Friday, June 12, 2015

Even more on Fine Tuning

The central portion of the Fine Tuning Argument (FTA) is valid: Assuming that $P(L|C)=1$ by definition, then $P(C|L)={P(L|C)P(C)}/{P(L)}≥P(C)$. Assuming that $0<P(C)<1$ and $P(L)<1$ then $P(C|L)>P(C)$. Observing that life exists definitely raises the ex ante probability that a creator exists for this world. Life is, in a sense, "evidence for" a creator.

However, to accept this argument as meaningful, we have to assume that $P(C)<1$. This assumption seems to require that we bite a serious philosophical bullet. If we take a frequentist ontology, then $P(C)<1$ means that there exist possible worlds where no creator exists. This contradicts one theistic definition of God as a modally necessary being: if God exists, God exists in all possible worlds. In this case, $P(C)=1$, and $P(C|L)=1$; because we were already certain, by definition, that God existed. Under a Bayesian ontology, $P(C)$ represents our subjective confidence that God exists, and $P(C)<1$ entails that we can reasonably be in doubt that God exists, which contradicts at least Plantinga's argument that God as properly basic: $C⇒P(C)=1$. Thus, if the FTA is rationally persuasive, then God must be neither modally necessary nor properly basic. But suppose we bite these philosophical bullets.

Another portion of the FTA is also valid: Assuming that $P(L|~C)<P(L)$, then $P(~C|L)={P(L|~C)P(~C)}/{P(L)}<P(~C)$: Observing that life exists definitely lowers the ex ante probability that no creator exists for this world.

But so what? We really want to know whether or not $P(C)>P(~C)$. The question is not whether observing life makes the race closer, we want to know if observing life changes the winner. We want to know if $P(C|L)>P(~C|L)$. Because we want to know if this particular inequality is true that we must condition on $L$. The FTA does not, however, help us answer this question. Regardless of how small $P(L|~C)$ is, we do not know whether or not it is less than $P(C)$, and the FTA requires that $P(C)>P(L|~C)$. If we simply assume that $P(C)>P(L|~C)$, then we can just as simply assume that $P(C)<P(L|~C)$ (neither assumption entails a contradiction), so the FTA works only on an arbitrary choice of assumptions, which essentially begs the question. And, according to Ikeda and Jeffries, we have a plausible argument that $P(C|L&F)<P(C|L)$, because $P(~F|C&L)>0$. If we assume that $P(F|C)=1$, then $P(F|C)=P(L|C)$ and we're back where we started.

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