# Ben123

Karma: 2
• I think the agent should take the bet, and the double counting is actually justified. Epistemic status: Sleep deprived.

The number of clones that end up betting along with the agent is an additional effect of its decision that EDT-with update is correctly accounting for. Since “calculator says X” is evidence that “X = true”, selecting only clones that saw “calc says X” gives you better odds. What seems like a superfluous second update is really an essential step—computing the number of clones in each branch.

Consider this modification: All N clones bet iff you do, using their own calculator to decide whether to bet on X or ¬X.

This reformulation is just the basic 0-clones problem repeated, and it recommends no bet.

if X, EVT = ¯100 =
0.99 × N winners × \$10
− 0.01 × N losers × \$1000
if ¬X, EVT = ¯100 =
0.99 × N winners × \$10
− 0.01 × N losers × \$1000

Now recall the “double count” calculation for the original problem.

if X, EVT = 9900 = 0.99 × N winners × \$10
if ¬X, EVT = ¯10 = −0.01 × N losers × \$1000

Notice what’s missing: The winners when ¬X and, crucially, the losers when X. This is a real improvement in value—if you’re one of the clones when X is true, there’s no longer any risk of losing money.

• Is that a general solution? What about this: “Give me five dollars or I will perform an action, the disutility of which will be equal to twice that of you giving me five dollars, multiplied by the reciprocal of the probability of this statement being true.”