“Naively in the actual Newcombe’s problem if omega is only correct 1⁄999,000+epsilon percent of the time…”
I’d like to argue with this by way of a parable. The eccentric billionaire, Mr. Psi, invites you to his mansion for an evening of decision theory challenges. Upon arrival, Mr. Psi’s assistant brings you a brandy and interviews you for hours about your life experiences, religious views, favorite philosophers, ethnic and racial background … You are then brought into a room. In front of you is a transparent box with a $1 bill in it, and an opaque box. Mr. Psi explains:
“You may take just the solid box, or both boxes. If I predicted you take one box, then that box contains $1000, otherwise it is empty. I am not as good at this game as my friend Omega, but out of my last 463 games, I predicted “one box” 71 times and was right 40 times out of 71; I picked “two boxes” 392 times and was right 247 times out of 392. To put it another way, those who one-boxed got an average of (40$1000+145$0)/185 = $216 and those who two-boxed got an average of (31$1001+247$1)/278=$113. ”
So, do you one-box?
“Mind if I look through your records?” you say. He waves at a large filing cabinet in the corner. You read through the volumes of records of Mr. Psi’s interviews, and discover his accuracy is as he claims. But you also notice something interesting (ROT13): Ze. Cfv vtaberf nyy vagreivrj dhrfgvbaf ohg bar—ur cynprf $1000 va gur obk sbe gurvfgf naq abg sbe ngurvfgf. link.
Still willing to say you should one-box?
By the way, if it bothers you that the odds of $1000 are less than 50% no matter what, I also could have made Mr. Psi give money to 99⁄189 one boxers (expected value $524) and only to 132⁄286 two boxers (expected value $463) just by hfvat gur lrne bs lbhe ovegu (ROT13). This strategy has a smaller difference in expected value, and a smaller success rate for Mr. Psi, but might be more interesting to those of you who are anchoring on $500.
“Naively in the actual Newcombe’s problem if omega is only correct 1⁄999,000+epsilon percent of the time…”
I’d like to argue with this by way of a parable. The eccentric billionaire, Mr. Psi, invites you to his mansion for an evening of decision theory challenges. Upon arrival, Mr. Psi’s assistant brings you a brandy and interviews you for hours about your life experiences, religious views, favorite philosophers, ethnic and racial background … You are then brought into a room. In front of you is a transparent box with a $1 bill in it, and an opaque box. Mr. Psi explains:
“You may take just the solid box, or both boxes. If I predicted you take one box, then that box contains $1000, otherwise it is empty. I am not as good at this game as my friend Omega, but out of my last 463 games, I predicted “one box” 71 times and was right 40 times out of 71; I picked “two boxes” 392 times and was right 247 times out of 392. To put it another way, those who one-boxed got an average of (40$1000+145$0)/185 = $216 and those who two-boxed got an average of (31$1001+247$1)/278=$113. ”
So, do you one-box?
“Mind if I look through your records?” you say. He waves at a large filing cabinet in the corner. You read through the volumes of records of Mr. Psi’s interviews, and discover his accuracy is as he claims. But you also notice something interesting (ROT13): Ze. Cfv vtaberf nyy vagreivrj dhrfgvbaf ohg bar—ur cynprf $1000 va gur obk sbe gurvfgf naq abg sbe ngurvfgf. link.
Still willing to say you should one-box?
By the way, if it bothers you that the odds of $1000 are less than 50% no matter what, I also could have made Mr. Psi give money to 99⁄189 one boxers (expected value $524) and only to 132⁄286 two boxers (expected value $463) just by hfvat gur lrne bs lbhe ovegu (ROT13). This strategy has a smaller difference in expected value, and a smaller success rate for Mr. Psi, but might be more interesting to those of you who are anchoring on $500.