0
$\begingroup$

I am working on my study guide for the upcoming test and ran into a problem on how to solve two of the probabilities.

1)You roll a fair $4$-sided die. If your roll is a $1$ or $2$, you will roll once more. If you roll a $3$ or $4$, you stop after just one roll. What is the probability that the total of your rolls is exactly $3$? In general I think I need to find the probability that first $2$ rolls were either $1$ or $2$ and third one was $3$ or $4$. I am just not sure how to do it.

2)For the previous problem, if you are given that you rolled a total of exactly $3$, what is the probability your $1st$ roll was a $1$? Should this be $\frac{1}{2}$ as if i rolled the die exactly $3$ times it means that first roll had to be $1$ or $2$ out of the first $2$ rolls?

$\endgroup$
1
$\begingroup$

Operating under the interpretation of the problem that a maximum of two rolls can occur we have the possible outcomes: $\{(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3), (4)\}$.

That is to say, we roll the die the first time. In the event that the first die's result is a $3$ or $4$, we stop rolling immediately. In the event that the first die's result is a $1$ or $2$, we roll the die a single additional time. (No more rolls occur regardless what the result of the second die is.)


The probability of arriving at the outcome $(3)$ is simply $\frac{1}{4}$, similarly the probability of arriving at the outcome of $(4)$ is also $\frac{1}{4}$. The probability of each of the rest of the outcomes, for example $(1,1)$ is $\frac{1}{4}\cdot \frac{1}{4}=\frac{1}{16}$.

The first part of the question asks us what the probability the total sum of the die result(s) is $3$ (not the total number of throws), this corresponds to the outcomes $(1,2),(2,1)$ and $(3)$. Since these outcomes are mutually exclusive to one another, we can add their respective probabilities, resulting in a total of $\frac{1}{16}+\frac{1}{16}+\frac{1}{4}=\frac{3}{8}$


The second part of the question asks us to find the conditional probability that the first die result was a $1$ given the total of the die or dice rolled was $3$. Approaching via Bayes' Theorem:

Let $A$ be the event that the first die result was a $1$ and $B$ the event that the total was $3$

$P(A\mid B)=\frac{P(A)P(B\mid A)}{P(B)}$

We calculated $P(B)$ in the first part of the question to be $\frac{3}{8}$. $P(A)$ is known to be $\frac{1}{4}$ since the die is fair. $P(B\mid A)$ is also going to be $\frac{1}{4}$ since given that the first die result is a $1$, the only unknown is what the result of the second die is and it would need to be a two to have the correct sum which occurs with probability $\frac{1}{4}$.

We have then $P(A\mid B)=\frac{\frac{1}{4}\cdot \frac{1}{4}}{\frac{3}{8}}=\frac{1}{6}$


Another different interpretation of the problem: there are an unlimited number of rolls. If we roll a $1$ or a $2$ we roll again (regardless how many rolls we have already done), and if we roll a $3$ or a $4$ we stop.

The question of finding the probability of having rolled a total of three times (where we don't care what numbers on the faces are or what they add up to, just how many rolls happened until we were told to stop) we are indeed then looking for the probability the first two rolls were low and the third roll was high. This appears to be what you interpreted the problem to be (though I disagree with this interpretation due to the phrasing "the total of the rolls" instead of "the total number of rolls").

Here, let $L_i$ be the event the $i$'th roll (if it occurred) was low (a $1$ or a $2$) while $H_i$ be the event that the $i$'th roll (if it occurred) was high.

We are looking then for $Pr(L_1\cap L_2\cap H_3)$ which by multiplication principle is $Pr(L_1)Pr(L_2\mid L_1)Pr(H_3\mid L_1\cap L_2)$ which by the assumption the die is fair and the results of any rolls that occur are independent gives us a result of $\frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}=\frac{1}{8}$

For the second part of the problem, the probability that the first roll was a $1$ given that there were a total of three rolls that occurred, your intuition is correct and the probability is $\frac{1}{2}$.

$\endgroup$
0
$\begingroup$

Hint:

Say we roll a $1$ in the first try. For a total of $3$, we need a $2$ or $1+1$ on successive rolls. If we roll a $2$, then we have to roll the die again (why?) , which gives a total $>3$. Similarly, if we keep rolling $1$, the rolls never stop but the total becomes $>3$. Similar is the case when we roll a $2$ in the first throw, a roll of $1$ on the next try does no good to have a total of $3$ (why?).

So, for a total of $3$, we have to roll a $3$ in the first throw itself. Hope you can take it from here.

$\endgroup$
  • $\begingroup$ if we have to roll a 3 on the first throw itself than the probability is 1/4. How about the second problem? or was my rationality correct and the answer is 1/2. $\endgroup$ – user3365751 Feb 22 '17 at 6:55
  • $\begingroup$ @user3365751 Yes, you are absolutely correct. So, can you deduce the answer for the second part? $\endgroup$ – Rohan Feb 22 '17 at 6:56
  • $\begingroup$ @user3365751 The only possible case to get a total of 3 is when 3 is rolled first itself. Then there is no probability that the first throw can be $1$. So, $$P=0$$ $\endgroup$ – Rohan Feb 22 '17 at 7:02
  • $\begingroup$ I disagree with this interpretation of the problem. I would not think that you continually roll again and again and again if you roll small numbers, but rather the total number of rolls caps at two. The possible outcomes are $(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3), (4)$ (note: these are not equally likely). You could then end with a total of three if you were to roll a $1$ followed by a $2$ (no more rolls are going to occur). That is to say I interpret it as "If your (FIRST) roll is a $1$ or a $2$, you will roll once more." $\endgroup$ – JMoravitz Feb 22 '17 at 7:04
  • $\begingroup$ @JMoravitz Well, at first even I thought like that, but I will let the OP clarify and then I will change. Thanks for your comment. $\endgroup$ – Rohan Feb 22 '17 at 7:09

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.