Previous formulation here. (There’s a link to the original formulation from there.)
I showed a related problem to someone and got back the objection “well, that’s a coordination problem- you need game theory to model the other players, and so you can’t simply declare a strategy correct.” At first I thought that was an evasion, and so reformulated it so you make all the decisions. It seems to me that this problem is isomorphic to the previous formulation, except switching from nay to yay is more clearly ridiculous (if you disagree, I’d like to hear it!). Here it is:
To simplify calculations, assume you are risk-neutral with regards to dollars at this scale. You provide me a byte of 8 bits, all zeros or ones. I flip a fair coin: if it lands heads, I select one bit from the byte you provided me uniformly at random; if it lands tails, I select seven bits from the byte you provided me uniformly at random.
If all bits selected are 0s, I pay you $21. For every selected bit that is a 1, I pay you $4.
You do some byte-level calculations before I flip the coin, and decide that the byte 00000000 is best, because it has an expected value of $21/2+$21/2=$21. The coin is flipped, and then without telling you the flip I give you the option to flip every bit that I will select (or you imagine this, and submit the opposite byte instead). Bit-level reasoning suggests you should flip all bits, as each bit impacts the total result in 8⁄16 cases, and in seven of those cases the coin came up tails. 7/8*$28+1/8*$4=$25>$21. Byte-level reasoning suggests you shouldn’t flip all the bits, because the byte 11111111 has an expected value of $4/2+$28/2=$16<$21. This is the conflict in the formulation linked above- before you know you’re a decider, you think “nay” is superior,but once you know you’re a decider, you calculate that “yay” is superior.
But this still has a coordination problem- the $21 payoff requires every bit to be a 0. What happens when we get rid of that?
Now, if I get heads, I pay out $21 for a 0 and $4 for a 1. If I get tails, I pay out $3 for each 0 and $4 for each one. E[00000000]=21, E[11111111]=16 (as before), but now E[0]=21*1/8+3*7/8=5.25 and E[1]=4*1/8+4*7/8=4. Now, by precisely the same amount as the byte-level analysis, E[0]>E[1]! Indeed, this appears to be general.1
Is there a way to formulate an Uncoordinated Psy-Kosh’s Non-Anthropic Decision Problem? (UPKNADP for short.) Or is the wrinkle in it solely that the individual analysis stumbles when it comes to dealing with coordinated action? Note the issue isn’t modeling other players- you’re making the moves for all players in the game, and can model yourself perfectly. The issue is how you count the rewards associated with coordinated action.
1.You provide me with n bits. I flip a fair coin: if it lands heads, I select one bit uniformly at random and pay out a for every 0; if it lands tails, I select n-1 bits uniformly at random and pay out b/n-1 for every 1. I can always add or subtract a constant amount to each prospect without changing the strategy for a risk-neutral player, meaning I can simplify the payout matrix down to just a and b.
The byte-level2 calculations suggest that E[n 0s]=a*1/2+(n-1)*0*1/2=a/2 and E[n 1s]=0*1/2+(n-1)b/2(n-1)=b/2. Your best strategy is to pick 0s if a>b and 1s if b>a (and you’re indifferent if they’re the same).
The bit-level calculations suggest that E[0s]=a*1/n+0*(n-1)/n=a/n. E[1s]=0/n+b/(n-1)*(n-1)/n=b/n. Again, the payoff ratio is the exact same- you should pick 0s if a>b and 1s if b>a (and you’re indifferent if they’re the same).
The byte-level and bit-level calculations agree, for all values of a and b and all n>1.
2. I suppose I shouldn’t use “byte” to refer to a set of n bits, but I’d rather have my cake and eat it too by both using byte and having a general set of n bits.
Revisiting Psy-Kosh’s Non-Anthropic Decision Problem
Previous formulation here. (There’s a link to the original formulation from there.)
I showed a related problem to someone and got back the objection “well, that’s a coordination problem- you need game theory to model the other players, and so you can’t simply declare a strategy correct.” At first I thought that was an evasion, and so reformulated it so you make all the decisions. It seems to me that this problem is isomorphic to the previous formulation, except switching from nay to yay is more clearly ridiculous (if you disagree, I’d like to hear it!). Here it is:
To simplify calculations, assume you are risk-neutral with regards to dollars at this scale. You provide me a byte of 8 bits, all zeros or ones. I flip a fair coin: if it lands heads, I select one bit from the byte you provided me uniformly at random; if it lands tails, I select seven bits from the byte you provided me uniformly at random.
If all bits selected are 0s, I pay you $21. For every selected bit that is a 1, I pay you $4.
You do some byte-level calculations before I flip the coin, and decide that the byte 00000000 is best, because it has an expected value of $21/2+$21/2=$21. The coin is flipped, and then without telling you the flip I give you the option to flip every bit that I will select (or you imagine this, and submit the opposite byte instead). Bit-level reasoning suggests you should flip all bits, as each bit impacts the total result in 8⁄16 cases, and in seven of those cases the coin came up tails. 7/8*$28+1/8*$4=$25>$21. Byte-level reasoning suggests you shouldn’t flip all the bits, because the byte 11111111 has an expected value of $4/2+$28/2=$16<$21. This is the conflict in the formulation linked above- before you know you’re a decider, you think “nay” is superior,but once you know you’re a decider, you calculate that “yay” is superior.
But this still has a coordination problem- the $21 payoff requires every bit to be a 0. What happens when we get rid of that?
Now, if I get heads, I pay out $21 for a 0 and $4 for a 1. If I get tails, I pay out $3 for each 0 and $4 for each one. E[00000000]=21, E[11111111]=16 (as before), but now E[0]=21*1/8+3*7/8=5.25 and E[1]=4*1/8+4*7/8=4. Now, by precisely the same amount as the byte-level analysis, E[0]>E[1]! Indeed, this appears to be general.1
Is there a way to formulate an Uncoordinated Psy-Kosh’s Non-Anthropic Decision Problem? (UPKNADP for short.) Or is the wrinkle in it solely that the individual analysis stumbles when it comes to dealing with coordinated action? Note the issue isn’t modeling other players- you’re making the moves for all players in the game, and can model yourself perfectly. The issue is how you count the rewards associated with coordinated action.
1.You provide me with n bits. I flip a fair coin: if it lands heads, I select one bit uniformly at random and pay out a for every 0; if it lands tails, I select n-1 bits uniformly at random and pay out b/n-1 for every 1. I can always add or subtract a constant amount to each prospect without changing the strategy for a risk-neutral player, meaning I can simplify the payout matrix down to just a and b.
The byte-level2 calculations suggest that E[n 0s]=a*1/2+(n-1)*0*1/2=a/2 and E[n 1s]=0*1/2+(n-1)b/2(n-1)=b/2. Your best strategy is to pick 0s if a>b and 1s if b>a (and you’re indifferent if they’re the same).
The bit-level calculations suggest that E[0s]=a*1/n+0*(n-1)/n=a/n. E[1s]=0/n+b/(n-1)*(n-1)/n=b/n. Again, the payoff ratio is the exact same- you should pick 0s if a>b and 1s if b>a (and you’re indifferent if they’re the same).
The byte-level and bit-level calculations agree, for all values of a and b and all n>1.
2. I suppose I shouldn’t use “byte” to refer to a set of n bits, but I’d rather have my cake and eat it too by both using byte and having a general set of n bits.