# Does Pascal's Wager fail archimedean property?

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?

• Why do you assume damnation is $-\infty$? 0 seems to work just as well. – Giskard Dec 2 '18 at 14:29
• You're right, but even without it the archimedean property is still violated. – Zhang_anlan Dec 2 '18 at 15:44
• Yes, you seem to be correct. The wager's lotteries also seem to violate continuity. – Giskard Dec 2 '18 at 16:28