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This question is from DEGROOT's PROBABILITY and STATISTICS.

Problem

Suppose that two dice are to be rolled repeatedly and the sum $T$ of the two numbers is to be observed for each roll. We shall determine the probability $p$ that the value $T =7$ will be observed before the value $T =8$ is observed.

Solution

The desired probability $p$ could be calculated directly as follows: We could assume that the sample space $S$ contains all sequences of outcomes that terminate as soon as either the sum $T = 7$ or the sum $T = 8$ is obtained. Then we could find the sum of the probabilities of all the sequences that terminate when the value $T = 7$ is obtained.

However,there is a simpler approach in this example. We can consider the simple experiment in which two dice are rolled. If we repeat the experiment until either the sum $T = 7$ or the sum $T = 8$ is obtained, the effect is to restrict the outcome of the experiment to one of these two values. Hence, the problem can be restated as follows:

Given that the outcome of the experiment is either $T = 7$ or $T = 8$, determine the probability $p$ that the outcome is actually $T = 7$. If we let $A$ be the event that $T = 7$ and let $B$ be the event that the value of $T$ is either $7$ or $8$, then $A ∩ B = A$ and

$$ p = Pr(A|B) = \frac{Pr(A ∩ B)}{Pr(B)} =\frac{Pr(A)}{Pr(B)} $$

But $Pr(A) = 6/36$ and $Pr(B) = (6/36) + (5/36) = 11/36$. Hence, $p = 6/11$.

Now, my doubts are

  1. Why does the author say

    We could assume that the sample space $S$ contains all sequences of outcomes that terminate as soon as either the sum $T = 7$ or the sum $T = 8$ is obtained. Then we could find the sum of the probabilities of all the sequences that terminate when the value $T = 7$ is obtained.

  2. How can we go from lengthy sequences of outcomes that terminate as soon as either the sum $T = 7$ or the sum $T = 8$ is obtained to just the outcome of the experiment for which either $T = 7$ or $T = 8$ ?

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  • $\begingroup$ Actually, I asked the question on math.se, but couldn't understand the answer. $\endgroup$
    – Silent
    Commented Oct 3, 2013 at 9:24
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    $\begingroup$ Sush, it is generally not accepted here to cross-post (to post the question on different sites simultaneously). You can flag your question on math.se and request it to be migrated here. $\endgroup$
    – COOLSerdash
    Commented Oct 3, 2013 at 9:50
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    $\begingroup$ Migrating the Math question here is strongly preferable because it would collect all the answers in one thread rather than scattering them across two sites. In the meantime, I respect the thought behind this question and agree that the text and the existing answers are inadequate (although correct): probability calculations are notoriously tricky and it's easy to get subtly wrong answers by invoking the "obvious" leaps of intuition exhibited here. A good answer will rely only on basic axioms and proven results of elementary probability theory, justifying every step with rigor. $\endgroup$
    – whuber ♦
    Commented Oct 3, 2013 at 15:22
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    $\begingroup$ Hey, I've completely revamped my answer and included proof for the second question. I only use basic conditional probability concepts and a sum formula so it should not be too hard to grasp (i.e. no Markov Chains). Hopefully this answers your question. $\endgroup$
    – Cristian
    Commented Oct 4, 2013 at 13:19
  • $\begingroup$ I thank you all for pardoning and making me obliged. Also, many thanks for providing so good answers. I will not break the tradition from now and will ask one question in just one StackExchange site, and if not satisfied, will migrate it. $\endgroup$
    – Silent
    Commented Oct 4, 2013 at 16:29

2 Answers 2

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Question 1

  • Why does the author say

    We could assume that the sample space $S$ contains all sequences of outcomes that terminate as soon as either the sum $T = 7$ or the sum $T = 8$ is obtained. Then we could find the sum of the probabilities of all the sequences that terminate when the value $T = 7$ is obtained.

