I assume most people have heard of Pascal's Wager, in case you have not: https://en.wikipedia.org/wiki/Pascal%27s_Wager

By the Stanford encyclopedia of philosphy:

"We have a decision under risk, with probabilities assigned to the ways the world could be, and utilities assigned to the outcomes. In particular, we represent the infinite utility associated with salvation as ‘∞’. We assume that the real line is extended to include the element ‘∞’, and that the basic arithmetical operations are extended as follows: For all real numbers r: ∞+r=∞. For all real numbers r: ∞×r=∞ if r>0."

Doesn't this contradict one of the axioms of the expected utility theory? In particular the archimedean property states:

Let l, l’ and l’’ be three lotteries in L such that $l \succ l’ \succ l’’$. Then there are $p,q \in(0,1)$ such that: $pl + (1-p)l’’ \succ l’ \succ ql + (1-q)l’’$

and it basically means that there are no infinitely preferred or infinitely despised lotteries.

But in the Pascal's Wager, $l$ = salvation = $ + \infty$, $l'$ = nothing (god doesn't exist), $l''$ = damnation = $ - \infty$, but then there does not exist a probability $p > 0$ such that any mixed combination of damnation and salvation is strictly preferred than "god doesn't exist".
What do you think?

  • 1
    $\begingroup$ Why do you assume damnation is $-\infty$? 0 seems to work just as well. $\endgroup$
    – Giskard
    Commented Dec 2, 2018 at 14:29
  • 1
    $\begingroup$ You're right, but even without it the archimedean property is still violated. $\endgroup$
    – plastico
    Commented Dec 2, 2018 at 15:44
  • $\begingroup$ Yes, you seem to be correct. The wager's lotteries also seem to violate continuity. $\endgroup$
    – Giskard
    Commented Dec 2, 2018 at 16:28


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