Q1: Is the answer the same if SB is left asleep on H+Mon, but wakened on H+Tue?
Q2: Is the answer the same if the day SB is left asleep, H+Mon or H+Tue, is determined by a second coin flip?
Q3: On Sunday Night, flip two coins. Call them C1 and C2. Coin C1 is the one SB is asked about, and C2 is the second on in Q2. So, on Monday wake SB if either coin is showing Tails. On Monday Night, turn C2 over to show the opposite side. And on Tuesday, again wake SB if either coin is showing Tails. Whenever awakened, what should SB answer about coin C1? But before answering that....
Q3A: At any point in time, what is the probability, to an outside observer, that the coins are showing HH? HT? TH? TT?
Q3B: Does SB know and agree with these probabilities?
Q3C: Does an awake SB have “new information” about this set of possibilities?
Q3D: Is the conditional probability that the coins are showing HT, the same as the conditional probability that coin C1 landed on Heads?
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A1: Yes.
A2: Yes.
Q3A: Each has a prior probability of 1⁄4.
Q3B: Yes.
Q3C: Yes, the combination HH is eliminated.
Q3D: Yes.
Q3: The conditional probability of {HT}, given {HT, TH, TT} is 1⁄3.
I really don’t know why you keep coming up with these alternative formulations of the same problem, an answer to which I’ve already given. Can’t you just apply the logic I’ve very explicitly described in the posts and derive the answer for your modified problem yourself? But surething, here is how one supposed to reason:
There are 4 equiprobable outcomes of the experiment for the Sleeping Beauty:
HH_HT; HT_HH; TH_TT; TT_TH
Meanwhile for the person who randomly observes one of the days of the experiment there are these 4 equiprobable outcomes:
HH; HT; TH; TT
When the person learns that the Beauty is awaken, outcome HH is eliminated and now they are 1⁄3 confident that the first coin is Heads.
For this problem the conditional probability of HT is indeed 1⁄3
This, however, gives no new information to the Sleeping Beauty herself as she is participating in a different probability experiment—the one with either one or two awakenings, where she observes them in order. She has different sample space and no outcome is eliminated by her awakening.
For this problem HT isn’t an event and, therefore, doesn’t have any probability and the answers to Q3C and Q3D are “No”.
Q1: Is the answer the same if SB is left asleep on H+Mon, but wakened on H+Tue?
Q2: Is the answer the same if the day SB is left asleep, H+Mon or H+Tue, is determined by a second coin flip?
Q3: On Sunday Night, flip two coins. Call them C1 and C2. Coin C1 is the one SB is asked about, and C2 is the second on in Q2. So, on Monday wake SB if either coin is showing Tails. On Monday Night, turn C2 over to show the opposite side. And on Tuesday, again wake SB if either coin is showing Tails. Whenever awakened, what should SB answer about coin C1? But before answering that....
Q3A: At any point in time, what is the probability, to an outside observer, that the coins are showing HH? HT? TH? TT?
Q3B: Does SB know and agree with these probabilities?
Q3C: Does an awake SB have “new information” about this set of possibilities?
Q3D: Is the conditional probability that the coins are showing HT, the same as the conditional probability that coin C1 landed on Heads?
+-+-+-+-+
A1: Yes.
A2: Yes.
Q3A: Each has a prior probability of 1⁄4.
Q3B: Yes.
Q3C: Yes, the combination HH is eliminated.
Q3D: Yes.
Q3: The conditional probability of {HT}, given {HT, TH, TT} is 1⁄3.
I really don’t know why you keep coming up with these alternative formulations of the same problem, an answer to which I’ve already given. Can’t you just apply the logic I’ve very explicitly described in the posts and derive the answer for your modified problem yourself? But surething, here is how one supposed to reason:
There are 4 equiprobable outcomes of the experiment for the Sleeping Beauty:
HH_HT; HT_HH; TH_TT; TT_TH
Meanwhile for the person who randomly observes one of the days of the experiment there are these 4 equiprobable outcomes:
HH; HT; TH; TT
When the person learns that the Beauty is awaken, outcome HH is eliminated and now they are 1⁄3 confident that the first coin is Heads.
See the Observer Problem.
For this problem the conditional probability of HT is indeed 1⁄3
This, however, gives no new information to the Sleeping Beauty herself as she is participating in a different probability experiment—the one with either one or two awakenings, where she observes them in order. She has different sample space and no outcome is eliminated by her awakening.
For this problem HT isn’t an event and, therefore, doesn’t have any probability and the answers to Q3C and Q3D are “No”.