4 ms·

This reminds me of the Monty Hall problem, but in reverse - you were playing the Monty Hall game and just won the car! You forgot whether or not you switched do

by acchow 11y ago

This reminds me of the Monty Hall problem, but in reverse - you were playing the Monty Hall game and just won the car! You forgot whether or not you switched doors during the second step - what is the probability you switched doors?

deleted 11y ago
[deleted]
- Nadya 11y ago
  I also related it to Monty Hall in that one option is two options bundled together. The options are not [Heads||Tails] but rather [Heads||Tails+Tails]. The Tails+Tails is like having the Car+Goat bundled as a single option. Sleeping Beauty knows the coin is 50/50 but that if she is woken there is more opportunity to be woken during a Tails flip. To show that it is biased in favor of Tails - the experiment must be repeated. It shows more readily if you increase the number of times she is woken if the coin lands on Tails, similar to how the Monty Hall problem becomes more intuitive when you increase the number of doors. Instead of being woken 2 times - let's have her woken 1,000 times. Every time she is woken she is asked if the coin was Heads or Tails, with no recollection of her previous answers. She is given a dollar for every time she is correct. Should she guess Heads or Tails? If she guesses Heads every time - she can only win $1. If she guesses Tails every time - she can win $1,000. If she alternates answers she can only win $500.
Bartweiss 11y ago
There's actually a strong analogy to an extended Monty Hall here. Specifically: 1. Monty offers the challenge, and I pick door B. 2. Monty opens door C, showing a goat. 3. Monty asks if I want to switch. My odds of winning are now P(A) = 2/3, P(B) = 1/3. 4. My mom flips on the TV to see me play, but has missed my answers and only sees the state of the doors. What are her odds of guessing right? Result: my mom's odds of winning are 1/2, even though the odds of a given door winning are not. There's a bias between the two doors, but her perspective is neutral - she's guessing which door has better odds, and that guess is unbiased. As you say, this question is the reverse: the odds on the coin flip are unbiased, but my odds of being asked the question are biased towards one of the two outcomes.
- CamperBob2 11y ago
  See my other post; I actually think that a Monty Hall comparison argues in favor of an unbiased 1:2 guess, because of how the terms of the two problems differ. In the SB question, your odds of being asked the question are 1:1. You just don't know how many times you'll be asked. You didn't know the night before (and neither did the researchers), and you still don't know, even though the researchers now do. In the MH game, you give the host some new information when you make your initial choice. He already knew what door not to open, but now he knows what door he must open, and the rules require him to communicate that to you. That's when you receive the new information (as you point out with the example of your mom walking into the room). In the SB problem, you don't get any new information before the question is asked, including whether or not you're going to be put back to sleep. So the only answer you can rationally give is 1:2. If the researchers used a d20 instead of a coin to determine how many times to wake you up, you could safely guess that any given awakening wasn't your first or your last. But you still can't give any answer about the number on the die, other than a random guess from 1-20. You need to store some information for later recall, and they're not letting you do that.
- darkmighty 11y ago
  I quite like your phrasing because it highlights that probability is a model of uncertainty for an observer. So the mother has no information on the door, so she is 'neutral' (p=1/2 of guessing right); you have a small amount of information (p=2/3); you can even include Monty which of course has total information (p=1). You can't ask what are the probabilities of finding goats behind each door without specifying an observer and the available information.
sidcypher 11y ago
The Monty Hall problem is actually much better for educating people, as it highlights the bias hidden behind the "50/50" oversimplification. This Sleeping Beauty problem, on the other hand, is phrased to confuse you, so that arguing between the "halfers" and "thirders" boils down to their different assumptions on what is asked, I think. Similarly, the question "if a tree falls in a forest and nobody's there, does it make a sound?" has two valid answers for two meanings of "sound", an objectivistic "pressure wave" vs. a (somewhat?)solipsistic "consciousness' hearing".

by acchow 11y ago

deleted 11y ago

[deleted]

Nadya 11y ago
I also related it to Monty Hall in that one option is two options bundled together. The options are not [Heads||Tails] but rather [Heads||Tails+Tails]. The Tails+Tails is like having the Car+Goat bundled as a single option. Sleeping Beauty knows the coin is 50/50 but that if she is woken there is more opportunity to be woken during a Tails flip. To show that it is biased in favor of Tails - the experiment must be repeated. It shows more readily if you increase the number of times she is woken if the coin lands on Tails, similar to how the Monty Hall problem becomes more intuitive when you increase the number of doors. Instead of being woken 2 times - let's have her woken 1,000 times. Every time she is woken she is asked if the coin was Heads or Tails, with no recollection of her previous answers. She is given a dollar for every time she is correct. Should she guess Heads or Tails? If she guesses Heads every time - she can only win $1. If she guesses Tails every time - she can win $1,000. If she alternates answers she can only win $500.

Bartweiss 11y ago

There's actually a strong analogy to an extended Monty Hall here. Specifically: 1. Monty offers the challenge, and I pick door B. 2. Monty opens door C, showing a goat. 3. Monty asks if I want to switch. My odds of winning are now P(A) = 2/3, P(B) = 1/3. 4. My mom flips on the TV to see me play, but has missed my answers and only sees the state of the doors. What are her odds of guessing right? Result: my mom's odds of winning are 1/2, even though the odds of a given door winning are not. There's a bias between the two doors, but her perspective is neutral - she's guessing which door has better odds, and that guess is unbiased. As you say, this question is the reverse: the odds on the coin flip are unbiased, but my odds of being asked the question are biased towards one of the two outcomes.

CamperBob2 11y ago
See my other post; I actually think that a Monty Hall comparison argues in favor of an unbiased 1:2 guess, because of how the terms of the two problems differ. In the SB question, your odds of being asked the question are 1:1. You just don't know how many times you'll be asked. You didn't know the night before (and neither did the researchers), and you still don't know, even though the researchers now do. In the MH game, you give the host some new information when you make your initial choice. He already knew what door not to open, but now he knows what door he must open, and the rules require him to communicate that to you. That's when you receive the new information (as you point out with the example of your mom walking into the room). In the SB problem, you don't get any new information before the question is asked, including whether or not you're going to be put back to sleep. So the only answer you can rationally give is 1:2. If the researchers used a d20 instead of a coin to determine how many times to wake you up, you could safely guess that any given awakening wasn't your first or your last. But you still can't give any answer about the number on the die, other than a random guess from 1-20. You need to store some information for later recall, and they're not letting you do that.
darkmighty 11y ago
I quite like your phrasing because it highlights that probability is a model of uncertainty for an observer. So the mother has no information on the door, so she is 'neutral' (p=1/2 of guessing right); you have a small amount of information (p=2/3); you can even include Monty which of course has total information (p=1). You can't ask what are the probabilities of finding goats behind each door without specifying an observer and the available information.

sidcypher 11y ago

The Monty Hall problem is actually much better for educating people, as it highlights the bias hidden behind the "50/50" oversimplification. This Sleeping Beauty problem, on the other hand, is phrased to confuse you, so that arguing between the "halfers" and "thirders" boils down to their different assumptions on what is asked, I think. Similarly, the question "if a tree falls in a forest and nobody's there, does it make a sound?" has two valid answers for two meanings of "sound", an objectivistic "pressure wave" vs. a (somewhat?)solipsistic "consciousness' hearing".