<p>I assume most people have heard of Pascal's Wager, in case you have not: <a href="https://en.wikipedia.org/wiki/Pascal%27s_Wager" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Pascal%27s_Wager</a>   </p>

<p>By the Stanford encyclopedia of philosphy:  </p>

<p><em>"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."</em></p>

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

<p><em>Let l, l’ and l’’ be three lotteries in L such that <span class="math-container">$l \succ l’ \succ l’’$</span>. Then there are <span class="math-container">$p,q \in(0,1)$</span> such that:
<span class="math-container">$pl + (1-p)l’’ \succ l’ \succ ql + (1-q)l’’$</span></em></p>

<p>and it basically means that there are no infinitely preferred or infinitely despised lotteries.</p>

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

CC BY-SA 4.0
2018-12-02T14:29:45.047Z
Why do you assume damnation is $-\infty$? 0 seems to work just as well.
Giskard
CC BY-SA 4.0
2018-12-02T15:44:58.760Z
You're right, but even without it the archimedean property is still violated.
plastico
CC BY-SA 4.0
2018-12-02T16:28:35.307Z
Yes, you seem to be correct. The wager's lotteries also seem to violate continuity.
Giskard
<decision-theory><expected-utility><rationality>
Does Pascal's Wager fail archimedean property?
plastico
# Does Pascal's Wager fail archimedean property?

(Asked by: username_0 on 2018-12-02)

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?

**Tags:** decision-theory, expected-utility, rationality
> **Comments:**
> **username_1**: Why do you assume damnation is $-\infty$? 0 seems to work just as well.  
> **username_0**: You're right, but even without it the archimedean property is still violated.  
> **username_1**: Yes, you seem to be correct. The wager's lotteries also seem to violate continuity.  

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## 0 answers

