digits on a screen, define $X_1$ to be the result contains $77$, and $X_2$ to be the result contains $76$.
$p(X_1)=2*1/10*1/10*9/10+1/10*1/10*1/10$
$p(X_2)=2*1/10*1/10$
if i wasnt mistaken in my calculation, we get $p(X_2)>p(X_1)$ can anyone explain this shocking outcome?

If I understand your question correctly, you have a machine that generates 3 digits (10³ possible outcomes) and you want to know the probability that you have event $X_1$ : either "7 7 x" or "x 7 7" and compare this with event $X_2$: either "7 6 x" or "x 7 6".
Notice that the 2 sub-events "7 7 x" and "x 7 7" are not disjoint (they have "7 7 7" as common outcome) while the 2 sub-events "7 6 x" and "x 7 6" are disjoint.
$P(X_1) = P($"7 7 x" $\cup $ "x 7 7"$) = P($"7 7 x"$) + P($"x 7 7"$) - P($"7 7 7"$)$
$P(X_2) = P($"7 6 x" $\cup $ "x 6 7") = $P($"7 6 x") + $P($"x 7 6"$)$
This explains why $P(X_1) < P(X_2)$.