Answer

Sample space $S$ has $m \rightarrow \infty$ sequences of length $n \rightarrow \infty$ that end in either $7$ or $8$. Out of these sequences we're interested in summing up the probabilities of all the series that end in a $7$. The probability of a sequence of precisely $n$ throws ending in a $7$ is:

$$ P_n(7) = \left(\frac{25}{36}\right)^{n-1} \cdot \frac{6}{36} $$

However, since $n$ can take any value up to infinity, the overall probability of ending a sequence of any length in a $7$ is the sum of the prob. of ending a seq. after one throw plus the prob. of ending a sequence after two throws, and so on. This is the geometric series:

$$ \Phi_7 = P_1(7) + P_2(7) + P_3(7) + ... + P_n(7) $$

which, as $n \rightarrow \infty$, sums up to (basic geometric sum formula)

$$ \Phi_7 = \lim_{n \rightarrow \infty} \frac{\frac{6}{36}\left(1-\left(\frac{25}{36}\right)^n\right)}{1-\frac{25}{36}} = \lim_{n \rightarrow \infty} \frac{6}{11}\left(1-\left(\frac{25}{36}\right)^n\right) = \frac{6}{11} $$

This is the probability of ending a sequence of throws in a $7$ without ever hitting $8$. It's the answer you're looking for using the first, "more complicated" method.

Question 2

  • How can we go from lengthy sequences of outcomes that terminate as soon as either the sum $T = 7$ or the sum $T = 8$ is obtained to just the outcome of the experiment for which either $T = 7$ or $T = 8$ ?

Answer

This will become clear if we rephrase the first method a little bit. Sample space $S$ has $m \rightarrow \infty$ sequences of length $n \rightarrow \infty$ which end in either a $7$ or an $8$. The probability of you running a sequence of length $n$ which ends in $7$ is the probability

$$ P_n(7)|(P_n(7) \cup P_n(8)) = \frac{P_n(7) \cap (P_n(7)\cup P_n(8))}{P_n(7)\cup P_n(8)} = \frac{P_n(7)}{P_n(7) \cup P_n(8)} $$

$$ \frac{P_n(7)}{P_n(7) \cup P_n(8)} = \frac{\left(\frac{25}{36}\right)^{n-1} \cdot \frac{6}{36}}{\left(\frac{25}{36}\right)^{n-1} \cdot \frac{6}{36} + \left(\frac{25}{36}\right)^{n-1} \cdot \frac{5}{36}} = \frac{6}{11} $$

This is a lot of LaTeX for not a very impressive statement but it is useful because we can now use it to prove by induction the jump from a sequence of length $n$ to a sequence of length $1$. If we run the same formula for $n-1$ we get

$$ P_{n-1}(7)|(P_{n-1}(7) \cup P_{n-1}(8)) = \frac{P_{n-1}(7)}{P_{n-1}(7) \cup P_{n-1}(8)} $$

where

$$ \frac{P_{n-1}(7)}{P_{n-1}(7) \cup P_{n-1}(8)} = \frac{\left(\frac{25}{36}\right)^{n-2} \cdot \frac{6}{36}}{\left(\frac{25}{36}\right)^{n-2} \cdot \frac{6}{36} + \left(\frac{25}{36}\right)^{n-2} \cdot \frac{5}{36}} = \frac{6}{11} $$

But this means that

$$ \frac{P_n(7)}{P_n(7) \cup P_n(8)} = \frac{P_{n-1}(7)}{P_{n-1}(7) \cup P_{n-1}(8)} $$

and it follows, by induction, that

$$ \frac{P_n(7)}{P_n(7) \cup P_n(8)} = \frac{P_{1}(7)}{P_{1}(7) \cup P_{1}(8)} $$

Therefore, whatever value $n$ takes, the probability of a sequence of that length ending in $7$ given that it ends in either $7$ or $8$ is equal to the probability of a sequence of length $1$ ending in $7$ given it is a part of $S$.

