My intuition is from the six points in Kahan's post. If the next flip is heads, then the flip after is more likely to be tails, relative to if the next flip is tails. If we have an equal number of heads and tails left, P(HT) > P(HH) for the next two flips. After the first heads, the probability for the next two might not give P(TH) > P(TT), but relative to independence it will be biased in that direction because the first T gets used up.

Is there a mistake? I haven't done any probability in a while.

I would two-box on this problem because of diminishing returns, and one-box on the original problem.

Your returns must be

very rapidlydiminishing. If u is your kilobucks-to-utilons function then you need [7920u(1001)+80u(1)]/8000 > [3996u(1000)+4u(0)]/4000, or more simply 990u(1001)+10u(1) > 999u(1000)+u(0). If, e.g., u(x) = log(1+x) (a plausible rate of decrease, assuming your initial net worth is close to zero) then what you need is 6847.6 > 6901.8, which doesn't hold. Even if u(x) = log(1+log(1+x)) the condition doesn't hold.If we fix our origin by saying that u(0)=0 (i.e., we're looking at utility

changeas a result of the transaction) and suppose that at any rate u(1001) <= 1001/1000.u(1000), which is certainly true if returns are always diminishing, then "two-boxing is better because of diminishing returns" implies 10u(1) > 8.01u(1000). In other words, gaining $1M has to be no more than about 25% better than gaining $1k.Are you

sureyou two-box because of diminishing returns?