this post was submitted on 02 Oct 2025
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That's very interesting to me (I am a bit mathematically illiterate when it comes to probability). Wouldn't it still have a lower chance of being a 1 if you said you want your second roll to be the one that counts beforehand? Or would different permutations screw with the odds, say rolling a 12 then a 1, rolling a 15 and a 1, etc, counting towards unfavourable possibilities and bringing it back to 1/20?
Because the outcome of a dice roll is an independent event (ie. the outcome of any given event does not impact subsequent events), it doesn’t matter if you said only your 2nd/3rd/4th etc. roll counted. Every roll has a 1/20 chance of rolling a 1 on a D20 die.
Consider this thought experiment, there are ~60.5m people, each rolling a 6-sided die. Only the people who roll a 6 can continue to the next round, and the game continues until there is only 1 winner.
After the first roll, only ~10m people remain in the game. After the second roll, ~1.7m people remain After the third roll, ~280K After the fourth, ~46.5K 5th, ~7.8K 6th, ~1.3K 7th, ~216 8th, ~36 9th, ~6 After the 10th and final roll, there should only be ~1 player remaining.
So even though initially there is only a 1-in-65m chance of rolling 10 6s back-to-back initially, each attempt still has a 1/6 chance of succeeding. By the time we get down to the final six contestants, they have each rolled a 6 nine times in a row - yet their chances of rolling it another time is still 1/6.