The thing is that, if you roll a 6 and then a non-6, in an “A” sequence you’re likely to just die due to rolling an odd number before you succeed in getting the double 6, and thus exclude the sequence from the surviving set; whereas in a “B” sequence there’s a much higher chance you’ll roll a 6 before dying, and thus include this longer “sequence of 3+ rolls” in the set.
Yes! This kind of kills the “paradox”. Its approaching an apples and oranges comparison.
Surviving sequences with n=100 rolls (for illustrative purposes)
A: The probability that you roll a fair die until you roll two 6s in a row, given that all rolls were even.
B: The probability that you roll a fair die until you roll two non-consecutive 6s (not necessarily in a row), given that all rolls were even.
This changes the code to:
A_estimate = num_sequences_without_odds/n
B_estimate = num_sequences_without_odds/n
And the result (n=100000)
Estimate for A: 0.045 Estimate for B: 0.062
I guess this is what most people where thinking when reading the problem, i.e., its a bigger chance of getting two non consecutive 6s. But by the wording (see above) of the “paradox” it gives more rolls on average for the surviving sequences, but you on the other hand have more surviving sequences hence higher probability.
Yes! This kind of kills the “paradox”. Its approaching an apples and oranges comparison.
Surviving sequences with n=100 rolls (for illustrative purposes)
[6, 6]
[6, 6]
[2, 6, 6]
[6, 6]
[2, 6, 6]
[6, 6]
Estimate for A: 2.333
[6, 6]
[4, 4, 6, 2, 2, 6]
[6, 6]
[6, 2, 4, 4, 6]
[6, 4, 6]
[4, 4, 6, 4, 6]
[6, 6]
[6, 6]
Estimate for B: 3.375
if you rephrase
A: The probability that you roll a fair die until you roll two 6s in a row, given that all rolls were even.
B: The probability that you roll a fair die until you roll two non-consecutive 6s (not necessarily in a row), given that all rolls were even.
This changes the code to:
A_estimate = num_sequences_without_odds/n
B_estimate = num_sequences_without_odds/n
And the result (n=100000)
Estimate for A: 0.045
Estimate for B: 0.062
I guess this is what most people where thinking when reading the problem, i.e., its a bigger chance of getting two non consecutive 6s. But by the wording (see above) of the “paradox” it gives more rolls on average for the surviving sequences, but you on the other hand have more surviving sequences hence higher probability.