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Why would the value of a call option go infinity as volatility goes to infinity?

I understand how you could solve this question by taking $\sigma \rightarrow \infty$ in the solution to the black scholes equation. However, I cannot understand this on a more heuristic level. Surely as volatility goes to infinity you also have a larger chance that the option finishes out of the money (and potentially by a long way)?

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  • $\begingroup$ The probability of finishing out-of-the money actually goes to one in the limit as the whole mass of the price distribution shifts towards zero. $\endgroup$ – LocalVolatility Apr 30 '18 at 12:22
  • $\begingroup$ @LocalVolatility exactly, so why would the price go to infinity? $\endgroup$ – Permian Apr 30 '18 at 12:44
  • $\begingroup$ Because of the asymmetry mentioned in the existing answer. Keep in mind that (when rates and dividends are zero), then the forward is equal to the spot. I.e. the average stock price is still the same as the current. A lot of mass near zero is balanced by a few extremely high values. $\endgroup$ – LocalVolatility Apr 30 '18 at 12:47
  • $\begingroup$ @LocalVolatility ok $\endgroup$ – Permian Apr 30 '18 at 16:13
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The stock price may fall a lot with high volatility, but you can only lose the price of the option if you bought the option. So the upside gets bigger but the downside is bounded.

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