probability vs combinatorics interpretation of problems I have some difficulty to pass from a combinatorics solution of a problem to a faster possible solution with just the use of probability . For example :
100 people seat at random on a plane with 100 seats while they also have a ticket with an assigned seat , what is the probability to seat on your assigned seat if you are the 34th to get on the plane?
If you think about it with combinatorics it is simple to come up with the right answer : there is just a place where you can seat and 99! ways to put the rest of people while the total number of possible arrangements is 100! so the answer is 1/100
But I find it more difficult to think about it in terms of probability and so to get the same answer with 'less' effort (if I'm the 34th to seat how can the probability stay the same as if I was the first one?)
 A: From a probabilistic point of view, the event that I, as the 34th passenger, sit in my correct seat is the same as the event that I sit in my correct seat, and nobody before me sits in my seat. Let $A_i$ be the event that passenger $i$ does not sit in my seat, and $B$ be the event that I do sit in my seat. Then we want:
$$P(A_1\cap A_1 \cap \cdots \cap A_{33} \cap B).$$
We can expand this as
$$P(A_1)P(A_2\mid A_1)\cdots P(A_{33}\mid A_1\cap \cdots \cap A_{32})P(B \mid A_1 \cap \cdots \cap A_{33}).$$
Since there are $100$ seats and $99$ of them are not mine, $P(A_1 ) = 99/100$. Given $A_1$ (i.e., passenger 1 did not sit in my seat), there are now $99$ unoccupied seats and $98$ unoccupied seats that are not mine, so $P(A_2 \mid A_1) = 98/99$. Similarly, $P(A_3 \mid A_1 \cap A_2) = 97/98$, and so on. The 33rd person has $68$ unoccupied seats to choose from, of which $67$ are not mine, so $P(A_{33} \mid A_1 \cap \cdots \cap A_{32}) = 67/68$.
Finally, we need $P(B \mid A_1 \cap \cdots \cap A_{33}$). We are given that $33$ people sat in seats that were not mine. So there are $67$ unoccupied seats, one of which is mine, so this probability is $1/67$. Putting it all together we have:
$$\frac{99}{100}\cdot \frac{98}{99} \cdots \frac{67}{68}\cdot \frac{1}{67} = \frac{1}{100}.$$
