Sleeping beauty problem and Monty Hall paradox I read this argument on the internet about how the solution to the sleeping beauty problem is $\frac{1}{3}$:
All these events are equally likely in the experiment : 


*

*Coin landed Heads, it's Monday and Beauty is awake

*Coin landed Heads, it's Tuesday and Beauty is asleep

*Coin landed Tails, it's Monday and Beauty is awake

*Coin landed Tails, it's Tuesday and Beauty is awake
All these are mutually exclusive and exhaustive and also equally likely. So, all four of these events have a probability $\frac{1}{4}$. But when Beauty is awakened, she knows that she isn't asleep. So, the second possibility can be ruled out. The rest three are still equally likely with a probability $\frac{1}{3}$. Hence the probability that the coin landed Heads is $\frac{1}{3}$.
But I remember something from the Monty Hall problem and this situation looks somewhat similar. The solution assumes that when possibility no. 2 is ruled out, the remaining three remain equally likely. This doesn't happen in the Monty Hall problem.
For example, there are 100 doors. A prize is behind one of them. Clearly, all the doors are equally likely to have the prize. We pick one random door. The probability that it has the prize is $\frac{1}{100}$. The probability that the prize is in one of the remaining doors is $\frac{99}{100}$. When doors from the set of remaining 99 doors are ruled out one by one, all the doors no longer remain equally likely. Our door still has the probability $\frac{1}{100}$ while the group of remaining doors still hold a  probability of $\frac{99}{100}$.
Could this be true for the Sleeping Beauty Problem too? I mean the possibilities 1. and 2. that I've listed collectively hold a probability of $\frac{1}{2}$ and even when possibility no.2 is ruled out, it's probability gets transferred to possibility no.1, so that it still has a probability of $\frac{1}{2}$.
EDIT: Suppose Beauty is the contestant on the Monty Hall Show. She is presented four doors in front of her, A, B, C, D. Clearly, all the door currently have a winning probability of $\frac{1}{4}$. But she knows that before the prize was put behind one of the doors, a coin was tossed. If it landed heads, the prize was placed in one of the doors A or B and in case it was tails, the prize was put in C or D. Beauty knows this. Now, the host rules out door B as a possibility (which is equivalent to Beauty ruling out possibility 2). Do the doors A, C and D remain equally likely to have the prize or is it safer to choose A?
I think it's safer to choose A because either you can assume the coin landed tails and further burden yourself in choosing between C and D or you can assume the coin landed heads and then choose A, the only remaining Heads door.
 A: What's happening in the sleeping beauty problem is much worse than what's happening in the Monty Hall problem. I claim that the solution to the sleeping beauty problem is that in a world where things like the sleeping beauty problem happen to you, there is no such thing as probability. 
One way to cash out what it means that you assign a thing happening some probability $p$ is in terms of bets: if someone were willing to offer you slightly more than $100(1 - p)$ dollars if the thing happens if you pay them slightly less than $100p$ dollars if the thing doesn't happen, you would take that bet, because the expected value would be positive, and $p$ is the largest number for which this is true. 
But in the sleeping beauty problem, because sleeping beauty will be 1) possibly awakened twice and 2) drugged so she doesn't remember the first awakening if the second one happens, there are effectively sometimes two copies of her when she's awakened twice. So which one of them do you offer the bet to?


*

*If you only offer bets to sleeping beauty the first time she awakens, sleeping beauty should accept offers of slightly more than $50$ dollars if the coin landed heads in exchange for paying slightly less than $50$ dollars if the coin landed tails; in other words, she should bet as if she believes the coin lands heads with probability $\frac{1}{2}$. 

*If you offer bets to sleeping beauty every time she awakens, sleeping beauty should accept offers of slightly more than $33$ dollars if the coin landed heads in exchange for paying slightly less than $66$ dollars if the coin landed tails (because in the heads world she gets to collect twice); in other words, she should bet as if she believes the coin lands heads with probability $\frac{1}{3}$.


My opinion is that once you notice this fact there is no longer any remaining question of what the probability "actually" is; there are only bets, and depending on the structure of the bets sleeping beauty will accept or reject accordingly. In summary:

In a world where you are effectively sometimes copied, it becomes ambiguous what it means to offer you a bet, because I have to decide which of your copies to offer the bet to. 

(Your copies also have to decide how much they care about each other; let's assume for simplicity that they're all perfectly altruistic wrt each other, although in general this is a further complication.) 
A: To add to Qiaochu's answer, the Monty Hall problem's solution depends on our understanding of the exact process the game follows. The assumption is that if you are standing in front of a door with the goat, Monty will always open the other door with the goat. What if instead the procedure that Monty will open either door you aren't standing in front of with equal probability and if the door has the prize, you lose (or win, for that matter)? What if the procedure is that he will open any of the three doors with equal probability (and if he opens your door you get what's behind it, and if he opens a door that isn't in front of you and the prize is behind it you lose/win). 
In either case there is no longer any advantage to switching. If you were on this game show and a door with a goat that you weren't in front of happened to open up, but you didn't know which procedure Monty was following, you would not know whether there was any value to switching$.^*$ Compare to Qiaochu's sleeping beauty.
$\;^*$ Well, I suppose one should view the fact that you were shown a goat behind a door that you aren't in front of as evidence that you are playing the canonical version of the game since this situation is most probable under this version. But then to calculate odds you would need to specify what the possible versions of the game are and prior probabilities for each of them.
