Beauty is put to sleep and a coin is tossed. On heads she is woken once, on tails twice, and each time her memory of any earlier waking is erased. When she wakes she doesn't know what day it is. What should her chance of heads be? Lock in your answer, then test it with bets, a simulation and the explanations.
Beauty is put to sleep and a coin is tossed. On heads she is woken once, on tails twice, and each time her memory is erased. Only a third of all awakenings follow heads, but half of all experiments do. Philosophers have argued since 2000: "thirders" count awakenings, "halfers" count experiments. Below you can see which question each camp is answering and whose bets pay off.
This is one of the few puzzles on the site where serious researchers still disagree. So instead of a single "right answer" we show where each number comes from and which question it actually answers.
Beauty is put to sleep. She knows all the rules in advance. While she sleeps, the researchers toss a fair coin.
On heads she is woken on Monday, asked a question and put back to sleep until the experiment ends on Wednesday.
On tails she is woken on Monday and on Tuesday. After Monday she is given a drug that erases the memory of that waking. Tuesday looks to her exactly like Monday.
Each time she wakes she is asked: "How confident are you that the coin landed heads?" She doesn't know whether it is Monday or Tuesday, or whether she was woken before. What probability should she give?
"The coin is fair and I learned nothing new. So 1/2"
That is how the "halfers" argue, the philosopher David Lewis among them. But count the awakenings: of three possible ones, only one follows heads. So the "thirders", like Adam Elga, say 1/3. Both numbers are correct answers to different questions, and bets show when you need which.
We write H for heads and T for tails. Monday and Tuesday: Mon and Tue.
Put yourself in her place: you have just woken and don't know what day it is. Lock in your answer before reading on.
A breakdown of your answer will appear here.
Hard to decide? Open the hints one at a time: the easy one first, then stronger ones.
Imagine the experiment repeated 1,000 times. Count two things: how many experiments there were and how many awakenings.
Heads gives one awakening, tails gives two. What share of all awakenings follows heads, and what share of all experiments?
500 heads experiments give 500 awakenings, 500 tails experiments give 1,000. Think about which share Beauty is asked about when she is woken and asked afresh each time.
We toss the coin and wake you once or twice. Each time you bet $1 on heads or tails. Two cashiers settle the bets differently. The "per awakening" cashier takes $1 at every awakening. The "per experiment" cashier takes $1 for the whole experiment, so on tails each of the two awakenings is worth 50 cents. See which strategy wins at which cashier.
| Strategy | Per awakening | Per experiment |
|---|
Your confidence in heads is the price at which the ticket neither gains nor loses for you. Average profit per ticket at each price, on your own experiments:
| Price | Bought at every awakening | One per experiment |
|---|
The computer runs experiments one after another and keeps two tallies: the share of experiments that were heads, and the share of awakenings that came after heads. Same coins, same awakenings; only the denominator differs.
The dispute isn't about arithmetic: everyone agrees that heads accounts for 1/3 of awakenings and 1/2 of experiments. The argument is over which of these numbers Beauty should give when asked for "the chance of heads".
On waking, Beauty could be in one of three situations: H-Mon, T-Mon, T-Tue. T-Mon and T-Tue are indistinguishable to her, so equally likely. And if she were told "it is Monday", heads and tails would have to be equal: the coin could even be tossed after the Monday waking, and a fair coin gives 1/2. So all three situations are equally likely, and heads gets 1/3. That was Adam Elga's argument in 2000.
On Sunday the chance of heads is 1/2. On waking, Beauty gets no information she didn't have on Sunday: she knew she would certainly be woken. Probabilities change only with new information, so it stays 1/2. That was David Lewis's reply to Elga in 2001.
The price of Lewis's view: once Beauty is told it is Monday, she must find heads more likely, 2/3, even though the coin may not have been tossed yet. Many find that strange, and that is where most of the debate concentrates.
| Experiments | Awakenings | |
|---|---|---|
| Heads | 500 | 500 |
| Tails | 500 | 1,000 |
| Heads share | 500/1000 = 1/2 | 500/1500 = 1/3 |
"In what share of experiments did the coin land heads?" The answer is 1/2. "After what share of awakenings is the coin showing heads?" The answer is 1/3. In 2008 the philosopher Berry Groisman argued that both camps are right and simply answer different questions. The words "chance of heads" allow both readings.
If Beauty bets twice on tails and once on heads, her tails bets multiply. Then the calibrated confidence is 1/3: at a price of 1/2 she loses 1/6 per ticket on average. If only one bet per experiment counts, the right price is 1/2. That is what you saw in round 2. This is why the philosopher Christopher Hitchcock argued in 2004 that a halfer can be beaten by a series of bets, while halfers reply that per-awakening bets simply count tails twice.
The share of awakenings after heads: heads gives 1 awakening, tails gives 2, each with chance ½.
