Without the final "subtract from 100" step, the team won't be
guaranteed a win.

Here's a counterexample:

The players have already numbered themselves 1-100.

Organizer assigns the number 2 to player 1, and assigns 1 to
all of the other players 2-100.

Player 1 receives a list consisting of 99 1's. He totals
the list to get 99, adds 1 (his player number) to get 100,
ignores all but the last two digits and submits 0 as his guess.
He's wrong since his number was 2 (not 0.)

Players 2-100 each receive a list containing 98 1's and a single 2.
Each of these list totals 100 (since 98*1 + 2 = 100.)

After players 2-100 add their player numbers to 100, they get totals
of 102, 103, 104, ..., 199, 200. Ignoring all but the last two
digits, they finish will 2, 3, 4, ..., 99, 0. If they submit these
numbers as their guesses, they are all wrong (since all of them
were assigned 1 by the organizer, but none submitted 1 as their guess.)


[If you add the final "subtract from 100" step, player 1 would subtract
0 from 100 and submit 100 as his guess. He's still wrong.
Players 2-100 would subtract 2, 3, 4, ...., 99, 0 from 100 and
submit 98, 97, 96,..., 1, 0 as their guesses. Player 99 would be
the only one submitting a correct answer...but that's a WIN for the
team!]

It's quite a generous game for the organizer to sponsor, since even with
NO strategy, the team will probably win. If the organizer's assignment
of numbers is random, and if the players ignore the lists and simply
guess at their numbers, they will win roughly 63% of the time.
(1-(99/100)^100=.63396765...)
The flawed strategy (missing the "subtract from 100" step), while not
perfect, increases the probability of success even more (but not all
the way to 100%.)

Stan is of course very right that the final subtract needs to be done. I confused myself when I saw that it didn't matter if the players add or subtract their position in the modulo arithmetics. Exchanging addition for subtraction is the same as putting the players in the reverse order since (X + N) mod 100 = (X + (N - 100)) mod 100.

Follow-up puzzle: How many players should the organizer allow to give them at most even odds if they only guess?

/Tomas

Thank you for the amplification, Tomas. Nonetheless, I think I need to go back to the kiddie pool until I learn to swim a little better...

Mike

First I admitt I have not the sligthest idea WHY stans strategy works. I simulated it yesterday (to proof that it doesn't and to find a counterexample) and found it works. I also found: with random guessing (without any strategy at all) they have a 64,6% chance to win, and with the strategy 'without the last subtract from 100 -step' this is nothing better. I did this because my first approach to the problem was something similar, I tried to make sure that players with identical assigned numbers are guessing differently. But obviously I was (and I am) very fare from understanding the problem...

Stan (or Thomas) can you give us some insight why this works?

Thanks,
Heinz

Heinz, Stan did explain why it works but I can try to make it clearer.

Imagine that you are a player and you KNOW the last two digits if you sum up ALL 100 numbers. Maybe you know that the number should end in 42. You get your list of 99 numbers and sum them up. You will then KNOW which number you should chose to make the sum end in 42. There can only be one such number between 1 and 100.

In reality no one knows what he sum of all 100 numbers end in but we know that it can only end in 00, 01, 02, ... , 98 or 99, which happens to be exactly 100 different cases. So let one player guess that the sum ends in 00. Let another guess that the sum ends in 01, etc. One (and exactly one) of them will be correct and will therefore guess his own number correctly.

/Tomas

Oh...

thank you!