Betting arguments—including the “expected value of the lottery ticket” I saw when skimming this—are invalid since it is unclear whether there is exactly one collection opportunity, or the possibility of two. You can always get the answer you prefer by rearranging the problem to the one that gets the answer you want.
But the problem is always stated incorrectly. The original problem, as stated by Adam Elga in the 2000 paper “Self-locating belief and the Sleeping Beauty problem,” was:
“Some researchers are going to put you to sleep. During [the experiment], they will briefly wake you up either once or twice, depending on the toss of a fair coin (Heads: once; Tails: twice). After each waking, they will put you to back to sleep with a drug that makes you forget that waking. When you are [awakened], to what degree ought you believe that the outcome of the coin toss is Heads?”
Elga introduced the two-day schedule, where SB is always wakened on Monday, and optionally wakened on Tuesday, in order to facilitate his thirder solution. You can argue whether his solution is to the same problem or not. But if it is not, it is the variation problem that is wrong. And it is unnecessary.
First, consider this simplified experiment:
SB is put to sleep.
Two coins, C1 and C2, are arranged randomly so that the probability of each of the four possible combinations, {HH, HT, TH, TT} has a probability of 1⁄4.
Observe what the coins are showing:
If either coin is showing Tails:
Wake SB.
Ask her for her degree of belief that coin C1 is showing Heads.
Put SB back to sleep with amnesia.
If both coins are showing heads:
Do something else that is obviously different than option 3a.
Make sure SB is asleep, and can’t remember option 3b happening.
Note that if SB is asked the question in option 3a, she knows that the observation was that at least one coin is showing Tails. It does not matter what would happen—or if anything would happen—in 3b. Her answer can only be 1⁄3.
We can implement the original problem by flipping these two coins for one possible awakening in the original problem, and then turning coin C2 over for another.
Betting arguments—including the “expected value of the lottery ticket” I saw when skimming this—are invalid since it is unclear whether there is exactly one collection opportunity, or the possibility of two. You can always get the answer you prefer by rearranging the problem to the one that gets the answer you want.
But the problem is always stated incorrectly. The original problem, as stated by Adam Elga in the 2000 paper “Self-locating belief and the Sleeping Beauty problem,” was:
“Some researchers are going to put you to sleep. During [the experiment], they will briefly wake you up either once or twice, depending on the toss of a fair coin (Heads: once; Tails: twice). After each waking, they will put you to back to sleep with a drug that makes you forget that waking. When you are [awakened], to what degree ought you believe that the outcome of the coin toss is Heads?”
Elga introduced the two-day schedule, where SB is always wakened on Monday, and optionally wakened on Tuesday, in order to facilitate his thirder solution. You can argue whether his solution is to the same problem or not. But if it is not, it is the variation problem that is wrong. And it is unnecessary.
First, consider this simplified experiment:
SB is put to sleep.
Two coins, C1 and C2, are arranged randomly so that the probability of each of the four possible combinations, {HH, HT, TH, TT} has a probability of 1⁄4.
Observe what the coins are showing:
If either coin is showing Tails:
Wake SB.
Ask her for her degree of belief that coin C1 is showing Heads.
Put SB back to sleep with amnesia.
If both coins are showing heads:
Do something else that is obviously different than option 3a.
Make sure SB is asleep, and can’t remember option 3b happening.
Note that if SB is asked the question in option 3a, she knows that the observation was that at least one coin is showing Tails. It does not matter what would happen—or if anything would happen—in 3b. Her answer can only be 1⁄3.
We can implement the original problem by flipping these two coins for one possible awakening in the original problem, and then turning coin C2 over for another.