Let's have two 'parallel' but uncommingled arguments going on, if you don't mind. So, please do not mingle this post (or SUB-THREAD) with the one we have been having so far. Let's call this sub-thread the quantyst's approach.
Would you please just tell me where I went wrong with my solution? Thank you.
Answer: Probability of Sum being Even is 4/7 and Probability of Sum Being Odd is 3/7.
Here's why:
Let p denote the probability of the sum being even when the game stops.
Upon the first throw, we have Case (O) an odd roll with probability 1/2, and Case (E) an even roll with probability 1/2.
Under Case (O), we have three subsequent cases:
Case (O-SO) where the second roll is the same as first (odd and the same) with probability 1/6, at which point the process ends with an even sum. So, we reach this ending point with a net probability of (1/2)*(1/6)=1/12.
Case (O-DO) where the second roll is different but odd with probability 2/6, from which point onwards the probability of getting an even sum when the process ends is p as the first two rolls, both being odd, have an even sum. Thus, the net probability for this point is (1/2)*(2/6)*p=2p/12.
Case (O-E) where the second roll is even with probability 3/6, from which point onwards the probability of getting an even sum when the process ends is (1-p) as the first two rolls, one being odd and the other even, have an odd sum. The net probability for this point is (1/2)*(3/6)*(1-p)=(3/12)(1-p).
Under Case (E), we have three subsequent cases:
Case (E-SE) where the second roll is the same as first (even and the same) with probability 1/6, at which point the process ends with an even sum. So, we reach this ending point with a net probability of (1/2)*(1/6)=1/12.
Case (E-DE) where the second roll is different but even with probability 2/6, from which point onwards the probability of getting an even sum when the process ends is p as the first two rolls, both being even, have an even sum. Thus, the net probability for this point is (1/2)*(2/6)*p=2p/12.
Case (E-O) where the second roll is odd with probability 3/6, from which point onwards the probability of getting an even sum when the process ends is (1-p) as the first two rolls, one being even and the other odd, have an odd sum. The net probability for this point is (1/2)*(3/6)*(1-p)=(3/12)(1-p).
Combining all the above cases we have:
p=(1/12)+(2p/12)+(3/12)(1-p)+(1/12)+(2p/12)+(3/12)(1-p).
Solving the above equation, we get p=4/7.
