The way this works is that in a one-step, two-state model, you can hedge. Prof. Joshi does this a bit in his book, but you see a bunch of this in Shreve's Volume I (which I'll be honest I never bothered to read). Here's how you might think about it. Note: We are given asset C as the riskless asset, and it's worth 1 both today and tomorrow so you can think of this as a case where interest rates are 0.
The way to approach this, in a general setting, is to take some initial wealth, determine how many shares of stock you need to buy to hedge the risk, and take the remaining amount and invest in the money market. Let X(0) be how much you have initially, g be the number of shares (since I don't have a delta on my keyboard; sorry, you'll just have to deal), S(0) the value of asset A at time 0, and S(1) the value of asset A at time 1.
Then our portfolio at time 1 is worth:
gS(1) + (X(0)-gS(0)) = X(0) + g(S(1)-S(0))
We want to choose X(0) and g so that X(1) models the payoff of our option.
Let's use your example, where asset A has value 110 with probability 99% (to save me from writing nasty decimals, thank you) and value 90 with probability 1%.
Then we have two cases:
X(0) + 10g = 10
X(0) - 10g = 0
Solving for this gives us g=1/2, and X(0) = 5.
Now to purchase a unit of asset A costs $100. As you see above, we need g=1/2, which would cost $50. We also saw that X(0) = 5; then it costs $50 to buy half a share of asset A and we would need to short 45 units of asset C. So what are your possible values of our portfolio at time 1?
In the up case, the value of asset A will be 110, so 1/2 a share will be worth 55. In the down case, half a share will be worth 45. So X(1) now models perfectly the payoff of this call option struck at 100 (it pays 10 on an up-move and pays 0 on a down-move). This suggests an option price of 5 should work.
So now what? We need to introduce probabilities into the picture. However, the reason why your real-world probability choice is irrelevant is because of this ability to hedge. The fact that we can hedge does, in effect, remove the risk associated with those probabilities (that's kind of what hedging means
Let us now take X(0) our initial wealth and g the number of shares we purchase. We want to determine X(0) and g needed to perfectly model this system, using probabilities. In an up-move, X(1) = X(0) + (110-100)g = 10 and in a down-move, X(1) = X(0) + (90-100)g = 0
If we take two numbers p and q with p+q=1, then we can actually take:
X(0) + g[p(110) + q(90) - 100] = p(10)-q(0)
(we find this by just multiplying X(1) by p in an up-move case and by q in a down-move case and add them up). If we choose p so that p(110) + q(90) - 100 = 0, then we have chosen p so that 5 = 10p, which shows that p=1/2 afterall.
In general, suppose you have a set amount of wealth X(0); then you can do a similar trick by determining how many shares of your asset you would need to purchase in order to hedge your risk. After doing so, you should be able to replicate your option by constructing a portfolio that matches its payoff, and then use the above trick to determine how much the option would cost by determining the expected value of X(1) which can be done using those values of p and q you determine above.
So to summarize, this is indeed fictitious, but in some sense it actually isn't, because the ability to perfectly hedge your risk allows you to remove that risk. That's why you can use risk-neutral pricing. Note that this is why Prof. Joshi said that you needed asset C; without a riskless asset to work with, you can't hedge your risk and therefore you can really talk about risk-neutral pricing.
Anyways, I think that more or less is what is going on. Sorry for answering questions on here, but I'm using this as a chance to test my own understanding also. Maybe Prof Joshi could tell me if I'm more or less right here?