Interesting. The natural approach is to imagine that you just have a 3-sided die with 2, 4, 6 on the sides, and if you do that, then I compute A = 12 and B = 6[1]. But, as the top Reddit comment’s edit points out, the difference between that problem and the one you posed is that your version heavily weights the probability towards short sequences—that weighting being 1/2^n for a sequence of length n. (Note that the numbers I got, A=12 and B=6, are so much higher than the A≈2.7 and B=3 you get.) It’s an interesting selection effect.
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.
To illustrate with an extreme version, consider:
A: The expected number of rolls of a fair die until you roll two 6s in a row, given that you succeed in doing this. You ragequit if it takes more than two rolls.
Excluding odd rolls completely, so the die has a 1⁄3 chance of rolling 6 and a 2⁄3 chance of rolling an even number that’s not 6, we have:
A = 1 + 1⁄3 * A2 + 2⁄3 * A
Where A2 represents “the expected number of die rolls until you get two 6′s in a row, given that the last roll was a 6”. Subtraction and multiplication then yields:
A = 3 + A2
And if we consider rolling a die from the A2 state, we get:
A2 = 1 + 1⁄3 * 0 + 2⁄3 * A = 1 + 2⁄3 * A
Substituting:
A = 3 + 1 + 2⁄3 * A => (subtract) 1⁄3 * A = 4 => (multiply) A = 12
Interesting. The natural approach is to imagine that you just have a 3-sided die with 2, 4, 6 on the sides, and if you do that, then I compute A = 12 and B = 6[1]. But, as the top Reddit comment’s edit points out, the difference between that problem and the one you posed is that your version heavily weights the probability towards short sequences—that weighting being 1/2^n for a sequence of length n. (Note that the numbers I got, A=12 and B=6, are so much higher than the A≈2.7 and B=3 you get.) It’s an interesting selection effect.
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.
To illustrate with an extreme version, consider:
Obviously that’s one way to reduce A to 2.
Excluding odd rolls completely, so the die has a 1⁄3 chance of rolling 6 and a 2⁄3 chance of rolling an even number that’s not 6, we have:
A = 1 + 1⁄3 * A2 + 2⁄3 * A
Where A2 represents “the expected number of die rolls until you get two 6′s in a row, given that the last roll was a 6”. Subtraction and multiplication then yields:
A = 3 + A2
And if we consider rolling a die from the A2 state, we get:
A2 = 1 + 1⁄3 * 0 + 2⁄3 * A
= 1 + 2⁄3 * A
Substituting:
A = 3 + 1 + 2⁄3 * A
=> (subtract)
1⁄3 * A = 4
=> (multiply)
A = 12
For B, a similar approach yields the equations:
B = 1 + 1⁄3 * B2 + 2⁄3 * B
B2 = 1 + 1⁄3 * 0 + 2⁄3 * B2
And the reader may solve for B = 6.