Pascal's Wager (the new version)

[Jackalope]

Let's try to polish this turd and see what happens.

Pascal's Wager presents an argument that is oversimplified to the point of uselessness.

Let's look at it from an Expected Value (EV) point of view. The EV of accepting could be expressed like this:

EV = (P(g) * V(reward)) + ((1 - P(g)) * (-V(l))

where

P(g) = the probability that god exists
V(reward) = value of the reward earned from belief
V(l) = cost of being wrong, or the value of living one's life without incorrect belief.

(i.e. P(g) of the time, we win V(reward), otherwise we lose V(l))

As Pascal asserts that if one wagers and there is no god, one has lost nothing (e.g. V(l) is always zero, and the equation can be simplified to P(g) * V(reward). Pascal, in my view, is wrong to assert that.

As we have discussed ad nauseum in numerous threads here, Pascal ignores other relevant factors (possibly a result of presupposing that there's only one god, and it's his). The probability of winning the wager is NOT simply the probability that there is a god - presumably, one much choose the correct god, one's belief must be sufficiently sincere, and one must worship correctly:

EV = (F(ce) * V(reward)) + ((1 - F(ce)) * (-V(l))

F(ce) is "Cthulhu's Equation" -

F(ce) = P(a) * ( 1 / N ) * P(s) * P(p)

P(a) = the probability that any god exists.
N = the number of possible gods
P(s) = the probability that one's belief is sufficiently sincere to merit reward
P(p) = the "piety factor" - the probability that one's worship is sufficiently pious to merit reward

Many Christians and Muslims would probably argue that V(reward) = infinity, V(l) = 0, P(a) = 1.0, N = 1, P(s) = 1.0, and P(p) = 1.0. YMMV.

Where's my Nobel Prize?