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edited Jan 10, 2018 at 16:27

A few remarks that have not been highlighted yet in the other answers.

As vol goes higher, the value of an ATMF call and the value of an ATMF put will increase; initially pretty linear in vol, until they approach their limits (forward priceS for the call, present value of strike for the put), then they'll taper off towards said limit.
Since the value of both call and put go up, the reasoning that "it's more likely that the call will end up in the money" is fallacious. It's rather that when it ends up in the money, it'll be way in the money.
The probability that the call lands in the money will actually decrease as vol goes up. In fact, the value of a ATMF high digital (paying 1$ if S(T)>K) goes to zero as vol goes up, while the value of the low digital goes to present value of 1$. (When thinking about this, remember that the forward is kept constant!)
Option pricing works by hedging, that is replicating the option value. Every time you re-hedge a call (or put), you lose a bit (because of gamma). The higher the vol, the further the stock will move typically, so the more you lose. Thus, it costs more to produce a call (or put) when vol is higher. That's why its BS price increases with vol (until the limits are approached - and notice that there's no more gamma then).

A few remarks that have not been highlighted yet in the other answers.

As vol goes higher, the value of an ATMF call and the value of an ATMF put will increase; initially pretty linear in vol, until they approach their limits (forward price for the call, present value of strike for the put), then they'll taper off towards said limit.
Since the value of both call and put go up, the reasoning that "it's more likely that the call will end up in the money" is fallacious. It's rather that when it ends up in the money, it'll be way in the money.
The probability that the call lands in the money will actually decrease as vol goes up. In fact, the value of a ATMF high digital (paying 1$ if S(T)>K) goes to zero as vol goes up, while the value of the low digital goes to present value of 1$. (When thinking about this, remember that the forward is kept constant!)
Option pricing works by hedging, that is replicating the option value. Every time you re-hedge a call (or put), you lose a bit (because of gamma). The higher the vol, the further the stock will move typically, so the more you lose. Thus, it costs more to produce a call (or put) when vol is higher. That's why its BS price increases with vol (until the limits are approached - and notice that there's no more gamma then).

A few remarks that have not been highlighted yet in the other answers.

As vol goes higher, the value of an ATMF call and the value of an ATMF put will increase; initially pretty linear in vol, until they approach their limits (S for the call, present value of strike for the put), then they'll taper off towards said limit.
Since the value of both call and put go up, the reasoning that "it's more likely that the call will end up in the money" is fallacious. It's rather that when it ends up in the money, it'll be way in the money.
The probability that the call lands in the money will actually decrease as vol goes up. In fact, the value of a ATMF high digital (paying 1$ if S(T)>K) goes to zero as vol goes up, while the value of the low digital goes to present value of 1$. (When thinking about this, remember that the forward is kept constant!)
Option pricing works by hedging, that is replicating the option value. Every time you re-hedge a call (or put), you lose a bit (because of gamma). The higher the vol, the further the stock will move typically, so the more you lose. Thus, it costs more to produce a call (or put) when vol is higher. That's why its BS price increases with vol (until the limits are approached - and notice that there's no more gamma then).

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created Jan 10, 2018 at 16:16

A few remarks that have not been highlighted yet in the other answers.

As vol goes higher, the value of an ATMF call and the value of an ATMF put will increase; initially pretty linear in vol, until they approach their limits (forward price for the call, present value of strike for the put), then they'll taper off towards said limit.
Since the value of both call and put go up, the reasoning that "it's more likely that the call will end up in the money" is fallacious. It's rather that when it ends up in the money, it'll be way in the money.
The probability that the call lands in the money will actually decrease as vol goes up. In fact, the value of a ATMF high digital (paying 1$ if S(T)>K) goes to zero as vol goes up, while the value of the low digital goes to present value of 1$. (When thinking about this, remember that the forward is kept constant!)
Option pricing works by hedging, that is replicating the option value. Every time you re-hedge a call (or put), you lose a bit (because of gamma). The higher the vol, the further the stock will move typically, so the more you lose. Thus, it costs more to produce a call (or put) when vol is higher. That's why its BS price increases with vol (until the limits are approached - and notice that there's no more gamma then).