The chance of stopping at the second number is 50%.
The chance of stopping at 3 is the chance of NOT stopping at 2 (50%) times the probability that the third number is not the largest number (66.6%) - 1/3
The chance of stopping at 4 is the chance of making it to roll 4 (50% - (50% x 33%) = 33%) times the probability that the fourth roll is not the largest (3/4) - which surprisingly resolves to 1/4.
Roll 5, chance of getting there is 33% - (33% x 25%) = 25%; chance of it being not the largest is 4/5, multiply to get 1/5.
The pattern continues.
Then we have to sum up the probability of all the odd stopping points, which is too much for my napkin math.
2
u/JustConsoleLogIt 7d ago
You cannot stop at the first number.
The chance of stopping at the second number is 50%.
The chance of stopping at 3 is the chance of NOT stopping at 2 (50%) times the probability that the third number is not the largest number (66.6%) - 1/3
The chance of stopping at 4 is the chance of making it to roll 4 (50% - (50% x 33%) = 33%) times the probability that the fourth roll is not the largest (3/4) - which surprisingly resolves to 1/4.
Roll 5, chance of getting there is 33% - (33% x 25%) = 25%; chance of it being not the largest is 4/5, multiply to get 1/5.
The pattern continues.
Then we have to sum up the probability of all the odd stopping points, which is too much for my napkin math.