Can You Roll The Die?

As there are infinite possible cases of getting 1 in even turns.

Let $\color{#3D99F6}{E_1} , \color{#D61F06}{E_2} , \color{#20A900}{E_3} \ldots , E_{\infty}$ denotes the events of getting 1 in 2 trials, 4 trials , 6 trials and so on.

Let P(E) denotes the probability of getting 1 in even turns. $P(E) = P(\color{#3D99F6}{E_1}) + P(\color{#D61F06}{E_2}) + P(\color{#20A900}{E_3}) + \ldots + P(E_{\infty})$ $P(E) = \color{#3D99F6}{\dfrac{5}{6}\times\dfrac{1}{6}} + \color{#D61F06}{\left(\dfrac{5}{6}\right)^3\times\dfrac{1}{6}} + \color{#20A900}{\left(\dfrac{5}{6}\right)^5\times\dfrac{1}{6}}+ \ldots \infty$ Using infinite sum formula , $P(E) = \dfrac{\dfrac{5}{36}}{1 - \dfrac{25}{36}} = \dfrac{5}{11}$

I think the easiest way to think of this problem is to consider rolling two dice together: a red die and a blue die. Repeat until a 1 comes up. Consider the red die to represent the odd-numbered rolls, and the blue die the even-numbered rolls. If the red die shows a 1, it's equivalent to the first 1 coming up at an odd roll. If the red die doesn't show a 1, but the blue one does, then it's equivalent to the first 1 coming at an even roll. (In the scenario that both dice show 1s at this roll, the red one represents an earlier roll in the sequence of single rolls, and so is the one that matters.)

What matters is the probability distribution of the two possibilities at the roll when the first 1 comes. There are 11 possible rolls of the two dice that include a 1. Of these, 6 have the red die showing a 1; therefore there's a chance of $\frac{6}{11}$ of this outcome. On the other hand, 5 of the possible rolls have the red die showing 2-6 and the blue one showing 1 - so the probability of this happening is $\boxed{\frac{5}{11}}$ .