Specifically I’m asking about this situation that came up during a D&D game:
Say I’m rolling a 20 sided die, and I do not want to roll a 1. I know that the odds of rolling a 1 are 1/20.
I know that the chances of rolling a 1 twice in a row is (1/20 * 1/20), which is far a lower occurrence.
Say then, before I rolled my “real” roll, I rolled the die again and again until I landed on a 1, then proceeded to roll my “real” roll, would I have reduced the odds of rolling a 1 to (1/20 * 1/20), given that I’ve just rolled a 1 prior?
This is the logic I’m having trouble reasoning about and I’d appreciate it if anyone could clarify what is or is not accurate about the assumptions being made in this scenario.
In: Mathematics
You’re correct that the odds of you rolling 2 1’s in a row are (1/20) * (1/20) but your setup means that it doesn’t help you. Your odds of any given roll landing on a 1 are 1/20.
>before I rolled my “real” roll, I rolled the die again and again until I landed on a 1
This means that you’re not wondering the odds of rolling two 1’s in a row, you’re only looking at the odds of rolling a 1 *after you know you have already rolled 1*, the odds of the getting two 1’s in a row in this case aren’t (1/20) * (1/20) they’re (1) * (1/20) since you’re only calculating odds after you’ve already rolled a 1, its a given so its probability is 100% (you can’t get here if it didn’t already happen)
If you really want to change your odds of rolling a 1, get an automated roller and check the bias of your dice. Most plastic D20 dice are accidentally weighted to one side or another and that changes the real roll distribution
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