Yay, a puzzle! I haven’t read any of the later responses. My assumption is that all offers have to made in increments of $100 (no change given).
My proposed division would be: $4800 to myself, $100 to the third oldest, $100 to the fifth oldest (i.e., youngest). Assuming everyone will try to maximize payoffs in subsequent rounds, this is the greatest payoff person 3 & person 5 can expect, so they should vote for the division, along with myself. 3 votes passes.
Dave Perkins agrees with you, from the past.
That’s what I’d do. I don’t believe in “perfectly rational agents”.
And I agree with Staff Sergeant, in the present!
That was my thought exactly! LOL
balut
1607
You give the creepy boy all $5000 in the hope he doesn’t devour your soul.
<sound of me spitting diet coke all over monitor laughing>
When I was interviewing for my tech job I was asked a perfectly reasonable logic puzzle, and I have used it in interviews. What sets it apart is it is clearly stated, and there is truly only 1 correct answer. But in working it out over an interview table face to face watching how the candidate approaches it and how they react to subtle hints when they get stuck can tell you a lot if you pay attention.
You have 25 horses, and a 5 lane race track. But you forgot to bring a stopwatch that day, so from any one race, you can only determine the relative speeds of the horses in that heat. What is the minimum number of heats you need to run to determine the 1st 2nd and 3rd fastest horse out of the group.
I can do it in 7. It might be possible to do it 6 I guess. Clearly can’t be done in less than 5.
To do it in seven, run five non-overlapping heats. Then run a sixth heat amongst the winners. For exposition, number the horses: 1 is the winner of the sixth heat, 2 through 5 are the other horses in horse 1’s first race in order of their placement in that race, 6 is the second place horse in the sixth race, etc.
We now know horse 1 is the fastest.
We can now eliminate horses 16 through 25 because 16 came fourth in the sixth race and 16 is faster than 17 through 20, and 21 can be no better than fifth fastest and is faster than 22 through 25. We can also eliminate 12 and 13, because we know 11 can be no better than third quickest. Similarly we can eliminate 8.
This leaves five horses with undetermined ranks: 2, 3, 6, 7, 11. Run a seventh race among these horses. The winner is the second fastest horse, the next horse is third.
Edit: Nevermind I’m dumb.
Qmanol
1614
Skedastic’s works. I had only gotten far enough to determining a relatively simple method for doing it in 8 heats. Basically, run 5 heats, eliminate the 4th and 5th horses from each heat, then run 3 heats, one of the winners, one of the horses that came second, and one of the third placed horses. That gives you more information than you need, but it quickly establishes a narrow range of 5-8 to work within.
Edit: Nevermind, I think I get it.
Raife
1616
Nuke the track from orbit. No heats.
I don’t think you can do it in 6 heats. Every horse HAS to appear in a race, or else it might be a top three horse. With 5 races, then, you’d have to race them in distinct batches of five. With your sixth race, which horses would you race? The winners from the other five races, perhaps? But this would not do the trick; one of the 2nd place horses from the first five races might be one of the three fastest horses.
I can’t think of a way to do it in six, though I can think of two ways to do it in eight and skedastic has one in seven.
baruk
1619
Race one - the 4th and 5th horses are eliminated
Subsequent races - The 3rd fastest horse from race one (lets call him Dave), is matched against 4 new horses. If Dave wins, then all 4 horses get eliminated.
The controversial part: since the question specifies minimum number of heats, we can assume that Dave conveniently wins every race. After 6 heats, we have eliminated (2+4+4+4+4+4) 22 horses, leaving us with the 3 fastest.
Kael
1620
Thats very interesting tricky logic.
