z2newton
Joined: Mar 22, 2014
• Posts: 62
June 25th, 2015 at 10:38:19 PM permalink
I was reading a blog where it talked about this probability puzzle. I'd never heard of the Sleeping Beauty problem before so I thought I'd post it here to see what the more math inclined have to say about it. From what I can gather there is not a consensus on the answer.

[http://en.wikipedia.org/wiki/Sleeping_beauty_problem]

Sleeping Beauty volunteers to undergo the following experiment and is told all of the following details: On Sunday she will be put to sleep. Once or twice, during the experiment, Beauty will be awakened, interviewed, and put back to sleep with an amnesia-inducing drug that makes her forget that awakening. A fair coin will be tossed to determine which experimental procedure to undertake: if the coin comes up heads, Beauty will be awakened and interviewed on Monday only. If the coin comes up tails, she will be awakened and interviewed on Monday and Tuesday. In either case, she will be awakened on Wednesday without interview and the experiment ends.

Any time Sleeping Beauty is awakened and interviewed, she is asked, "What is your belief now for the proposition that the coin landed heads?"

I guess they knock her out before she can ask what day it is.

Since this is a gambling forum here is an interesting twist I found. The experiment is the same except Sleeping Beauty is given the chance to make a bet before the experiment begins. She is given 3 to 2 odds that the coin will come up heads. Instead of the original question, when she is awakened she is asked if she wants to cancel her bet.

To make it interesting lets say she bets \$1000 on heads.
RS
Joined: Feb 11, 2014
• Posts: 8626
June 25th, 2015 at 10:53:47 PM permalink
What are we trying to solve?
FleaStiff
Joined: Oct 19, 2009
• Posts: 14484
June 26th, 2015 at 1:25:18 AM permalink
whether she should have a greater than even guess as to the proper answer to give.
Canyonero
Joined: Nov 19, 2012
• Posts: 509
June 26th, 2015 at 7:08:38 AM permalink
She should always guess "tails". She will be right on Mondays and Tuesdays and will be wrong only on Mondays. 66% correct.
z2newton
Joined: Mar 22, 2014
• Posts: 62
June 26th, 2015 at 9:13:22 AM permalink
Quote: Canyonero

She should always guess "tails". She will be right on Mondays and Tuesdays and will be wrong only on Mondays. 66% correct.

Heads is 1/3? So it's like the monty hall problem.
RS
Joined: Feb 11, 2014
• Posts: 8626
June 26th, 2015 at 11:37:05 AM permalink
Not like monty hall. Well maybe?

But if it's tails, then you'll get interviewed more often than if it's heads.

I can't wait until Alan and his dice crew start saying it's 50/50 because there are 2 sides of a coin...
MangoJ
Joined: Mar 12, 2011
• Posts: 905
June 27th, 2015 at 12:57:49 AM permalink
Quote: z2newton

I guess they knock her out before she can ask what day it is.

What is your question anyway ?

If she gets the interview and doesn't know anything about previous interviews or the day, 2/3 will be tails and 1/3 will be heads.
You can solve all those questions with Bayesian inference.
SDSDNSR
Joined: Apr 27, 2015
• Posts: 12
July 9th, 2015 at 8:03:33 PM permalink
Let's take a specific example. On Sunday Sleeping beauty is offered to bet on a coin flip. If the result is heads, she wins \$64,000. If it is tails, she loses \$128,000. She will not learn the result, however, until after she has gone through the sleeping and waking process. If the result is heads, she will be put to sleep and awakened on Monday and asked if she would like to cancel the bet, then put back to sleep until Wednesday. If the result is tails, she will be put to sleep and awakened on Monday and asked if she would like to cancel her bet, given the amnesia inducing drug, put back to sleep, and awakened on Tuesday with the same question about canceling the bet. Then she will be awakened on Wednesday and the bet resolved.

If the result of the coin toss is tails, she is queried twice, the bet being canceled if she so requests on either or both occasions and remaining active if she does not cancel it on either occasion. And of course if the result is heads and she requests cancellation on Monday, it is also canceled.

