Expected value of dice game (with options)

Setup

The question (modified from an old Jane Street interview):

(i) Two people throw a 6-sided dice. The person with a larger number wins $1. If the dice values are equal, neither person wins a reward. How much are you willing to pay to play this game?

(ii) A followup: say you now have the option to pay $0.25 to add 2 to the number you get. (You can choose this option after rolling your dice.) How much are you willing to pay to play this game now?

Part (i) seems easy: $E[game] = \frac{1}{6}(0 + \frac{1}{6} + \frac{2}{6} + \frac{3}{6} + \frac{4}{6} + \frac{5}{6}) = 15/36$. I was less sure about part (ii).

Question: what's the solution to part (ii)?

My initial hunch was that I should always exercise the option, even if I rolled a $6$ (since this would guarantee I win a dollar, as opposed to a $5/6$ chance). This would suggest $E[game]= \frac{1}{6}(\frac{2}{6} + \frac{3}{6} + \frac{4}{6} + \frac{5}{6} + \frac{6}{6} + \frac{6}{6}) - \frac{1}{4} = \frac{17}{36}$ is the game' value.

However, other answers I've seen online claim "when you roll 1, 2, 3, 4, 5, you should exercise the option; when you roll to 6, you shouldn't." This would suggest $E[game]= \frac{1}{6}(\frac{2}{6} + \frac{3}{6} + \frac{4}{6} + \frac{5}{6} + \frac{6}{6} + \frac{5}{6}) - \frac{5}{6} * \frac{1}{4} = \frac{35}{72}$. I'm unsure why/if this claim is true.

If you have rolled a $6$ and do not pay to add $2$, your expectation is $\frac 56 \approx 0.83$ because you win $1$ unless your opponent has also rolled a $6$. By paying $0.25$ you increase this to $1$, so your expectation of rolling a $6$ and buying the $2$ is $1-0.25=0.75$. As this is less than $0.83$ you should not buy the $2$.

Both calculations of the game's value with the option are correct for the strategy employed. The fact that the second is greater than the first is another way to show that the second strategy is superior.