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Probability Riddle--are you a "halfer" or a "thirder"?
- Raphael 18y agoI am a halfer. 1/2 1/2 / \ | Mon Tue Mon 1/2 1/2 1 T T H I think I have identified the thirder problem. They are assuming that it's Monday: Mon Tue 3/4 1/4 / \ | 2/3 1/3 T | | H T
- GavinB 18y agoYour first diagram makes it look like something is split between Monday and Tuesday, when it's not split at all. You get both of those "tails" interviews, in sequence. So you get asked two times for the answer when it is tails, and only once when it is heads. Try doing the experiment several times and see how well you do picking heads every time you "wake up." There's no need to have memory alteration if you set your strategy beforehand. You'll only be right 1/3 of the time if you bet on heads.
- GavinB 18y agoOn Heads, sleeping beauty gets asked once. On Tails, she gets asked twice. If I walk into the room and see her being asked, it's a better bet that the coin landed tails. If this experiment is repeated many times, "Tails" will be the right answer twice as often. You can try this with a friend. As someone pointed out later on in the page, you just answer the first question freely and if it's tails are required to answer the second question the same way as you did the first. Or just answer heads every time--no memory wiping required. If it's really 50/50 then answering heads should get you your answer. Because the interviewer asks questions in an unbalanced manner, the interviewee actually gets more information just by being asked.
- alex_c 18y agoThe way I read it, the problem is ambiguous. 1/2 is the answer to "What is the probability that the coin landed Heads?" 1/3 is the answer to "What is the average probability of you getting it right if you say the coin landed Heads?" So I guess that makes me a "Can you rephrase the question?-er"
- psyklic 18y agoDefinitely. If your goal is to maximize the number of correct guesses across all questionings, then always answer 1/3. If your goal is to guess the current coin flip as accurately as possible, then always answer 1/2. (I interpret the question as this.)
- emmett 18y agoTry thinking about it this way instead: If the coin comes up tails, you will be asked 1,000,000 times in a row. If the coin comes up heads, you have a 1/1,000,000 chance of being asked. I wake you up. If you had to guess if the coin was heads or tails, what would you say?
- GavinB 18y agoThe reason that people are confused is because of the word "probability." I guess they don't realize that probability is simply a tool that tells you what to guess.
- psyklic 18y agoYou are proposing the 1/3 argument -- by answering tails, she will get it right more often across trials. But, 1/2 of the time she will be awoken once, and 1/2 she will be awoken 1,000,000 times. So, let's suppose she is awoken. There are 1,000,001 possible awakenings. The probability of her being in the first of these is 1/2 (Heads). The probability of her being in any particular other is 1/2*(1/1,000,000) (Tails). Hence, she should answer 1/2 if she wants to guess the current coin flip correctly, since the chance of the coin being Heads on any arbitrary awakening is 1/2.
- sysop073 18y agoI feel like I must be missing something, although I suppose all good paradoxes give you that feeling. If it's heads, she's asked once (Monday) what it was, if tails, she's asked twice (Monday and Tuesday). Therefore, if she has no idea which of the three states she's currently in, there's a 1 in 3 chance it was heads
- psyklic 18y agoThere are three states, but they are not equiprobable. One-half of the time she'll be asked once, and one-half of the time she'll be asked twice. Hence, the one-half argument is because she'll only be in the "once" (Heads) state 1/2 of the time.
- zasz 18y agoThey are equiprobable. When the coin is flipped, it is 50% likely that it will be heads or tails. When it's tails, Beauty must be interviewed again. Therefore, the number of times she's in the state Monday+tails is equal to the number of times she's in the state Tuesday+tails. However, the number of times she's in the state Monday+tails is equal to the number of times she's in the state Monday+heads. No other possible states can occur, so these three states are equiprobable. The answer is that at any point in time, absent other information, the coin is equally likely to be heads or tails, but as soon as Beauty receives the new information that she's being interviewed, it changes the known information, making it 1/3 instead of 1/2.
- psyklic 18y agoThe three states are not equiprobable. Suppose that she must say instead which particular state she is most likely to be in: M+T, T+T, or M+H. Well, we know that Heads is chosen 1/2 of the time, so she should choose M+H! After all, the probability she is in M+T is only 1/4, and the probability she is in T+T is only 1/4. (Since Tails is chosen 1/2 of the time.)
- Eliezer 18y agoI don't know.
- deleted 18y ago[deleted]
- emmett 18y agoYou cannot be woken up on tuesday with heads; that's part of the problem definition. So there are 3 ways to be woken up, and two of them are in the tails case.
- deleted 18y ago[deleted]
- ars 18y agoAre you reading a different riddle? We're reading the second green one on that page.
- ars 18y agoNo so simple, if it's Heads she does not get woken up on Tuesday at all. So it's thirds: There are two way to be woken up for tails, and one way for heads.
- deleted 18y ago[deleted]
- hugh 18y agoIn a slightly more evil thought experiment (Harvey Dent might approve), we could make one slight change: if it's heads we interview her once and then let her go, but if it's tails we'll keep on waking her and interviewing her every day for a thousand years. I think this version makes it clearer: if it's heads she'll be interviewed once, if it's tails she'll be interviewed thousands of times. At any given interview she should assign a very small probability to the possibility of heads.
- cousin_it 18y agoWanna fry your brains? Some time ago I was intrigued to learn about the non-probabilistic variant of the two envelopes problem: http://en.wikipedia.org/wiki/Two_envelopes_problem#A_non-probabilistic_variant http://en.wikipedia.org/wiki/Two_envelopes_problem#A_non-pro...