Does setting records in a sport become less likely (under certain assumptions)? Problem
Let $X_{j}$ be the number of seconds $j$-th swimmer takes from one end of the pool to the other, where $X_{1}, X_{2}, ...$ are i.i.d. with a continuous distribution. We will say that the $j$-th swimmer sets a records if $X_{j}$ is greater than all of $X_{j-1}, ..., X_{1}.$
Is the event that "the 110th swimmer sets a record" independent of the event that "the 111th swimmer sets a record"?
Approach: Prove $P(I_{111}=1, I_{110}=1) = P(I_{111}=1)P(I_{110}=1)$ where $I_{j}$ is an indicator random variable for the $j$-th person setting a record.
$P(I_{j} = 1) = \frac{1}{j}$, since by i.i.d. properties, all of the first $j$ swimmers are equally likely to set the record.
$P(I_{111}=1, I_{110}=1) = \frac{109!}{111!}$, since we fix the 111-th and 110-th swimmers in record setting positions.
Then, 
$$P(I_{111}=1, I_{110}=1) = \frac{109!}{111!} = \frac{1}{111 * 110} = P(I_{111} = 1) * P(I_{110} = 1).$$
Thus, the events in question are independent.
Intuition
Suppose a million swimmers participate in the competition. If the $999999$-th swimmer sets a record, the swimmer probably completed the task in an incredible short period of time, since $999998$ swimmers before him had chances to set records. This means that for the $1000000$-th swimmer to set a new record, they will probably need to complete the task in an extremely unlikely amount of time. So, the events in question are dependent. 
Question
I am having trouble reconciling my intuition with the result obtained after doing the computation. I have a suspicion the key is in the fact that $X_{j}$ are i.i.d., Any pointers?
Note
The question comes from strategic practice and homework 4 of Stat110.
 A: Here is another intuitive argument:


*

*One of the $110$ initial swimmers went faster than the other $109$.  Let's look at who was fastest and how long they took

*By symmetry, each of them was equally likely to be fastest, each with probability $\frac{109!}{110!}=\frac1{110}$, since each of the $110!$ orders of these swimmers are assumed equally likely

*By a similar symmetry argument, the time taken by the fastest is independent of which one of the $110$ was fastest 

*For the $111$th swimmer to set a new record, they have to go faster than all the previous $110$ swimmers, which has probability $\frac{110!}{111!}=\frac{1}{111}$ by considering the $111!$ possible orders assumed to be equally likely

*The event of the $111$th swimmer setting a new record is therefore independent of the $110$th swimmer setting the previous record


I suspect this fits better with your simulation.  
The conditional probability that the $111$th swimmer sets a new record given that the $110$th  swimmer has set the previous record is $\dfrac{\frac{1\times 1 \times 109!}{111!}}{\frac{1\times 109!}{110!}}=\frac{1}{111}$, which is the same as the unconditional probability that the $111$th swimmer set a new record.
