OT: A Friday brain problem...

wrote

What if the oddball (joke?) is heavier OR lighter?

That's the correct form of the problem, the OP got it wrong. Although, it doesn't change the answer (3) at all, or the procedure.

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Yes. The oddball is a different weight - heavy or light, unknown - and: your task is to identify it; you are not required to determine whether heavy or light.

Finally, modify the problem: either one ball is odd, or they are all true. Solve.

I don't know about "wrong." 8 balls, one of them too heavy, is also a valid problem. It also tricks people because they look at the 8 and think that the answer must be to weigh 4 vs 4, then 2 vs. 2, and finally 1 vs. 1. So they guess the wrong answer of 3, instead of the correct answer of 2.

This sounds trickier, although I think I could get somewhere by weighing 123 vs. 456. Then, assuming I don't get lucky enough for them to balance, I'd weigh 14 vs. 25. I think that path would get me to a solution in 3 weighings.

This sounds a bit like the previous one. Start with 123 vs. 456. The only new case would be down the path where they balance. But then you'll still have two weighings to deal with 7 and 8. First, weigh 7 and 8 against each other. Then, if they don't balance, weigh 7 against 1.

actually you could include the other two balls in the second step 147 vs 258

third step is 783 vs 156

and that finds the bad ball with no decision tree needed for the weighings.

what's more all thses can be done in the same number of weighings with

for the first case 123 vs 456 147 vs 258

(if both balance ball 9 is the culprit)

for the second and third with 9 balls.

weigh 123 vs 456 then 456 vs 789 then 147 vs 258 three predeterimed weighings suffice for 12 balls with 1 or 0 bad balls

weigh 123A vs 456B then 456A vs 789B then 147A vs 258C

I wonder if there a way to do 13 balls in three weighings.

Yes. PN2222A