Consider rolling a 6 sided die until you roll a six. The expected number of rolls to get a six is 6.
True, but incomplete
Only because you deleted the next line that I wrote:
But if you only roll 4 times, you have about a 51% of seeing a six, even though you've rolled less than the expected number.
The whole point I am making is that there is a difference between "the expected number of rolls to see a 6" and "the number of rolls you need to have a 50% chance of seeing a 6".
Take your die: You want to roll a six. You roll 6 times. Whats the probability of not seeing the six?
5/6 ^ 6 = 33.48% So a third of the time (with six observations) you will not see the 6 at all.
Read about what "expected value" means.
It sounds like you think it means "most likely value", but it doesn't.
Take your 1000 sided die. Let's play a game. If you roll a 6, you win a million dollars, but if you roll anything else, you lose a dollar.
What's the expected value of rolling the die? Almost every time you roll (99.9% of the time), you lose a dollar. So you 'expect' to lose almost every time.
But that's not what expected value means. Your expected value for each roll is +$999.001:
(1 * 1e6 - 999 * 1) / 1000 = 999.001