Advanced Man Mathematics

[kayaker]

@BigT

Yeah, it does change on whether they enter at the same time or separately. Separately, it’ll be 7, then either 2 or 3 (as no one uses 1 if their are other choices), leaving the third man with 5. Together, it would be 7-5-3. Together, they all to to the end. Separately, the “furthest away” rule becomes an issue.

And, yes, if they come in 2+1, it’s still properly 7-5-3, while 1+2 it’s 7-2-4/5, as the guy on 7 will be done soon enough so that everyone can still avoid 1. It’s not like peeing takes a lot of time or anything.

How much time you spending studying men peeing?:dubious:

[ZenBeam]

@MegaBee

Is the seventh urinal one of those half-height ones?

It might be either #1 or #7, but it’s definitely one of those two, and it’s definitely there.

[MegaBee]

#3 or #5 depending on whether #1 or #7 is the half-height urinal.

[tdn]

Easy. #1.

Because if it’s #2, I’m going to a stall.

[Labrador_Deceiver]

@crucible

Two gentlemen of color standing side by each at urinals…“Man, the water in this bathroom is cold!”, “Deep, too”, says the other.

Two Inuit men walk into this thread… “Man, that fucking joke is older than I am!”. “Tired, too”, says the other.

[MrDibble]

@BigT

no one uses 1 if their are other choices

Say what, now? Whyever not?

[DragonAsh]

The correct answer is, he stands by the mirror and fiddles with his hair until 1 or 7 opens up.

[Dead_Cat]

@Lord_Mondegreen

Agreed. If you take 3 or 5 then there should be a 100% chance that the next person to come in will not stand beside you.

If you take 4, then the next person in will have to stand next to exactly one person, so if he makes a random choice there is a 50% chance it will be you.

I voted for 3 or 5 using the same reasoning, but having read the rest of the thread I think it may be just as likely that first man takes #7, second man takes #5 (because how far apart do you need to be, really?). Even in this case, however, #3 is a good choice. But more research (with fewer variables) is definitely needed.

I have a similar “problem” at work, where in one of the toilet areas we have four cubicles which are all totally self-contained (i.e. proper walls and doors, basin is in there as well). My algorithm for this is as follows: if it is available, always take #4 (furthest from the door) UNLESS #3 is the only occupied stall, in which case take #1. If #4 is occupied always take #1 (maximum separation, unless #2 is also occupied in which case you only have one neighbour). If 1 and 4 are both occupied, take #3 if available (occupant of #4 should finish before occupant of #1, all else being equal, provided they both used the same logic as me).

[cmyk]

This is the best fill order of urinals, as men 1, 2, then 3 enter, in this situation:

1 2 3 4 5 6 7 ——>(door)

X X X X X X X

1 X X X X X X

1 X X X X X 2

1 X X 3 X X 2