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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?


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.

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  • $\begingroup$ Your equations seem to indicate order doesn't matter. All $P(E|K)=P(E)=0.9$ for any condition $K$. Where does the order come in? $\endgroup$ Sep 12, 2018 at 2:56
  • $\begingroup$ @herbsteinberg Hmmmmm what do you think of this... I think one way to solve the problem is you condition on the father choosing a dish first, thereby making "order matter" and then you use $P(E) = P(E \mid F) P(F) + P(E \mid F^c) P(F^c)$ which will solve the problem. But then order doesn't need to matter to solve this problem... so I suppose order doesn't matter unless you define it that way in how you arbitrarily condition it? $\endgroup$ Sep 12, 2018 at 3:00
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    $\begingroup$ Order matters if $P(E|F)=1$, that is the son chooses something other than what the father chooses. When the choices are independent, order does not matter. $\endgroup$ Sep 12, 2018 at 15:17
  • $\begingroup$ @herbsteinberg that is very helpful. When thinking about the sum of a two dice roll, you could condition on the outcome of the first roll (and call it the first outcome) but the outcome of each roll are independent so "order doesn't matter" for the sum of two dice $\endgroup$ Sep 12, 2018 at 17:50

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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.

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  • $\begingroup$ That is helpful... is this true: father choosing first is a different experiment then the son choosing first, but either way the final probability is the same and the son/father are just arbitrary labels. $\endgroup$ Sep 12, 2018 at 3:02
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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 $\frac9{10}$

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  • $\begingroup$ Thanks for reading through my thought process. I suppose if you look at all the set of possible outcomes then $9$ out of $10$ will be in the event that they choose different dishes. In this manner it's easy to see order isn't really part of the question... I think it's also a valid way to solve the problem by conditioning on the dish that one of them chooses, though perhaps order doesn't matter there. $\endgroup$ Sep 12, 2018 at 2:57
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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.

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