# Is there a distinction between one iteration and multiple iterations of Sleeping Beauty Problem?

Take two set-ups of the Sleeping Beauty experiment

1. Set-up 1 The experiment is performed once. What is the probability that a random awakening corresponds to Heads?

2. Set-up 2 The experiment is repeated n times. The memory is erased every day througout the experiment. What is the probability that a random awakening corresponds to Heads?

From wikipedia, I take the following re-formulation of the problem:

Another way to see the two different questions is to simplify the Sleeping Beauty problem as follows. Imagine tossing a coin, if the coin comes up heads, a green ball is placed into a box; if, instead, the coin comes up tails, two red balls are placed into a box. We repeat this procedure a large number of times until the box is full of balls of both colours. A single ball is then drawn from the box. In this setting, the question from the original problem resolves to one of two different questions: "what is the probability that a green ball was placed in the box" and "what is the probability a green ball was drawn from the box".

If we do the above box experiment with just one coin toss, then the answer to "what is probability that a random ball drawn from the box is green?" is 0.5

If we do the above box experiment with n coin tosses, the answer to "what is the peobability that a random ball drawn from the box is green? is 0.33.

This time, we have not changed the question, unlike in the quote from Wikipedia. Instead, the same question gives different answers depending on whether we did the coin toss once or repeatedly.

Should there be a difference in Sleeping Beauty's credence that a random awakening corresponds to Heads, depending on whether we do the single coin toss or the repeated coin toss experiment?

Edit I would like to elaborate on both the probabilities using repeated trials:

1. We do the single coin toss experiment a large number of times, and draw a random ball from each of the experiments. We can see that 50% of boxes will give a green ball.

2. We do the multiple coin tosses experiment a large number of times, and draw a random ball from each of the experiments. Now around 33% of the boxes will give a green ball .

So we can see that the probabilities I gave can be defended in a frequentist interpretation.

• Feb 24 at 0:44
• @Conifold i have written an answer where i also address the betting in the end. Can you please tell your thoughts on it? Feb 24 at 6:51
• Can elaborate on 2? I'm not sure if this is exactly right: the total duration of the experiment is 2N days. On day 1 a coin is flipped: on heads: SB is allowed to sleep through day 2, but is woken (and put back to sleep) on day 3. On heads SB is woken and put back to sleep on Day 2, and also (like in the other case) woken and put back to sleep on day 3. This cycle repeats with the coin flips happening every odd day. In this case the coin flips are determining whether SB is woken&slept on the even numbered days.
– Dave
Feb 26 at 16:37
• If SB is guaranteed to be be awakened one final time then that one instance might need to be accounted for as long as N<\infty, or one could specify that SB sleeps forever after the period of 2N days.
– Dave
Feb 26 at 16:40

No, the credence one should apply should be derived from large sample distribution. The idea is that any given particular instance of the sleeping beauty experiment corresponds to one of the cases of putting red/green balls into the bin. Since we don't know which one of those trials in some figurative sense "matches up with" the current sleeping beauty case in question, we just need to use the overall statistics.

You might be confusing this argument, which is specifically for the 2 day sleeping beauty experiment from the N day sleeping beauty experiment (heads = woken up, tails = woken up and put back to sleep N times over successive days). A variation that has also been used to probe this kind of thought experiment.

• I have edited in the "large sample distribution" calculation for both 50% and 33% Feb 24 at 2:41

After some thinking, I made two variations each of the "single coin toss" and "multiple coin tosses" experiment. The conclusion is that the credence in a multiple coin tosses experiment is 0.33 in both variations, while the credence in a single coin toss experiment is upto debate

1. Single coin toss experiment

Variation (a): We take a box and toss a coin once. If Tails, we place in two Green balls labelled 'T'. If Heads, we place in one Green ball labelled 'H' and one black ball labelled 'H'

A random ball is drawn and it turns out to be green. What is the probability that it has label 'H'? This is 0.33, as we gain information by knowing that the drawn ball is not black.

We can confirm this in a frequentist approach. Repeating the experiment with n boxes, among the boxes where the drawn ball is green, around 33% of the balls will have label 'H'

Variation (b): Same as (a), except we only place a green ball labelled 'H' if the coin lands Heads, and no black ball.

