Throwing a surprise party: A math paradox
rzwitserloot posted in Uncategorized on June 30th, 2006
Here’s the plan: It’s sunday, and its your birthday. You know your work buddies will throw you a surprise party during lunch one day in the upcoming workweek.
Now, lunch always starts at 12:00 sharp, and you are absolutely certain that one of the five upcoming workdays will feature the surprise party.
One of your workbuddies calls you later today and hints at the surprise party. However, you confidently explain that they cannot possibly throw one. You proceed to explain exactly why this is impossible. Then, on tuesday, they throw the party and you are surprised anyway, mostly at the sudden fallability of your math skills.
Here’s the explanation you give:
Let’s say its thursday, 12:01. In that case you know with certainty that the party has to be on friday, but then it wouldn’t be a surprise when you walk into the lunch area, so the surprise party couldn’t possibly be on friday.
Knowing that, however - let’s say it wednesday, 12:01. In that case its either thursday or friday. But you know it can’t be friday, so it’ll have to be thursday. Except, then, it won’t be a surprise, again.
yadayada - using the exact same reasoning you eliminate all 5 days as a possible opportunity to throw a surprise party - because you’ll be certain that it has to be that day BEFORE you walk into the lunch area.
Quite elated at your advanced understanding of the theory of induction, you smugly walk to lunch on tuesday, and stand flabbergasted as your work buddies throw you a surprise party.
How’d that happen?
I don’t actually know. Discuss.
September 6th, 2006 at 17:14
“Knowing that, however - let’s say it wednesday, 12:01. In that case its either thursday or friday. But you know it can’t be friday, so it’ll have to be thursday. Except, then, it won’t be a surprise, again.”
this is wrong. it’s just impossible to be a surprise party on friday if it’s thurday 12:01.
if it’s wednesday, the condition (if it’s thursday) doesn’t apply.
September 6th, 2006 at 20:44
Not quite. The inductive bit first asserts it CANT be friday, ever, even if it’s now still monday, because if it was friday, then the birthday fellow will not actually be surprised friday, he already knew. He doesn’t now, but that’s immaterial.
Given that it couldn’t possibly be friday, the induction loop kicks off. Except it doesn’t, because if it would, surprise parties are impossible whereas clearly they aren’t.
My own ‘theory’ (read: ramblings) goes a little something like this:
Because by induction a surprise party is never possible, each day is equally possible. Division by 0, more or less. This strategy leads to indeterminate state and is not a useful way to figure out anything concrete.
October 9th, 2006 at 20:31
I think your logic is correct.
I think the flaw is that it’s based on your workmates using this logic to determine when to throw (or not throw) your surprise party.
And of course they don’t do this - they just pick a day and - Happy Birthday!