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Well, I'll be damned. What are the chances of these two rolls occurring? Can't lie, Roll God; you had me a bit nervous.
Watch me do just that, LOL.Next time roll three 6’s in a row
DAMNNNN THE MATH THO. Thank you for taking the time to do it, aha! That's insane. I'll definitely be keeping this one in the books.Using mathematical probabilities, rolling 4 out of 150 has a 1/150 probability of occurring.
Therefore, in order to be defeated by a 4-150 roll, one must roll 3 and below.
The probability of rolling 1, 2 and 3 is 1/150 individually. Accordingly, the combination of the three probabilities is 3/150.
Therefore, the probability of this occurring (where the second roller losses against a 4-150 roll) is (1/150)*(3/150).
This is equivalent to 0.013333333%.
The probability of the second roller exactly rolls 3 while the original roller rolls 4 is (1/150)^2.
This is equivalent to 0.00444444444%.
"HAH, NERD. IMAGINE DOING MATH, COULDN'T BE ME."
yeet, not gonna lie, that's a really rare occurrence. Keep the screenshot, it'll be a good memory!
You've summoned the devil!waas out of 20 but
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