-1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Do you know the die is fair (i.e. initially all six faces are equally likely and each throw is independent of the others)? $\endgroup$ – Henry Oct 3 '17 at 9:36
  • $\begingroup$ Yes, it's a fair die $\endgroup$ – Karl Oct 3 '17 at 9:43
  • 1
    $\begingroup$ Then memorylessness suggests $\frac16$ $\endgroup$ – Henry Oct 3 '17 at 9:45
  • $\begingroup$ 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. $\endgroup$ – Karl Oct 3 '17 at 10:03
  • $\begingroup$ 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 $\frac16$. $\endgroup$ – drhab Oct 3 '17 at 10:17
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thank you all for your answers. I have found all useful in trying to understand. amd, your answer has especially provided insight for me. Actually am looking into these types of problems for analysis of Greedy Pig game to help out 10th grade student -- thought it would be fairly simple, but beginning to see that such a simple game can have extremely complicated analysis: i.e. optimization of strategy. $\endgroup$ – Karl Oct 5 '17 at 0:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.