Idiotest spoiler alert.
This past week on Idiotest, two teams reached a tie at the end of three rounds for the first time in Idiotest history. The tie was $120 for each team and each team solved the exact same puzzle correctly within the same second back in the first round.
Here are the Idiotest tie-breaker rules:
1) When tied at the same cash amounts at the end of the regular game, the tiebreaker is whichever team solved the most number of puzzles correct, wins.
2) When Tiebreaker #1 fails (since each team both solved one puzzle), the next tiebreaker is that whoever solved the puzzles in the least amount of seconds wins.
3) When Tiebreaker #1 and #2 fail, the final tiebreaker is whoever buzzed in first by tenths of seconds wins.
Opinion-based, these three tiebreakers are not the greatest on Idiotest's part. The first tiebreaker will almost never occur because it is not probable that two teams will tie the same amount of earnings with different puzzles. In the second tiebreaker, if two teams lock in with the same earnings, that means they locked in within the same second. Therefore, the second tiebreaker makes absolutely no sense. The third tiebreaker is too precise, too hard to judge and it is easily possible that two teams can lock in at the exact same time (unless they want to get down to hundredths of seconds).
I really do not think it is right to judge the winner of a game by a few tenths of a second or less. The easiest tiebreaker would be to have another puzzle with seconds and money counting down. But, Idiotest is trying to be cost effective with their current tiebreaker rules. It would probably be best for Idiotest to have a tiebreaker puzzle that only involves seconds and not money.
To wrap up, the Idiotest tiebreaker is pretty bad and should be readjusted with a no-cost tiebreaker puzzle for future seasons.