2/3rds. If she's awake, one of three things happened:
1. The coin came up heads, and it's Monday.
2. The coin came up tails, and it's Monday.
3. The coin came up tails, and it's Tuesday.
Hence, 2:1 odds.
I'm not quite convinced yet. Those three things aren't equally likely to happen. Suppose the following scenario:
Instead of being interviewed twice with a mind-wipe inbetween, she's placed into an eternal Groundhog Day loop of mondays forever, her memories erased every night. (If the coin comes up heads, she's woken up and interviewed once, without time shenanigans, as before.) Now, there are infinite cases where the coin came up tails, and just one where it came up heads, so the probability of it having come up tails when Sleeping Beauty is woken is 1. The probability would, by this logic, remain the same if instead of a coin toss, we rolled a d100 and put her to a time loop on a one. Or a d10^100. As long as there is a nonzero chance that she's been put into a time loop, she can say with perfect certainty that she is in a time loop.
Applying this logic outside the thought experiment, we can conclude that the most likely explanation for it being Friday today in the real world is that the world is stuck in an infinite time loop of Fridays. >:]
Or to invert the example: An infinity-sided die is rolled. On any number but one, Sleeping Beauty is put to sleep and awakened on the next day, and the experiment ends. On one, Sleeping Beauty is put to sleep and awakened on the next day, mind-wiped, put to sleep again and reawakened on the next day. Since there are three cases
1) The die came up one, and it is day 1
2) The die came up one, and it is day 2
3) The die came up something other than one and it is day 1
The probability for the literal infinity-to-one chance having happened is 2/3.
Also I spent too long typing this and goblolo said the same thing.
Whatevs.