A random ball is drawn and turns out to be green. What is the probability that it is labelled 'H'? This is 0.5. Knowing that it's green adds no information.

We can confirm this in a frequentist approach. Repeating this with n boxes, around 50% will draw a Heads labelled ball.

2. Multiple coin tosses experiment

Variation (a): We take a box and toss a coin a large number of times. Whenever it lands Tails, we place two Green Balls with labels 'T'. Whenever it lands Heads, we place one black ball and one green ball with labels 'H'.

A random ball is drawn and turns out to be green. What is the probability that it's labelled 'H'? This is 0.33. The initial chances of drawing an 'H' label or a 'T label' were 0.5 each, but knowing that the drawn ball is green gives us information.

Variation (b) Same as (a), but we only place a green ball upon heads.

A random ball drawn is green. What is the probability that it's labelled 'H'? This is 0.33.

Knowing that it's green adds no information, but there were more balls lebelled 'T' to begin with, so the initial probability of drawing 'H' was 0.33

We can see that there is no dispute in a multiple coin tosses experiment. Both variations give 0.33, doesn't matter whether drawing a green ball gives or does not give information

Coming back to the single coin toss experiment : Beauty is asked to participate in an experiment with a single coin toss (which is the conventional formulation).

The equivalent of "drawing a ball" is "opening your eyes/waking up". The dispute is if we get information by knowing "I am awake" (i.e. knowing "the ball is green"), or equivalently, knowing "I am not asleep" (i.e. knowing "the ball is not black".

Now, "waking up" in a single coin toss experiment must correspond to "drawing a ball" in Variation (b) of the single coin toss experiment. This is because outcome of "opening your eyes" can never be "I am alseep", just like the outcome of "drawing a ball" in variation (b) can never be a Black ball. So one gains no information by knowing "I am not asleep", as it was never a possibility.

All in all, the credence should be 0.5 in a single coin toss experiment according to the analysis in Variation (b).

P.S. About the betting in a single coin toss experiment, Beauty should bet on Tails because she is practically given twice the money on saying "Tails", as the question gets asked twice. This has no effect on credence.

• If the coin was tails SB makes two draws from box, not one.
– Dave
Feb 26 at 16:49

Should there be a difference in Sleeping Beauty's credence that a random awakening corresponds to Heads, depending on whether we do the single coin toss or the repeated coin toss experiment?

A repeated measurement of a fair coin should confirm that the probality of the coin landing heads = probability of the coin landing tails or 50/50.

What is the minimum amount of information that Sleeping Beauty needs to determine (100% accuracy) the state of the coin? All she needs is the day of week and whether she is put back to sleep or goes home:

• Monday = Heads (Go home)
• Monday = Tails (Go back to sleep)
• Tuesday = Tails (Go home)

Without knowledge of the day of the week and whether she goes home or back to sleep, when she wakes up, all she will think is "Is it Monday or Tuesday and that's based on the toss of a fair coin: 50/50.

If the coin is Tails. an interesting twist occurs:

1. Sleeping Beauty is awakened the first time.

If Sleeping Beauty goes home then she will know it's either Monday or Tuesday. If she knows she's going back to sleep then she knows the coin is tails before she goes back to sleep and can tell the experimenter with 100% confidence.

Its unclear from the way the problem is stated if she is aware she is returning to sleep or was back to sleep without her knowledge. But that knowledge changes the problem from 50/50.

Sleeping Beauty can use this information to only place a bet on Tails ( in between sleeps) and ignore the other outcomes which remain 50/50. She will win 100% of the time.

With 12 coinflips (assuming perfect 50/50 H T H T H T H T H T H T etc.):

Coin Monday Interviews Tuesday Interviews
H H
T T H
H T
T T T
H H
T T H
H T
T T T
H H
T T H
H T
T T T

On Monday (with this extra information) the % probility is 25/75 in favor of Tails: 3 Heads and 9 Tails

On Tuesday, the extra information is now lost, so on Tuesday the odds return to 50/50. If you want to bet based on just the number of interviews:

18 interviews = 6 Heads and 12 Tails or