Probably probable

Note that the last digit of a power is only influenced by the last digit of the base.

The last digit of $2016^n$ will always be $6$ , as any power of $6$ will end $6$ .

Finding the last digit of $2013^{n+4}$ is the same as finding the last digit of $3^{n+4}=3^n*3^4=3^n*81$ . $81$ will not affect the last digit, as it ends in $1$ . $3^n$ must therefore end in $9$ to solve the problem. $3^n$ is cyclical, with a period of $4$ . Only one of these powers ends in $9$ , so the probability is $\frac{1}{4}$ .

The solution that I have shown, shows the repetition of the numeral values in the unit place of a number.

For eg. here, in 2013 , 3 is in unit's place, and for a number with three in its unit's digit. The numbers which will occur will be:

• 3
• 9
• 7
• 1

Now same can be done for 6.

After knowing these things, the result can be drawn out easily as shown.

$NOTE :$ I have shown the solution from the starting. And it will follow the same pattern for $n>1000$ also, this was just for an insight of the problem. I gave $n>1000$ because the first $FIVE$ terms in the figure, will disturb the probability fraction. So to isolate the problem from that, I increased the number $n$ .

Now there will be four numbers in the unit digit always and those are $9, 5, 3, 7$ .

Probability for $5$ to occur= $\boxed{0.25}$

Hope this helps .