As you're walking through the haunted woods behind your house, a troll waddles up and threatens to eat you. He will let you go however, if luck favors you. He tells you to pick a number from 1 to 6.
He puts 6 numbered blocks in a bag. (Trolls carry numbered blocks at all times.) He tells you that he will pull out a number from the bag - if the number is the number you had picked then you get eaten. If the number he pulls out is not the number you have picked then he will continue. The number does not get put back in the bag.
He will only pull 2 numbers out of the bag. What are the odds that you will be eaten?
Is this logic right?
$\frac16\to$ First round of numbers - this is the odds of getting eaten
$\frac15\to$ Second round of numbers - (since we're not replacing - our odds of getting picked goes up)
So odds are: $\frac16+\frac15=\frac{11}{30}$
Yes? No?
The logic is incorrect; probabilities don't add like that.
The probability that you will not get eaten is $$\frac56×\frac45=\frac23$$ where the $\frac56$ comes from the fact that five of the six numbers in the bag are not the one you picked, and similarly for the $\frac45$. Then the probability you are eaten is $1-\frac23=\frac13$.