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stanalger Special user St. Louis, MO 996 Posts 
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 1100. Organizer assigns the number 2 to player 1, and assigns 1 to all of the other players 2100. 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 2100 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 2100 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 2100 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%.) 

TomasB Inner circle Sweden 1143 Posts 
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.
Followup puzzle: How many players should the organizer allow to give them at most even odds if they only guess? /Tomas 

MPHanson New user 18 Posts 
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 

Heinz Weber New user Austria 83 Posts 
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 

TomasB Inner circle Sweden 1143 Posts 
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 

Heinz Weber New user Austria 83 Posts 
Oh...
thank you! 

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