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I calculated the Romeo and Juliet meet for a date exercise.

I thought that since the role of the Romeo and Juliet are equivalent, I can just assume 1 of them to be 1st, calculate the probability of the second arriving in time, and since the symmetric case has the same ratio of good/all events, the probability of both cases together is the same: $2g/2a=g/a$

I got $7/32$, but the solution is twice of that: $7/16$.

The solution is there and I understand it. The reason I am asking is beyond this exercise. I am interested in why this line of thinking (symmetry argument) of mine caused a bad result.

Parcly Taxel
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juma
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    Presumably in your calculation of good events you had a particular one arriving first, and of all events you had either arriving first. If in the picture in the linked question the square is divided into $32$ equally likely triangles, you can either say a good result is $14$ out of $32$ triangles, or using your restricted approach a good result is $7$ out of $16$. They do produce the same result – Henry Jul 21 '18 at 09:05

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When you apply a symmetry reduction, you shrink both the space of desired events and that of all events. In this case, you would have computed the probability of the second person arriving in time to meet the first, but that only applies if the designated first person actually arrives first, which occurs half the time. That person may arrive second, in which case the symmetry argument applies to the situation obtained by switching the people.

Thus, the space of all events should be halved from what you did, leading to the correct answer.

Parcly Taxel
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