On 1/9/2013 2:14 PM, Quentin Anciaux wrote:
We start each with 100$ that we use to make the first bet, the column contains the $ we have in our pocket after the bet depending on the result.


I don't know what "Me" and "Brent" mean in this?  betting on survival or death?

bet n°  Experimenter survive    Experimenter die
bet n°  ME      BRENT   ME      BRENT
1       200$ (win 100$)         0$ (lost 100$)  0$ (lost 100$)  200$ (win 100$)
2       300$ (win 100$)         -100$ (lost 100$)       100$ (lost 100$)        
100$ (win 100$)
3       400$ (win 100$)         -200$ (lost 100$)       200$ (lost 100$)        
0$ (win 100$)
4       500$ (win 100$)         -300$ (lost 100$)       300$ (lost 100$)        
-100$ (win 100$)
...                             


All bets on the column 'experimenter die' are finals, no more bets can be put after because the experimenter is dead and won't revive.

Yes, that's why in the equation

    E = 0.99(-$100) + 0.01($100 + E)

there is no "+E" in the first parentheses; 99% of time there is no continuation.



Only on the first bet, do I lose money (yes it 99% of the resulting world *after and only* the first bet. But after that first bet if the experimenter has survived *all* next bet are winner bet (in *all* worlds weither the experimenter lives or not making it a final bet).

But his survival is rare, so all those good looking 200$, 300$,...are rare. You write outcomes, but with not probabilities - that's not how to calculate expected values. I stand by my analysis.

Brent


Regards, Quentin

2013/1/9 meekerdb <meeke...@verizon.net <mailto:meeke...@verizon.net>>

    On 1/9/2013 11:52 AM, Quentin Anciaux wrote:


    2013/1/9 meekerdb <meeke...@verizon.net <mailto:meeke...@verizon.net>>

        On 1/9/2013 3:10 AM, Quentin Anciaux wrote:

            Hi,

            let us start with the proposed QS experiment by Tegmark, a QS 
machine with
            a 99/100 chance of a *perfect* kill (so let's put aside HP failure 
or
            whatever so to have either the experimenter is killed with the given
            probabilities or it is not, no in between, so in 1/100 he is not 
killed and
            perfectly well, 99/100 he is killed).

            You are a witness of such experiment, and you're asked to make a 
bet on the
            experimenter surviving (or not).

            So you bet 100$, if you bet on the experimenter surviving, if he 
survive,
            you'll get 200$, if he does not you'll lose your bet, likewise if 
you bet
            on him die.

            What you should do contrary to what seems reasonable, is to bet on 
the
            experimenter will survive for the following reason:

            If MWI is true:

            1st Test: in 99/100 worlds you lose 100$ (and the bet ends here, 
there is
            no experimenter left for a second round), in 1/100 worlds you win 
200$
            2nd Test: well... you cannot play again in the 99/100 worlds where 
you did
            lose 100$, so you start already with 200$ in your pocket for this 
2nd test,
            so you should do the same, no here in 99/100 worlds, you did make a 
draw
            (you put 100$ in 1st test + 100$ win on the 1st test - 100$ you did 
lose
            now because the experimenter is dead), in 1/100 you win again 200$, 
that
            make 300$ in your pocket.

            >From the 3rd test on, you can only get richer, weither the 
experimenter
            lives from your POV or not.

            In QM+collapse, if the guy luckily survive two tests, you win 
money...
            you'll only lose money if he is killed at the first test.


            So contrary to what you may think, you should bet the experimenter 
should
            live, because in MWI, it is garanteed that you'll win money in a lot
            branches after only two succeeded test, and as in QM+collapse, only 
the
            99/100 of the first test lose money, all the others either make no 
loss or
            win money.


        Did you bother to calculate the expected value of playing this game?  
It's
        $98/0.99 whether you bet on survival or death.  And since $98/0.99<$100 
you had
        to start with, it's better not to play at all.


    ??

    you only lose on first bet if the experimenter die, which in MWI happens in 
99% of
    the worlds... so discounting that *first* and only bet where you lose, you 
win 100$
    every time till the experimenter die.

    On 2nd bet, you win nothing if the experimenter die (100$ (from first bet) 
+100$
    (from winning first bet)-100$(from losing second bet).

    At the third bet, you win 100$ if the experimenter die... and 100$ more 
every time
    you see the experimenter survive. Only on the first bet when the 
experimenter die
    you lose 100$ (and in that case, there is no more bet possible as there is 
no more
    experimenter).


    Let E=expected value of playing the game by always betting on survival

        E = 0.99(-$100) + 0.01($100 + E)

    Solve for E ==> E=98/0.99  Let F=expected value of playing the game and 
always
    betting on death

        F = 0.99($100) + 0.01(-$100 + F)

    solution is left as an exercise to the reader.

    Brent




    But after the second bet, all worlds following that 2nd bet if MWI is true,
    contains *only* winner witness.

    Quentin


        Brent


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