A puzzle

[Googolplex]

You have good handwriting, I'll give you that. However, I completely understand the situation so the explaination isn't necessary. I just don't agree with it. Supposedly it IS a paradox, but I just think people are overthinking and overcomplicating it. The end result is that there are two options, so the first pick can be disregarded from the conclusion.
1 / 3 = 0.33 = 33% = chance of getting right pick first round (but you have no idea whether you got it right)
2 - 1 = 1 = remove one of the wrong picks to remain with 1 right and 1 wrong.
1 / 2 = 0.5 = 50% = chance of getting right pick second round.

Thanks for my handwriting. But you think wrong. The first pick will not be affected by removing one door. Here is an other example: In a lottery you have a chance of 1/140.000.000 to win the jackpot. It's like you say one number between 1 and 140.000.000. For example you take the 10. Your chance is 1/140.000.000. Now I will remove all wrong numbers except the 11. Now there are only two possible numbers remaining. Do you really believe you win the jackpot with your first pick? Wouldn't it make sense to switch to the 11? Your chance isn't 50/50 now. It's still 1/140.000.000, but when you switch you will have a chance of 1/139.999.999. That's the logic behind this puzzle.

But it makes no sense to me.
If the host always reveals a wrong door before the final choice, then that door shouldn't even exist.

If there are 3 doors, but my final choice is between 2 of them, why does the third exist, when it has nothing to do with my choice?


The percentages that is shown in the video are from when there are 3 doors to pick from. When there are 3 doors, there are 3 different results. That means the 66% of winning the car, are 66% of 3 different results.

But Mugbill and I don't see 3 doors. We see 2 doors. With 2 doors there are 2 results. That gives us 50% chance of 2 results.


Taking your picture into account



End result
You are only choosing between 2 doors. One with a prize and one without. The last wrong choice is non-existant.

Id does no matter, if a door get removed or not. When you pick a door, you know that one of the others must be wrong. So what does the showmaster do? He just remove that wrong door. Your chance is still 1/3.

My sketch is correct. You couldn't summarize these two choices. These are not the same. Either you pick the first door, the second or the third. The sketch shows what happens then.

Yes, compare this. One of the doors can be excluded from the equation as it's always removed, leaving us with 50/50.

No, 50/50 is wrong. The correct answer is 1/3 and 2/3. You would have 50/50 when you would mix the two remaining doors and pick again.