What strategy can she use to create the largest possible positive expected value of the bet, and what will be that expected value? Because of the amnesia, she must use the same strategy every time she wakes up.

Whenever she wakes up, she flips a fair coin twice and requests the bet to be canceled unless the coin comes up heads twice. That is, she requests that the bet be canceled with probability 3/4. The expected value to her of the bet is \$4,000, which you can easily prove is the highest possible value.
Zcore13
Joined: Nov 30, 2009
• Posts: 3803
July 9th, 2015 at 10:39:58 PM permalink
It does not say who is flipping the coin.

Is is possible Sleeping Beauty previously spent 7 years practicing coin flipping influence and maybe even attended a class and can now know the outcome based on how she holds the coin prior to flipping? Or did she get to choose the person flipping the coin as a show of fairness, in which case she would obviously choose the famous Coin Flipping Captain.

ZCore13
I am an employee of a Casino. Former Table Games Director,, current Pit Supervisor. All the personal opinions I post are my own and do not represent the opinions of the Casino or Tribe that I work for.
MathExtremist
Joined: Aug 31, 2010
• Posts: 6526
July 9th, 2015 at 11:26:17 PM permalink
Quote: Zcore13

It does not say who is flipping the coin.

Is is possible Sleeping Beauty previously spent 7 years practicing coin flipping influence and maybe even attended a class and can now know the outcome based on how she holds the coin prior to flipping?

"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
surrender88s
Joined: Jun 23, 2013
• Posts: 291
July 10th, 2015 at 12:45:30 AM permalink
editted...

Replying to the OP directly, she made a good bet, and she should never cancel it. She's getting 3/2 on a coin flip. Am I wrong?

If each time she wakes up, she is given a bet as to whether or not it is heads, I believe that should pay 2/1. But that bet before the experiment begins should pay 1 to 1.
"Rule No.1: Never lose money. Rule No.2: Never forget rule No.1." -Warren Buffett on risk/return
24Bingo
Joined: Jul 4, 2012
• Posts: 1348
July 10th, 2015 at 2:30:44 AM permalink
I'm pretty sure it's 50/50, since she's been given no new information. Or coming at it from another angle, waking up, she's half as likely to have woken up in either of the worlds where she will have been woken up twice. Picture an item that fell out of a box, where one item in each box is special in a way not immediately identifiable. Doesn't the chance you've gotten the specified item depend on the number of items in the box? Now say you've got boxes of eight marbles, forty, and three hundred. First one box is chosen at 1/3 probability of each, then one marble from that box. Does the fact that the larger box has more marbles make it more likely that that marble comes from that box?

She certainly shouldn't cancel the bet if she's only allowed deterministic strategies, since then her only options are "always cancel" and "never cancel," so it comes down to a 3:2 bet on a coin flip. If she's allowed probabilistic strategies, it's another story, but that's because she'll be deciding twice, not because the probability of the original flip changes.
The trick to poker is learning not to beat yourself up for your mistakes too much, and certainly not too little, but just the right amount.
RS
Joined: Feb 11, 2014
• Posts: 8626
July 10th, 2015 at 2:47:42 AM permalink
Quote: 24Bingo

I'm pretty sure it's 50/50, since she's been given no new information. Or coming at it from another angle, waking up, she's half as likely to have woken up in either of the worlds where she will have been woken up twice. Picture an item that fell out of a box, where one item in each box is special in a way not immediately identifiable. Doesn't the chance you've gotten the specified item depend on the number of items in the box? Now say you've got boxes of eight marbles, forty, and three hundred. First one box is chosen at 1/3 probability of each, then one marble from that box. Does the fact that the larger box has more marbles make it more likely that that marble comes from that box?

She certainly shouldn't cancel the bet if she's only allowed deterministic strategies, since then her only options are "always cancel" and "never cancel," so it comes down to a 3:2 bet on a coin flip. If she's allowed probabilistic strategies, it's another story, but that's because she'll be deciding twice, not because the probability of the original flip changes.

In one of the cases, she is awakened twice. The other case she will be awakened once.