The share of experiments with heads: each experiment counts once, however many awakenings it has.
Elga: the three situations are equally likely, because T-Mon can't be told from T-Tue, and on Monday the coin may not even be tossed yet.
Lewis: heads ½, and tails is split between the two days, ¼ each. Told that it is Monday, the halfer gets:
If tails brings k awakenings, the share of awakenings after heads is:
None of these formulas is wrong. The difference is what counts as "one case": an awakening or an experiment. A Beauty who is asked afresh each time naturally answers about awakenings. A Beauty who only cares about the coin itself answers about the experiment.
From one coin to questions that look quite different: how many people are yet to be born, when to leave the highway, and how to survey patients. Everywhere the same fork: count worlds or observers, people or visits.
The philosopher Nick Bostrom proposed an extreme version: on tails Beauty is woken a million times. Choose how many awakenings tails brings and watch the two camps drift apart.
With a million awakenings the thirder is nearly sure of tails: the chance of heads is 1/1,000,001. A fair coin, and she bets on tails at a million to one. The halfer holds 1/2, but once told "this is your first awakening" she must give heads 1,000,000/1,000,001, becoming almost certain of heads because of the word "first".
Each camp ends up with its own awkward conclusion. Some, Bostrom among them, defend "double halfing": 1/2 on waking and 1/2 after hearing "Monday", at the cost of dropping the usual rule for updating probabilities. No version has convinced everyone yet.
The astrophysicist Brandon Carter in 1983, and later the philosopher John Leslie, reasoned like this. You are roughly the hundred-billionth human ever born. Compare two futures: 200 billion people in total (early doom) or far more (a long future). If you treat yourself as a random person from your world, being the hundred-billionth is far more likely in the small world. That is halfer logic: each world counts once, and you are a random point inside it.
Nick Bostrom named the two approaches: SSA (treat yourself as a random observer in your world) and SIA (a world with more observers is in itself more likely to contain you). SSA matches the halfers and produces the doomsday shift. SIA matches the thirders: it weights each world by its number of people, and that weight cancels the shift exactly. Under SIA the chance of early doom equals the prior.
So the one-coin Sleeping Beauty question is a model of whether your own birth rank can tell you anything about humanity's end.
In 1997 Michele Piccione and Ariel Rubinstein described a driver on a night highway with two exits, unable to tell the first from the second. Turning off at the first scores 0. At the second, 4: that is home. Passing both scores 1. The driver can't remember passing an exit, so behaves the same at each: drive on with probability p.
The best plan made at home: p = 2/3, score 4/3. But at an exit the driver asks: "Where am I, first or second?" If he gives the first a 3/5 chance and optimises just this moment, he gets p = 1/3. It is the same fork as Beauty's: count the drive or each exit. The same year Robert Aumann, Sergiu Hart and Motty Perry showed that if the driver remembers his decision acts at both exits, the 2/3 plan stays best.
A clinic has 1,000 patients. Half come once a year, half twice. Survey whoever is in the waiting room over the year, and each visit yields one questionnaire. The share of once-a-year patients among questionnaires is 500/1500 = 1/3; among patients it is 1/2. That is exactly Beauty: the rare visitor is heads, the frequent one is tails with two awakenings.
Statisticians call this length-biased or frequency-biased sampling and fix it with weights: a frequent visitor's questionnaire counts with weight 1/2. The same effect explains why the bus you wait for seems later than the timetable says: the inspection paradox.
The problem was devised by the American philosopher Arnold Zuboff, who taught at University College London. It circulated in unpublished work from the mid-1980s and was printed in his 1990 paper "One Self". His version had many days: on one coin result a waking every day, on the other on just one random day.
Piccione and Rubinstein published the absent-minded driver in the game theory journal Games and Economic Behavior. The same paper has an example very close to Sleeping Beauty, three years before Elga's paper.
Adam Elga published a five-page paper in the journal Analysis and answered 1/3. He credited the name "Sleeping Beauty" to Robert Stalnaker, who had learned of such examples from Zuboff's unpublished work. A year earlier the puzzle had been argued over in the rec.puzzles newsgroup.
The next year, in the same journal, David Lewis, one of the most influential philosophers of the 20th century, replied to Elga: 1/2. More than a hundred papers have followed, and the camps are still called thirders and halfers.
Nick Bostrom, author of a book on anthropic reasoning, proposed the version with a million awakenings on tails. It sharpens both positions: the thirder bets a million to one against a fair coin, and the halfer, once told "Monday", is almost sure of heads.
Peter Winkler, a well-known collector of mathematical puzzles, went through every camp's arguments in "The Sleeping Beauty Controversy" for the American Mathematical Monthly. He titled his talk on it "The Math Puzzle that Spawned 100 Philosophy Papers".
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