Because each roll is independent, you multiply the probability of success for each individual trial. With 11 possible rolls, the probability of not rolling a specific value 49 times in a row and then rolling that specific value is as follows:
P(49 non-35's then a 35) = q^49 x p
P = (10/11)^49 x (1/11)
P = 0.00768 (or 0.768%)
100 / 0.768 = 130.2
So, odds of it taking 50 attempts to roll 35fcr is approx 1 in 130.2 chance.
Note that the above calculation differs from the previous (including the x 1/11) because the question states the 50th attempt DID roll a 35 exactly.
Whereas if we were only calculating the odds of NOT rolling 35fcr 49 times in a row we'd have:
P(49 non-35's) = q^49
P = (10/11)^49
P = 0.084475 (or 8.4475%)
100 / 8.4475 = 11.838
So, if we're not gauranteeing the 50th roll IS a 35, and only calculating the odds of not getting 35 49 times in a row, we're looking at approx 1 in 11.8 chance.
Essentially, if we assume were missing the 35fcr 49 times in a row, we'd need to do that 11 times to probabalistically hit exactly 35fcr on the next roll.
Food for thought ;)
Wait so 1 in 11.8 chance to not roll 35fcr 49 times in a row, but also 1 in 11 chance to roll it on first try
That’s fked up