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  • $\begingroup$ Thanks a lot!! You have cleared my many doubts with your answer(not this answer helps me for this question but also helps me solving exercises). But still, shall I ask one question? $\endgroup$
    – Silent
    Commented Oct 4, 2013 at 17:00
  • $\begingroup$ Sure, ask away, but note that comments should not be used for lengthy discussion. I can clarify things in my answer but if it's something different you should simply ask another question. $\endgroup$
    – Cristian
    Commented Oct 4, 2013 at 17:02
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    $\begingroup$ In $S$ there are $m$ sequences of various lengths $n$ that end either on a $7$ or $8$. Each sequence has a specific probability. If you add all the probabilities of all the sequences you'll get $1$. But since we're only interested in those that end in a $7$ we only count those ones and basically ignore the ones that end in an $8$. This gives us the proportion out of $S$ that is occupied by $7$ series and hence the probability of hitting a $7$ before an $8$. $\endgroup$
    – Cristian
    Commented Oct 4, 2013 at 17:09
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    $\begingroup$ Because we're interested in the prob. of hitting one before the other. If we stop on a $7$ we know for sure that we did not hit an $8$. If we stop on an $8$ we know for sure that we can't hit a $7$ without hitting an $8$. If we stop on $6$ we learn nothing about the probability of hitting a $7$ before hitting an $8$. $\endgroup$
    – Cristian
    Commented Oct 4, 2013 at 17:14
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    $\begingroup$ Then you're not answering the original question anymore. Much like the other point about stopping before hitting either, you gain no useful information for this question by going on after $7$. You'd only keep going after $7$ if you had a question that said something like "what's the p of hitting one $x$ after we've already hit one $7$" or any other question of that kind. This is not your current question however. $\endgroup$
    – Cristian
    Commented Oct 4, 2013 at 17:44
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  1. Is a definition of what the probability of an event $A$ is e.g.

    $$ P(A) = \frac{\text{number of ways $A$ can happen}}{\text{total number of things that can happen}} $$ Like when you try to figure out what is the probability that you roll a two given that you know you rolled an even number is $1/3$ since there is just one way that a two can come up, and 3 ways that an even number could have come up.

    So, using this definition, to get $P(\text{we roll a 7 before we roll an 8})$ is out of all the sequences of two dice rolls that end in either 7 or 8, how many of them have a 7 occur before 8.

    The author notes that such a strategy for determining the probability is a bit too complex and continues to describe a simpler strategy

  2. This can be justified with noting that $$P(\text{you roll 7 before 8}| \text{you rolled either a 7 or an 8}) = P(\text{you roll 7 before 8})$$ that is the event of rolling a 7 before an 8 is independent of the event of rolling either a 7 or an 8 (i.e. If I told you the result one one die was either 7 or 8, it will not impact the probability that a 7 happens before an 8)

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    $\begingroup$ In this application the "total number of things that can happen" is infinite. How, then, do you apply this "definition" of probability? $\endgroup$
    – whuber ♦
    Commented Oct 3, 2013 at 17:37
  • $\begingroup$ That is exactly why you would consider a simpler method that the author of the solution outlined. $\endgroup$
    – bdeonovic
    Commented Oct 3, 2013 at 18:01
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    $\begingroup$ I believe that misses the point: the O.P. is looking for a rigorous justification of the simpler method. But there is a fundamental problem with your answer, no matter what: this is not how probability is defined, nor does it directly apply to this question. In fact, this "definition" is inherently ambiguous because it depends on how you characterize the "things." As Fermat remarked in 1654, you first have to establish that these things are exhaustive, mutually exclusive, and have equal chances (which is why your definition is inherently circular). $\endgroup$
    – whuber ♦
    Commented Oct 3, 2013 at 18:28
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    $\begingroup$ @whuber, you understand my doubt very well.Thanks for reminding me too that this "Classical definition of probability" lacks preciseness. $\endgroup$
    – Silent
    Commented Oct 4, 2013 at 16:35

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