With 100 doors swapping wins 99 out of 100 times; the only time you lose is when your initial door (1 in 100) contained the prize.

With 100 doors switching should give you a 99% win rate.

You're essentially concentrating the entire thing into this one vs not this one, and when you initially chose there was a 99% chance it was not this one.

After Monty opens all the other doors, the odds that the right answer is not this one is still 99% except that now the entirety of not this one is represented by that single other door. The Grand Prize has nowhere else to be, and the odds that you picked it first is still only 1%.

So, to bring it back down, with three doors, the odds that the right answer is not this one 66%, and we end up exactly where we expected to be.

but I don’t understand the math of how to get that answer

There's four total outcomes of the problem:

Scenario 1: you originally pick the winning door (1/3) and don't switch (1/2), therefore winning. Probability = 1/6

Scenario 2: you originally pick the winning door (1/3) and did switch (1/2), therefore losing. Probability = 1/6

Scenario 3: you originally pick a losing door (2/3) and don't switch (1/2), therefore losing. Probability = 1/3

Scenario 4: you originally pick a losing door (2/3) and do switch (1/2), therefore winning. Probability = 1/3

Now consider scenarios 1 and 3 together, these two are when you don't switch. P(S1) is 1/6 and P(S3) is 1/3, meaning S3 is twice as likely than S1. So if you don't switch, you are twice as likely to lose. And now consider scenarios 2 and 4 together. P(S4) is 1/3 and P(S2) is 1/6, meaning if you switch you are twice as likely to win than to lose.

You can also consider this problem in terms of conditional probability like this:

P(win as long as no switch) = P(win and no switch) / P(no switch) = P(S1)/(1/2) = (1/6)/(1/2) = 2/6 = 1/3

P(win as long as switch) = P(win and switch) / P(switch) = P(S4)/(1/2) = (1/3)/(1/2) = 2/3

P(win as long as switch) > P(win as long as no switch)

The "second round" of the game is always just, "flip your odds of winning if you swap". That's all it is.

Monty will always open the proper doors to ensure this happens every time. Did you pick the winning door in the first round? Monty will eliminate all other doors but leave one of the losers. Did you pick a losing door in the first round? Monty will eliminate all the other losers and only leave the winner. It's always the opposite of what you picked. Therefore, if you swap, you will simply get the opposite odds of the first round.

100 doors to pick from, only 1 winner? 1/100 chance to win if you just picked at random and ended it there. Now Monty offers a swap. Without the swap, you have 99 different ways to lose this. But with the swap, all 99 of those ways become winners, because Monty will always swap the opposite with you.

I have no clue what this actually is about. But I always remember watching Deal or No Deal and thinking "If it was down to two boxes, £1 and £250K, I would absolutely swap my box." There is no way I would believe that all along I've been holding the 250K box. In my mind it makes more sense that I'm holding the £1 box and I need to swap.

But by staying on your door you're still making a choice relying on that ½ chance...