The Sleeping Beauty 'Problem'

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Why the ‘Sleeping Beauty Problem’ Is Keeping Mathematicians Awake

A thought experiment that’s dividing mathematicians can help illuminate how belief shapes rational decisions

www.scientificamerican.com

The condensed version:
You put Sleeping Beauty to sleep on Sunday, and flip a fair coin. If it lands heads, you wake them on Monday, ask them the question (stated below), put them to sleep, and wake them Wednesday. If tails, you wake them Monday, put them to sleep, wake them Tuesday, put them to sleep, and then wake them Wednesday, choosing between Monday or Tuesday to ask the question. When Sleeping Beauty wakes up, they have no memory of whether they've been woken previously, are not told the results of the coin flip, nor told what day it is. They are asked for the probability that the coin showed heads.

As you may note from the scare quotes, I think the answer is obvious, and nor for any reasoning presented in the article from SA. I'll put my solution in spoiler tags in the next comment.

 

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Spoiler: My take on the answer

This problem becomes easier once you specify how the decision is made between Monday or Tuesday, should tails appear.

If tails-Monday is even with tails-Tuesday, then you have 50% chance of tails * a 50% chance of Monday, and a 50% chance of tails * a 50% chance of Tuesday, so the odds that Sleeping Beauty awakens on tails-Monday is 25%, and similarly for tails Tuesday. Add them together, and there is a 50% chance of tails, and therefore 50% of heads.

If you are more likely to choose Monday, e.g., 80% Monday versus 20% Tuesday if you flip tails, that becomes 50% chance of tails * a 80% chance of Monday = 40% chance of tails-Monday, and a 50% chance of tails * a 20% chance of Tuesday is a 10% chance for tails-Tuesday. 40% + 10% = 50% chance of tails, and 50% chance of heads.

No matter how you split up the probability of Monday/Tuesday, the odds of heads will be 50%.