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What began as a simple question at first glance slowly grew more interesting than I initially thought.

The problem goes as such: Franky and Sal fashioned a dart board to play darts after school. It just so happened that no two sectors were of the same size, leading to all sectors having different odds of getting hit with the dart. Franky threw a dart and hit one of the sectors. What is more probable, Sal hitting the same sector on his throw, or his hitting the adjacent sector, moving clockwise around the center of the board?

That's it---nothing is said about the sector Franky hit or the others. Maybe this is a lack of sight on my part, or maybe this funny dartboard inquiry has rightfully trumped me. Of course, this is a question with only two answers. Let's just keep it to one attempt, though---no guessing.

What's the proof and subsequent answer for this problem?

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  • $\begingroup$ homework question? what is your own approach? $\endgroup$ Commented Dec 16, 2023 at 20:29
  • $\begingroup$ At first, I thought the same as user2661923 below but I wanted to try looking at it a bit more pointedly. I drew a picture and made three sectors, one for the one Franky hit, one for the adjacent one, and one that represents all other sectors on the board. However, I couldn't seem to go from there---it seems the information in the question is a bit too sparse to draw more conclusions, or maybe that's just me. $\endgroup$
    – JJD
    Commented Dec 16, 2023 at 20:44

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Consider a dartboard with $n$ sectors and respective sector probabilities $p_0, \dots, p_{n-1}$.

The probability of hitting sector $i$ twice in a row is $p_i^2$, and the combined probability of hitting the same sector twice in a row is $q := \sum_{i=0}^{n-1} p_i^2$.

Let $[i+1]_n$ denote $(i+1)$ modulo $n$. The probability of hitting sector $i$ and then sector $[i+1]_n$ is $p_i p_{[i+1]_n}$. The total probability of such an event occurring is $$r := \sum_{i=1}^n p_i p_{[i+1]_n} \le \left( \sum_{i=1}^n p_i^2 \sum_{i=1}^n p_{[i+1]_n}^2 \right)^{1/2} = \sum_{i=1}^n p_i^2 = q,$$ where we used the Cauchy-Schwarz inequality. The inequality holds with equality if and only if $p_i p_{[i+1]_n} = p_i^2$ for all $i$ which is equivalent to requiring that $p_i$ is the same for all $i$.

So it is more likely to hit the same sector twice than to hit the two different sectors as specified in the question.

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  • $\begingroup$ Ah, I see. Thank you! We were just recently looking at that inequality, too---no surprise that you used it here! $\endgroup$
    – JJD
    Commented Dec 16, 2023 at 21:04
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    $\begingroup$ you're welcome. If your question is answered can you mark the answer accordingly? $\endgroup$ Commented Dec 16, 2023 at 21:14
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This is a conditional probability problem. You are supposed to infer that the sector that was hit probably has more area than the alternate sector, since it was hit. This is a reasonable inference.

This means that it is more likely that you will hit the same sector than the alternate sector.

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  • $\begingroup$ a more precise answer to this question is possible $\endgroup$ Commented Dec 16, 2023 at 20:31
  • $\begingroup$ @JorisBierkens I agree. And now that you raise the issue, I think that the problem actually deserves a more explicit answer. Unfortunately, the last time that I opened a Probablility text book was about 50 years ago, and I don't want to make a fool of myself. If you can expand my idea into a separate, more explicit answer, I think that that would be a good idea. $\endgroup$ Commented Dec 16, 2023 at 20:37
  • $\begingroup$ Yes, this is something I did run across. I didn't think it was concrete enough, though. $\endgroup$
    – JJD
    Commented Dec 16, 2023 at 20:41
  • $\begingroup$ I am afraid your mathematics is incorrect in several ways. What do you mean by success probability? Consider for example a dart board with two sectors of relative size $p=9/10$ and $1-p = 1/10$. Can you compute the probability of the different outcomes in this case? $\endgroup$ Commented Dec 16, 2023 at 21:23
  • $\begingroup$ @JorisBierkens It would be $~\frac{0.81}{0.81 + 0.09} = 0.9.~$ Then, in general, the relative probability would be $~\dfrac{p^2}{p^2 + p(1-p)} = \dfrac{p^2}{p} = p.~$ I initially considered this, but then I remembered the old problem of having a success on the $~(n+1)~$ trial, given that an event with unknown probability has it's first $~n~$ trials successful. If I remember correctly, the probability is $~\dfrac{n}{n+1}.~$ That being said, I am going to delete the added portion of my answer, because I am way too ignorant in formal probability theory to have a confident opinion here. $\endgroup$ Commented Dec 16, 2023 at 21:33

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