Beginner probability question about the phrase "order doesn't matter"

Question

This is a lame question but it's been bugging me for a while...

A father and his son are at a diner and each make one selection (randomly and independently) from a list of $10$ different dishes on a menu. What is the probability they choose different dishes?

Let $E$ be the event that they choose separate dishes. Let's condition on the father ordering first (call the event $F$) then,

$$P(E) = P(E \mid F) P(F) + P(E \mid F^c) P(F^c)=\left(\frac{1}{2}\right)\left(\frac{9}{10}\right)+\left(\frac{1}{2}\right)\left(\frac{9}{10}\right) = \frac{9}{10}$$

Question 1: I don't think you can say "order doesn't matter" for this problem. I think order does matter. Is that correct?

You could say "regardless of the order for this situation the conditional probabilities happen to be symmetrical, $P(E|F) = P(E|F^c) = 9/10$, and since the probability of the conditioning event is one half, the probability of the event in question is equal to just one of the conditional probabilities." That is $P(E) = P(E|F)$. Is that the right way to think about this problem?

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Let's say I condition on the dish ordered by the first person

$$\sum_{i=1}^{10} P(E \mid T=i)P(T=i) = \sum_{i=1}^{10} \frac{9}{10}\frac{1}{10} = 9/10$$

Question 2: Is this the same situation where "order does matter" it's just that the $P(E)$ is equal to either $P(E \mid T=i)P(T=i)$ summed for all $i$ whether the father or son goes first?

Question 3: If order does matter, then how does the problem change if you say they choose them exactly simultaneously? Thanks.

Answer 1

I don't understand what you are trying to express.

There are two seperate events: the father choose a dish, and the son choose a dish.

It doesn't matter whether they do it simutalneously, or one after the other, the probability is always $\frac9{10}$

Answer 2

We can say that the order doesn't affect the probability. Of course, father choosing then son would be a different experiment from its opposite. Father and son are just labels and do not affect the probability of the event.

If they do it simultaneously, then there are $10\times 10 = 100$ different outcomes, with $10$ where they choose the same dish. So $\frac{10}{100} = \frac{1}{10}$ of choosing the same dish, which is equivalent to $\frac{9}{10}$ of chosing different.

Answer 3

What is the chance that Father and Son choose the same dish? For that, we need to fix one dish, then compare the other. Say we know what Father chose, now the question basically is: Let the son pick a number from 1 to 10 and see if it's the one number chosen by Father: $\frac{1}{10}$ chance (and a $\frac{9}{10}$ chance of picking something else).

It doesn't matter if the Son or the Father 'picks first'; what matters here is that we take one of the outcomes as a reference, then compare the other to it. The results do not vary if we change the reference point.