Hoop hop showdown

This Riddler puzzle is a shout-out to this YouTube video of a game called “Hoop hop showdown”.

Here is an idealized list of its rules:

  • Kids stand at either end of N hoops.
  • At the start of the game, one kid from each end starts hopping at a speed of one hoop per second until they run into each other, either in adjacent hoops or in the same hoop.
  • At that point, they play rock-paper-scissors at a rate of one game per second until one of the kids wins.
  • The loser goes back to their end of the hoops, a new kid immediately steps up at that end, and the winner and the new player hop until they run into each other.
  • This process continues until someone reaches the opposing end. That player’s team wins!

You’ve just been hired as the gym teacher at Riddler Elementary. You’re having a bad day, and you want to make sure the kids stay occupied for the entire class. If you put down eight hoops, how long on average will the game last? How many hoops should you put down if you want the game to last for the entire 30-minute period, on average?

Here is a derivation of how I solved the problem:
[Show Solution]

And if you just want to see the solutions:
[Show Solution]

11 thoughts on “Hoop hop showdown”

  1. I had a different interpretation of the last segment. Assumed there would be one additional RPB battle after Orange wins your k=1 case and then one more second to jump to victory. My answer: Tavg = N^2+7N+1 seconds. This results in N=8, 121 secs. N=39, 30 mins (minus 5 secs). General approach still the same, but very different answers to the 538 Riddler question.

  2. Hi Laurent,
    I like your solution — I got almost the same except I stupidly solved the linear algebra instead of summing the columns and therefore found the algebraic form only by looking-at-answers.
    But I think your answer is too large by one factor of 3/2 and should be (N^2 + 7 N – 3)/2. I think you double-count the first RSP game, since you count the game once in f(k) and once in T_{avg}.
    To see this quickly, consider N=1. The players hop to the middle (1 sec) and play RSP (3/2 sec). The winner has won the game, so that took 5/2 sec. But your formula gives 4 sec.

      1. This depends on your interpretation of whether there is a final RPS challenge before jumping out of the last hoop and whether the last jump takes another second. If yes, I still think N=3 hoops would require an avg T = 31 secs and in general Tavg = N^2+7N+1. The illustration under the puzzle appears to support this view.

        1. I wasn’t able to glean much from that illustration… and the description in the wording of the problem is ambiguous/incomplete. I guess we’ll just have to wait and see how Oliver intended the problem to be interpreted!

          1. Looks like your interpretation was right. Congrats again for the call out in the solution!

  3. Hi Laurent,

    Thank you for your nice description of your solutions, I am learning a lot.
    I have only difficulties with how you simplify your expressions, could you please elaborate little bit more about how you were simplifying that Tavg expression?

    Thank you.

    1. Sure, would be happy to elaborate. The trickiest part is the fourth-to-last line where
      \left( \left\lfloor \frac{2m+1}{4} \right\rfloor + \left\lfloor \frac{2m+2}{4} \right\rfloor \right)
      \] is simplified to $m$. I have an additional explanation of that step right after the derivation. If there is a specific part you’re struggling with, just let me know and I can elaborate on it.

      1. I didn’t understand how you got from line 3 to 4, than how you got to to line 5 with m variable.
        That simplification from 5 to 6 is very nice, only I don’t get how you introduced that m = 2q+r.

        Thank you.

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