If she always says tails, then she will win 2/3 of the time. If she says heads, she will win 1/3 of the time.
SDSDNSR
Joined: Apr 27, 2015
• Posts: 12
July 10th, 2015 at 10:17:48 AM permalink
Concerning the OP's 3:2 odds bet of \$1000:

If the bet is not canceled, she will win \$1500 if the coin comes up heads and will lose \$1000 if the coin comes up tails. When awakened on Monday and perhaps Tuesday she may cancel the bet. If she cancels on Monday but is awakened and decides not to cancel on Tuesday, the bet remains canceled.

This is a a good bet for her. If her strategy is never to cancel, her expected winnings are \$250. But she can do better.

Her best strategy is to cancel with probability 1/4 whenever she is awakened. Her expected winnings in this case are \$281.25.
24Bingo
Joined: Jul 4, 2012
• Posts: 1348
July 10th, 2015 at 7:37:32 PM permalink
Quote: RS

If she always says tails, then she will win 2/3 of the time. If she says heads, she will win 1/3 of the time.

She'll be winning 2/3 of the times she speaks, but only in half the experiments. It's just that she gets to be right/wrong more often in one. If every time a red card hits the felt, you put it back into the shoe and pull it out again, does that mean twice as many red cards are dealt?
The trick to poker is learning not to beat yourself up for your mistakes too much, and certainly not too little, but just the right amount.
SDSDNSR
Joined: Apr 27, 2015
• Posts: 12
July 11th, 2015 at 9:37:31 AM permalink
The OP asks two questions about the Sleeping Beauty scenario. The first is what she believes about the coin flip when she wakes up, and the second is if she should accept, and how should she handle, a bet of \$1000 on heads with 3:2 odds (winning \$1500 on heads and losing \$1000 on tails if the bet is not canceled).

Concerning the belief:
Most people would agree that as she is put to sleep her belief is 50% that the coin will show heads and 50% that it will show tails. Now the only reason one should change one's belief is if one obtains new information. Whenever she is awakened she obtains no new information, therefore she should not change her belief.

Concerning the OP's bet:
I pointed out earlier that for the OP's specific bet, her best strategy is to cancel the bet with probability 1/4 whenever she wakes up, resulting in an expected value to her of \$281.25. (Recall that if she cancels on Monday, but doesn't cancel on Tuesday, the bet remains canceled.)

Concerning a bet in general:
Consider a bet in which, if the bet is not canceled, she receives H if the coin shows heads and T if the coin shows tails. (Here H and T can be negative, so for the OP's bet, H = \$1500 and T = -\$1000.) Her best strategies and the results are as follows:

If 0 < H < -2T, then she should cancel with probability (H + 2T) / (2T) whenever she wakes up and thus earn an expected return of H*H / (-8T),
Otherwise, if H + T > 0 she should never cancel and thus earn an expected return of (H + T) / 2.
Otherwise she should simply not take the bet, or equivalently, always cancel whenever she wakes up and thus earn an expected return of 0.

Note that it may be possible for her to turn what looks like a bad bet into a good one by judiciously canceling. In a previous example of mine, H = \$64,000 and T = -\$128,000, which looks like bad bet. But if she cancels the bet with probability 3/4 whenever she is awakened, her expected return is \$4000.

The betting result is obtained by noting that the expected return is a quadratic function of the probability of canceling (or more easily in terms of the probability of not canceling). Maximizing a quadratic function over an interval simply requires checking the returns at the endpoints and the return at the inflection point of the quadratic (if it is in the interval).
z2newton
Joined: Mar 22, 2014
• Posts: 62
July 11th, 2015 at 8:37:12 PM permalink
The way I understood it and meant to write is there are 2 different experiments. One is her belief of heads and the other the bet. Maybe I read it wrong but she's not being asked both question during the experiment. Also for clarity once she cancels her bet it's canceled regardless what day it is.

What I can gather is you have to decide is it halves or thirds.

1/3 2/3
500 - 666.66 = -166.66
1/2 1/2
750 - 500 = 250.00

I hope my math is right.
z2newton
Joined: Mar 22, 2014