You failed to see that the independent events are conditionally dependent on having a particular outcome. Since you like the dice example:
Say I have a cup with 2 fair 6 sided dice. I roll them and turn over the cup so you can't see them. I peek under the cup, look at both dice, and without changing either die I slide out one of the dice to reveal that is a 6. Do you think the odds of the other dice being a 6 is still 1/6? No, it's 1/11 and here's the breakdown:
Odds of not rolling any 6s: 25/36 (I can't reveal a 6, so these cases are impossible as I showed you at least one 6)
Odds of rolling exactly 1 6: 10/36 (I reveal the 6 and the other die is not a 6)
Odds of rolling exactly 2 6s: 1/36 (I reveal either 6 and the other die is a 6)
It's like a mini Monty Hall problem. The key is that I have advanced knowledge of the outcome all the events and selectively reveal information. The only way the die I revealed is a 6 and the in the cup is a 6 is if I rolled double 6s to begin with (odds 1/36), and by revealing one of the a dice as a 6 I eliminate 25 of the possible cases to make the remaining odds of double 6s 1/11, which is still worse than 1/6 on an independent roll.