Dice: Probability of rolling a 1 on the nth throw where k 1's have already been rolled

You have a six sided die and are rolling it for the nth time. In all the previous (n-1) throws, a total of k 1's have already occurred. What is the probability that you will throw a "1" on the nth throw. I don't know if it is correct or not, but is the shorthand way of writing this P(A:,n,k) where event A is rolling a 1 on the nth throw where k 1's have been rolled (in any order) in the previous (n-1) throws? (Question not about the notation). Thanks for any help.

Karl

• Do you know the die is fair (i.e. initially all six faces are equally likely and each throw is independent of the others)? – Henry Oct 3 '17 at 9:36
• Yes, it's a fair die – Karl Oct 3 '17 at 9:43
• Then memorylessness suggests $\frac16$ – Henry Oct 3 '17 at 9:45
• I am surprised by this! Assume after n-1 throws you threw a total of n-1 "1's. In this case, the probability would be very low that you will throw another 1 (assuming large n). Therefore I believe the probability has to be a strong function of k. But I could be wrong. – Karl Oct 3 '17 at 10:03
• But after throwing $n-1$ times 1 you pick up the die to go for the $n$-th throw. Do you really think the die has thoughts/memory and mumbles in itself something like: "mmm, to keep the balance not a 1 this time..."??? Believe me Karl, if it is fair then the probability that it will come up with a 1 is $\frac16$. – drhab Oct 3 '17 at 10:17

I think that the key to understanding this is that the first $n-1$ rolls have already happened. This changes the event space that you need to look at.
A simple definition of probability is the fraction of “successful” outcomes out of the total number of possible outcomes. When you started off all of this die-rolling, all $6^n$ possible sequences of die rolls were in play. The event that consists of rolling a 1 every time is just one of these many possible outcomes, so its probability at that point is very small indeed.
After having rolled $n-1$ times, however, you’ve eliminated most of the possibilities with which you started. Now, there are only six possible outcomes instead of $6^n$, and so the probability of rolling another one is $1/6$. In fact, it doesn’t matter at all what the preceding $n-1$ rolls were: they’ve eliminated most of the possible outcomes that you started with, leaving only six.
Here’s an informal way to get a feel for this: When you start off, the chances of your rolling a one $n$ times are quite small, but with each successful roll the chance that you’ll actually do it improves. At step $k$, you only have to roll another $n-k+1$ ones, which is certainly more likely than rolling $n$ of them.