1
$\begingroup$

This example comes from the book "A First Course in Probability" by Sheldon Ross, Chapter 3, Section 3.5, Example 5C, Page 95. The example is given a solution but there is a part that I don't really understand. I hope someone can explain the logic to me.

The original question is as follow:

Independent trials, each resulting in a success with probability $p$ or a failure with probability $q = (1-p)$, are performed. We are interested in computing the probability that a run of $n$ consecutive successes occurs and before a run of $m$ consecutive failures. We are given the solution as follow:

Solution.: Let E be the event that a run of $n$ consecutive successes occurs before a run of $m$ consecutive failures. To obtain P(E), we start by conditioning on the outcomes of the first trial. That is letting H denote the event that the first trial results in a success, we obtain $$P(E) = pP(E|H)+qP(E|\bar H)$$ Now, given that the first trial was successful, one way we can get a run of $n$ successes before a run of $m$ failures would be to have the next $n-1$ trials all result in successes. So, let us condition on whether or not that occurs. That is, letting F be the event that trials 2 through n all are successes, we obtain $$P(E|H) = P(E|FH)P(F|H)+ P(E|\bar{F}H)P(\bar{F}|H)$$ On the one hand, clearly, P(E|FH) = 1; on the other hand, if the event $\bar{F}H$ occurs, then the first trial would result in a success, but there would be a failure some time during the next $n-1$ trials. However, when this failure occurs, it would wipe out all of the previous successes, and the situation would be exactly as if we started out with a failure. Hence, $$P(E|\bar{F}H) = P(E|\bar{H})$$ Pause !! This is the part where I don't really understand. I agree that when this failure occurs, it would wipe out all of the previous successes. But the situation for E to occur shouldn't be exactly the same as if we started out with a failure because if $F$ is the event that trials $2$ to $n$ are successes, then $\bar{F}$ is the event where not all $2$ to $n$ trials are successes. That being said, the last next $n-1$ trials could be successes. As an example, if the last two of $n-1$ trials is a failure then a success; then the situation for E to occur should be exactly the same as $P(E|H)$. So, can someone explain to me why $P(E|\bar{F}H) = P(E|\bar{H})$ ?

$\endgroup$
  • $\begingroup$ Shouldn't $P(\overline{F}H)$ be $P(\overline{F}|H)$ instead, in the second equation? You condition $P(.|H)$ on $F$. Otherwise the whole second product would reduce to $P(E)$. $\endgroup$ – Henno Brandsma May 16 '17 at 8:02
  • $\begingroup$ @HennoBrandsma Thanks for noticing the error. I have edited the second equation. $\endgroup$ – M.A.N May 16 '17 at 8:29
2
$\begingroup$

Since we are looking at an infinite number of trials, after removing a finite prefix of trials, we still have an infinite number of trials. Therefore, we can simply ignore this prefix for the probability of $E$ (even the conditions for $E$ would already be fulfilled by the prefix) and therefore $P(E) = P(E \mid B)$ for any event $B$ that restricts the outcomes of the trials only for a finite prefix.

This hopefully answers your question, but the partial solution you gave above seems rather complicated to me. Thankfully, I was able to find a pdf version of the book via Google. There, the author gives a rather complicated probability for $P(E)$ (not included in your question). That surprised me, because intuitively I thought, $P(E) = 1$.

Thinking about that in more detail, I am now convinced that, indeed, $P(E) = 1$: Let $A$ be the event, that we consecutively have $n$ successes and $m$ failures in $n + m$ trials. Clearly, $P(A) ≠ 0$, because $P(A) = p^n ⋅ q^m$ (assuming $0 < p < 1$).

Now let us iterate these $n + m$ trials and let $E'$ be the event that event $A$ occurs in the first block of $n + m$ trials, or in the second block of $n + m$ trials, or in the third block, etc. Since $P(A) ≠ 0$ and therefore $P(\overline{A}) ≠ 1$ it is easy to see that $P(\overline{E'}) = 0$, since we are looking at an infinite number of trials. Therefore we have $P(E') = 1$.

Since $E' ⊆ E$, it follows that also $P(E) = 1$.

For short: In an infinite sequence of trials you are able to find any finite pattern with probability $1$.

Therefore I suspect that there is a mistake in the solution in the book – unless I made a mistake myself or misunderstood the problem.

