What Does a Sample Space Look Like When Rolling a Die Until a 3 Appears?

[jcosmet]

Sample space of rolling a 6 sided die

I've just had my first tutorial for an introductory probability class taught by the Econometrics department and I'm having trouble understanding why my solution is wrong. The tutor didn't write up complete solutions(just pictures to help visualise the problems) as he went through the questions on the board, so I found it hard to follow. I tried to speak to him after class but I found it difficult to understand him(English is not his first language and he has a thick accent).

Homework Statement

(c) An experiment consists of rolling a die until 3 appears. Describe the sample space
and determine:
(1) how many elements of the sample space correspond to the event that 3
appears on the k-th roll of the die;
(2) How many elements of the sample space correspond to the event that the 3
appears no later than the k-th roll of the die

The Attempt at a Solution

My solution for (c) was: The sample space consists of the collection of finite sequences ( x_{1} x_{2}... x_{k}) such that the k-th term in each sequence is 3 and all other terms are one of 1,2,4,5,6.

It also consists of infinite sequences of the form (x_{1} x_{2}...) such that x_{i}≠3.

The tutor did not describe the sample space, however I was told that we could not have infinite sequences as we would not then be able to calculate the probability of anyone event occurring. Could someone elaborate?

(1) Solution was 5^{k-1}. The tutor also got the same answer.

(2)My solution was \Sigma^{k}_{1}5^{k-1}, since we simply add up all the possible out comes from the 1st roll to the k-th roll.

The solution the tutor gave was: 6^{k}-5^{k}!? He said it had something to do with the total number of outcomes for k rolls, which is 6^{k} and then subtracting the number of outcomes!?
Can someone show/explain how he got the equation?

 

[voko]

Your answer seems correct for your sample space.

Your tutor's answer seems to imply that sample space consists of fixed k-length sequences, in which case, indeed, 6^k is the total number of outcomes, and 5^k is the number of outcomes when 3 is never obtained. But such a sample space would require that the experiment should continue even after 3 is obtained and stop even if 3 is not obtained, both of which contradict the specification.

 

[Ray Vickson]

I've just had my first tutorial for an introductory probability class taught by the Econometrics department and I'm having trouble understanding why my solution is wrong. The tutor didn't write up complete solutions(just pictures to help visualise the problems) as he went through the questions on the board, so I found it hard to follow. I tried to speak to him after class but I found it difficult to understand him(English is not his first language and he has a thick accent).

Homework Statement

(c) An experiment consists of rolling a die until 3 appears. Describe the sample space
and determine:
(1) how many elements of the sample space correspond to the event that 3
appears on the k-th roll of the die;
(2) How many elements of the sample space correspond to the event that the 3
appears no later than the k-th roll of the die

The Attempt at a Solution

My solution for (c) was: The sample space consists of the collection of finite sequences ( x_{1} x_{2}... x_{k}) such that the k-th term in each sequence is 3 and all other terms are one of 1,2,4,5,6.

It also consists of infinite sequences of the form (x_{1} x_{2}...) such that x_{i}≠3.

The tutor did not describe the sample space, however I was told that we could not have infinite sequences as we would not then be able to calculate the probability of anyone event occurring. Could someone elaborate?'

*******************************
It depends on the level of rigour you want to use. Many mathematicians would say that such problems need a measure-theoretic foundation in order that you can be sure of what you are doing, so your sample space could consist of all _infinite_ sequences of integers from 1 to 6. Any individual sample point does, of course, have probability 0, but an event such as {get 6 for the first time at toss k} is a subset of the set of all sequences and, as such, has a perfectly finite probability. On the other hand, some excellent probabilists, such as the late William Feller, can write entire books containing numerous deep results about problems such as yours, without ever using measure-theoretic methods at all (that is in Volume I of his famous book "Introduction to Probability Theory and its Applications", Wiley---by far my favorite probability book). Feller treats such problems as your c(1) by saying that the sample space is the set of finite sequences of integers from 1 to 6 and that have just one "6" at the end. Your c(2) would involve the set of all sequences of length at most k.

******************************************(1) Solution was 5^{k-1}. The tutor also got the same answer.

(2)My solution was \Sigma^{k}_{1}5^{k-1}, since we simply add up all the possible out comes from the 1st roll to the k-th roll.

The solution the tutor gave was: 6^{k}-5^{k}!? He said it had something to do with the total number of outcomes for k rolls, which is 6^{k} and then subtracting the number of outcomes!?
Can someone show/explain how he got the equation?

The question about the 3 appearing on or before toss k is, perhaps, ambiguous. If the word "the" is interpreted literally, it could mean that just one "3" occurs, and that is on or before toss k. Then the probability would be (1/6)\sum_{j=1}^k (5/6)^{j-1}. On the other hand, if you really mean that "a 6 appears on or before toss k" that would usually be interpreted (in standard English) as "at least one 6 appears ... ", so one or more "3" would be counted. In that case the easiest computation would be P\{\text{at least one 3}\} = 1 - P\{\text{no 3}\} = 1 - (5/6)^k. In both cases you could then get the number of sample points in the event by multiplying the probability by the total number of sample points altogether in k tosses, which is 6^k. So, it seems that your tutor is using the second interpretation ("a 3"), while you are using the first ("the 3") .

RGV