Two players decided to play many rounds of the same two-player game.
One of the players checked their winning percentages after every round and found they had winning percentages of exactly 30%, 40%, 50%, 60%, & 70% at some point, at least once and not necessarily in this order. What's the minimum number of rounds they played?
Extension: What if the set of percentages included all of 10% through 90%?
Adapted from an article in The Guardian.
The numerators and denominators each need to be increasing, but not strictly so. Starting by using the numerators that are relatively prime to 10:
30 50 70
3/10 10/20 21/30
40% = 1/5, could be 3/15 and 60% = 3/5 could be 15/25. Only 30 games need to have been played.
30 40 50 60 70
3/10 3/15 10/20 15/25 21/30
Expanding
10 30 50 70
1/10 6/20 15/30 28/40
We still need 40% and 60% (2/5 and 3/5 when reduced).
40% could be 6/15; that's fine, with no games won in games 16 through 20.
60%, or 3/5, could be 15/25; again we assume games 26 through 30 were lost.
10 40 30 60 50 70
1/10 6/15 6/20 15/25 15/30 28/40
We still need 20% = 1/5 and 80% = 4/5 and 90% = 9/10.
90% can be 45/50. Can we squeeze in the other two?
Well, 7/35 wouldn't work for 20% as we already have 15/25. We'd need 45/225.
I don't know if the other numbers can be rearranged to get fewer than 225 games.
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Posted by Charlie
on 2022-05-10 12:50:43 |
not necessarily minimal