$\endgroup$
  • $\begingroup$ Would you mind posting the pdf of the book you found? $\endgroup$ – Jason May 16 '17 at 15:07
  • $\begingroup$ zalsiary.kau.edu.sa/Files/0009120/Files/… $\endgroup$ – Toni Dietze May 16 '17 at 15:24
  • $\begingroup$ Thanks, I appreciate it. I've had a quick read and there seems to be numerous errors, and the problem is not very well defined. Your answer seems much better. I'll have a closer read and possibly post my own answer. $\endgroup$ – Jason May 16 '17 at 16:44
0
$\begingroup$

I have received a clear explanation and I thought it will be useful to share it out here. $$P(E|\bar{F}H) = P(E|\bar{H})$$ Because, we are not conditioning on the the results of trials 2 through n but rather just on whether they were all successes. That is, we are using

$$P(E|H) = P(E|FH) P(F|H) + P(E|\bar{F} H) P(\bar{F}|H)$$

so in determining $P(E|\bar{F} H)$ we do not know the results of trials 2 through n but only that they were not all heads, that is, that at least one was a tail. Because we know nothing about the sequence following that tail $$P(E|\bar{F} H) = P(E|\bar{H})$$

$\endgroup$
-1
$\begingroup$

@Toni Your answer is correct if the question is asking the probability of $n$ consecutive successes ever occurs before $m$ consecutive failures, i.e. trials do not stop after condition is satisfied. I think the question, may it be phrased ambiguously, was asking if either $n$ consecutive successes of $m$ failures appear, the game stops. If we play such games a large number of times, what's the probability that the game ends with n consecutive successes.

@OP To clarify further, let's draw from the example OP provided where the last two trials of $2^{nd}$ to $n^{th}$ were failure and then success respectively. You argued that if the next $n-1$ trials are success, this is equivalent to $P(E|H)$.

The problem of this argument is that conditioning on $\bar{F}$ does not mean conditioning on the full realization of the $2^{nd}$ to $n^{th}$ trials, but only on if all or not all are successes, i.e. the conditioning doesn't mean the game has proceeded to the point where the $2^{nd}$ to $n^{th}$ trials are all tried. If we ever get a failure within the $2^{nd}$ to $n^{th}$ trials at some point, we are already in the event space defined by $\bar{F}$, then $P(E|\bar{F}H)$ in this case would be conditioning on from that failure. Since whatever happens before that failure is now irrelevant, that's the reason why $P(E|\bar{F}H)$ is exactly the same as $P(E|\bar{H})$.

$\endgroup$
-2
$\begingroup$

There are several errors which suggest a fundamental misunderstanding of probability on the behalf of the author of this textbook. For starters, this problem is not properly defined - how many trials are there in total? Are we talking about precisely $n+m$ trials, or perhaps some finite number $N>n+m$, or an infinite sequence? The worked solution is inconsistent in which assumption it is using, and arrives at a conclusion which is incorrect regardless of which assumption we use. Let's go through it:

  • "One the one hand, clearly, $P(E|FH)=1$" - nope. $E$ is the event that we have $n$ successes followed by $m$ failures; $H$ is the event that the first trial is a success; $F$ is the event that the second through $n^\text{th}$ trials are successes. If we are assuming precisely $n+m$ trials, then given $F,H$, the event $E$ occurs if and only if trial $n+1$ through $m$ are failures. This clearly has probability less than $1$. Similarly, this will be less than $1$ (but larger than $P(E)$) if we have $N>n+m$ trials. If we have an infinite number, then yes, this probability will be $1$; but this is no more "clear" than the original problem.

  • "Hence, $P(E|F^cH)=P(E|H^c)$" - what?! You gave a very simple counterexample which shows this is not the case. The instance of the first failure indeed "resets" the count in a sense, but we might have only a single failure on the second trial, in which case we still have $n-2$ successes in a row. If we are assuming $n+m$ trials, the probability is already zero; if we are assuming $N>n+m+1$, this situation leaves us with a positive probability (whereas a later failure may make the probability zero) and if we are assuming infinitely many trials, then as before, each side is equal to $1$. (As Tony Dietze points out, since $E$ is an event only interested in the existence of a certain string, $P(E|C)=1$ for any event $C$ which depends only on finitely many trials.)

  • "Now, $GH^c$ is the event that the first $m$ trials all result in failures, so $P(E|GH^c) = 0$" - why? This is true if we have only $n+m$ trials, but then, $P(E|C)=0$ for any event $C$ which includes a failure in the first $n$ trials. In general, this statement is only true if we have $N<n+2m$ total trials. The event $G$ is also an extremely strange event to condition on - why should we care if all the first $m$ trials are failures? Isn't one enough? Additionally, the author makes the same weird error as before, namely asserting that $G^cH^c$ involves a success wiping out all the previous failures and so claiming that $P(E|G^cH^c)=P(E|H)$. Absolute nonsense.

  • "Thus, $P(E)=\ldots=\frac{p^{n-1}(1-q^m)}{p^{n-1}+q^{m-1}-p^{n-1}q^{m-1}}$" - clearly not true. If there $n+m$ trials, then $E$ is precisely the event where the first $n$ are success, and the rest are failures, so $P(E)=p^nq^m$. If there are infinitely many trials, the reasoning Tony Dietze provided shows that $P(E)=1$ (intuitively speaking, if we run trials for long enough, eventually any finite string will appear). If there are $N>n+m$ trials, $P(E)=P_N(E)$ must depend on $N$ with $P_N(E)\to1$ as $N\to\infty$.

It really blows my mind that such bad mathematics could appear in a textbook. I have not read any other part of this book, but if this example is anything to go by, you should get rid of it immediately and find a new textbook on probability.

$\endgroup$

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.