15 Questions

  1. 51627 Kerfluffles 1:0
  2. 03489 soyleche 1:1

Now I’m wondering: Does the “correct position” start from the back or the front if there are more digits guessed than in the actual number? If the number was ABC, and ABCDE was guessed, would that be 3:0 or 1:2?

it’d be 3:0, go left to right

I’m starting to think the random # of digits doesn’t really matter since you’d always just guess the max number of digits anyway and can get the right answer that way.

Am I thinking about this right…

This is starting to become a game of “how can we make this game hard enough so that it’ll take around 15 guesses to win”

Guss it would help if I bothered to read… :slight_smile:

Do you mean by using a guess to get the right number of digits? Like this?

:grinning_cat_with_smiling_eyes:

1234567890

Not quite. Since I say the digits can be 1-5, you should proceed by just assuming it 5 digits, and I think you’ll get the right answer regardless of how many digits there are.

I think

I think you’re probably right, because if we guessed
12345 and then 67890 we could get the total # of digits, it would just take 2 guesses.

Which Kerfluffles and soyleche did, with their first two guesses, just not in that order.

539

:bump:

  1. 51627 Kerfluffles 1:0
  2. 03489 soyleche 1:1
  3. 539 Celalta 0:1

314

  1. 51627 Kerfluffles 1:0
  2. 03489 soyleche 1:1
  3. 539 Celalta 0:1
  4. 314 Celalta 3:0

DING DING DING!! YOU GET SOME PI

okay that was too easy

It really was, but you made it easier by choosing pi :joy:

I have a number.

you wanna change the rules up a bit? The goal is to make it hard enough that it takes around 15 guesses for the average actuary.

I’m not listing rules. Figure it out :slightly_smiling_face:

31425

  1. Guess A:B:x (guesser)
  2. 31425 0:3:0 (JSM)