0

Textbook math problem: I give each person in the class a pot of $5$ dice, which they are to roll repeatedly until they have thrown straight sixes, then they can go home. After how many rolls will half the class still be there?

Textbook solution: About 5390 rolls. The probability of rolling straight sixes is $\left(\frac{1}{6}\right)^5$. For half the class to still be there after $t$ rolls we would need that $\left[1 - \left(\frac{1}{6}\right)^5\right]^t = \frac{1}{2}$. Solving for $t$ gives approximately 5390.

My question: Why not say that after $t$ rolls we would need $50\%$ success, so that $\left[\left(\frac{1}{6}\right)^5\right]^t = \frac{1}{2}$? In this case I see that $t$ would be approximately $0.077$ which makes no sense, but I do not see the flaw in my reasoning.

E2R0NS
  • 256
  • 1
  • 9
  • 2
    You are asking that the player roll straight sixes $t$ times in a row, which is not required. The point is, the player only needs one success, so it is better to look at the complement (all failures). – lulu May 31 '23 at 22:36
  • To do it with successes, you'd have to consider the case of exactly one success, exactly two, and so on, and then sum. Perfectly correct mathematically, but it's a lot easier to work with the complement. – lulu May 31 '23 at 22:40
  • Oh, it represents a single student. So we need a single student to fail about half the time in 5390 rolls. – E2R0NS May 31 '23 at 22:58
  • How does a single student's roll of the pot of dice fit in with the entire class? – E2R0NS Jun 01 '23 at 04:45
  • Linearity of expectation. If each student has a $\frac 12$ chance of being present at the end, then we expect half the group to be present. – lulu Jun 01 '23 at 09:42
  • But $t$ represents the number of rolls that the entire class has taken, not just a single student (it says so in their solution). So if there were three students, where the first one rolled $4$ times, the second one rolled $5$ times, and the third student rolled $2$ times, then $t$ would equal to $11$. – E2R0NS Jun 01 '23 at 19:03
  • I doubt that that is what $t$ means. I think every person rolls $t$ times (stopping of course if they ever get straight sixes). You aren't told the size of the group. If there were a million people then one turn per person is already a million rolls. So with your interpretation the problem can not be answered. – lulu Jun 01 '23 at 19:13
  • "For half the class to still be there after $t$ rolls we would need that ..." is a direct quote. – E2R0NS Jun 01 '23 at 19:15
  • so what? Your interpretation leads to an unsolvable problem. And it gets much worse if, as you suggest, you need to keep track of how many people get eliminated at each round. The sensible interpretation, and the one that matches the official solution, is to assume that $t$ counts the number of rounds. In a single round, everybody who hasn't been eliminated tosses their dice. – lulu Jun 01 '23 at 19:17
  • and, to be clear, we are only talking about the expected number of rounds. In reality, it could take many fewer or many more. The textbook solution addresses the average number of rounds it takes. – lulu Jun 01 '23 at 19:19
  • Still confused. I'm struggling with the scenario and the way we are framing the problem I think. And not sure what "linearity of expectation" means. So if $t \approx 5390$, then the probability of a single student failing to get at least one success in $5390$ rolls is $1/2$? – E2R0NS Jun 01 '23 at 19:31
  • 1
    So, google Linearity of Expectation. It is a very powerful concept, and directly relevant to this problem. The probability of a single student getting a single fail in $1-\left(\frac 16\right)^5=0.999871399$ so the probability of a single student failing $5390$ times in a row is $(0.999871399)^{5390}=0.499972086$ so, yes. – lulu Jun 01 '23 at 19:35
  • The only concepts that have been introduced are independence of discreet events and the product law for discreet events. It's the first section on probability in a calculus book. So this problem requires concepts not introduced yet? Or maybe it's just being intentionally fuzzy to foster learning. – E2R0NS Jun 01 '23 at 20:55
  • Perhaps your text regards it as obvious that, if, as here, you have $N$ independent coin tosses ,then you expect $\frac N2$ Heads. Personally, I'd regard that as a statement worth proving, but it really is kind of obvious. – lulu Jun 01 '23 at 20:57
  • I understand it, but I find it unsettling. I prefer to stick closely to the definitions when learning, but they seem to be insufficient in this section at least. So yes, I agree, it should be proved. – E2R0NS Jun 01 '23 at 21:01
  • If you prefer, you can directly compute the expected value for any binomial distribution. That's easy in this case (since $p=1-p=\frac 12$). Personally, I'd regard that as more work, not less. But it only uses basic properties of the distribution. – lulu Jun 01 '23 at 21:10
  • here is that computation. Like I say, not easier...but arguably more elementary. – lulu Jun 01 '23 at 21:27

0 Answers0