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As volatility goes to infinity, the delta of a call option goes to 1. The delta approximates the probability that the option expires in the money. So it seems that the probability of expiring in the money is very close to 1.

However, the price of the put option approaches the constant function at the strike price. This is what one would see if the probability of the call expiring out of the money is close to 1.

This seems to suggest that the probability that the call option expires in the money and out of the money is both close to 1.

Put another way, as the volatility increases, the probability of a call expiring in the money increases as well. But so does the price of a put option. So it would seem that the price of a put increases, even though the probability of the put expiring in the money decreases.

How to make sense of this?

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Where you are right, because of call-put parity $$C-P=DF(F-K)$$ with $F$ representing the forward price, the difference between the European call and a European put price is independent of the volatility.

This suggests that when $C$ increases due to increasing volatility, $P$ should therefore increase by the same amount all other things equal. And indeed the maximum put price is $DF\times K$ at which point the call is worth $DF\times F$.

Where you are wrong is that, under the risk-neutral measure: $\Delta$ (or rather $N(d_1)$ in the BS equation) is not the probability of expiring in the money for a call, $N(d_2)$ is.

See also these questions: Probability of exercise in the Black-Scholes Model and Yet another question about the risk-neutral measure. Why is the risk-neutral probability of an infinitely volatile GBM 0?

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