P( A | the first throw is 2) = P(A).

This becomes more obvious if A is interpreted as “the first 6 comes before the first odd number”. In other words, A is independent of the event that the first throw is 2, and therefore conditioned on A, the probability that the first throw is 2 is 1/6. Hence, conditioned on A, in each trial, 2, 4 and 6 are not equally likely. In fact, 6 is more likely than 2 and 4.

]]>Now consider the expected time it takes for a random sequence of dices to first reach one of the values in {1,3,5,6}. This value is clearly 3/2 since its the expected value of a geometric random variable with probability of success 4/6. Furthermore, this value is the average of the previous expectations, which finishes the proof. ]]>