## Experimenting on Humans

The subject of the experiment, whom we will refer to as “Chris” was privy to all the details of the experiment in advance, and participated willingly.

One Sunday evening, Chris went to sleep at the observation labs. In another part of the lab a fair coin-toss was conducted, yielding a result of heads or tails. Chris was aware that the coin-toss was made, but not the results of the toss. However, he was made aware that the following events would transpire, dependent on the results of the toss:

If the toss came up heads, Chris would be awoken at 6AM on Monday morning and asked what the probability was that the coin had come up heads.

If the toss came up tails, Chris Would be awoken at 6AM on Monday morning and asked what the probability was that the coin had come up heads. BUT, on Monday evening he would be given an amnestic, putting him to sleep and totally clearing his memory of the previous 24 hours. Then, on Tuesday morning he would be awoken at 6AM and asked what the probability was that the coin had come up heads.

The amnestic had no side effects other than removing the previous days memories and resetting his body’s internal clock back 24 hours. Upon awaking, Chris had no certain way of knowing whether the amnestic had been administered, or what day it was.

Now would we answer 1 in 2, or 1 in 3 if we were in Chris’s place? And why….

(I believe this puzzle to be by an MIT graduate student named Adam Elga.)

June 21st, 2011 at 5:20 pm

No idea if this is right (or if there even is a right answer), but I’ll give it a shot.

First, what day is it?

If heads (50%): 100% Monday

If tails (50%): 50% Monday; 50% Tuesday

Overall day: 75% Monday; 25% Tuesday

Second, what was coin result?

If Monday (75%): 50% heads; 50% tails

If Tuesday (25%): 100% tails

Overall result of heads: 75% * 50% = 37.5%

Chris should answer 3 in 8.

June 22nd, 2011 at 3:57 am

No matter what colourful story is provided, the odds of flipping a fair coin and getting a heads is 1 in 2.

In more detail: after the first flip, Chris is woken regardless of H or T. If it was H, then the experiment is over. If it was T, then the next day, regardless of H or T being flipped, Chris is awoken. On esch occasion the odds are 1 in 2.

June 22nd, 2011 at 4:51 am

Uh I think SP was saying that if you know you are awoken on Tuesday then surely the coin result was tails.

I’ve already seen that problem and its answer, but I’ll admit I didn’t quite understand it.

June 22nd, 2011 at 5:51 am

Hi Karys. I wasn’t responding to SP’s comment. In the posed problem you don’t know what day it is.

However, if you somehow knew that it was Tuesday, then you’d only know that the coin flipped to tails on the Monday, but not what it flipped to on the Tuesday.

Unless I’ve misread the question, or Karl has modified the source question, then I’m sticking with 1 in 2 of it being a heads at each and every flip.

June 22nd, 2011 at 8:26 am

A complication is: which coin toss is Chris assigning a probability to?

June 22nd, 2011 at 2:07 pm

its 1 in 3

There is only one coin flip

June 22nd, 2011 at 6:01 pm

Hi Jordan, please explain your conclusion.

I say it’s 1 in 2 that there’s one coin flip and 1 in 2 that there’s two coin flips – not that either is what’s being asked for.

June 22nd, 2011 at 7:40 pm

My understanding is that the coin is only flipped one time: on Sunday evening.

June 22nd, 2011 at 8:36 pm

Well spotted SP . I had supposed that the coin was tossed a second time (on Monday evening) if it had come up heads the first time (on Sunday evening). Now I won’t be able to sleep.

June 22nd, 2011 at 9:39 pm

I’ll play your dastedly fiendish probability game …

Chris will answer 1 in 2, as that’s the chances of it coming up heads. Day etc has no effect, a coin toss is 50:50 unless it’s rigged … which come to think of it, it may be, so Chris can be ‘played’ with. So it’s probably actually less likely heads, and more likely tails

June 23rd, 2011 at 10:14 am

1 in 2

June 24th, 2011 at 5:44 am

I shall post the answer as given, and a link to a pdf on the question. The answer will probably create further debate!

The answer as given:-

The heart of this paradox is epistemic in nature- how should Chris make a statement of fact about this probability.

Rather than tackle the problem head-on, let’s step back a moment and ask a more abstract question: What is Chris’s goal in answering the question presented to him?

This may sound odd- presumably his goal is to give the correct answer. But how is his correctness to be measured? Either answer seems defensible, depending on your perspective. While half the time the coin toss comes up heads, at two thirds of the interrogations, the coin toss has come up tails.

This is almost the same scenario as the Newcomb Problem. But with one important twist – the Newcomb Problem has a goal – to answer correctly. Why does this make a difference? Because we are asking Chris to defend a probability without defining how the probability will be measured.

Is it based on the percentage of runs of the experiment where the coin comes up tails? Or is it based on the percentage of interrogations where the coin comes up tails?

In short, the answer is that no probability can be given, because the experimenters have failed to define how probability is measured.

A link to a thesis on the question and answer:-

http://www.google.co.uk/search?client=safari&rls=en&q=adam+elga&ie=UTF-8&oe=UTF-8&redir_esc=&ei=AdoCTqmwE4HRhAfg8YWzCg

A link to the Newcomb Problem

http://www.kiekeben.com/newcomb.html

June 24th, 2011 at 7:23 am

Here’s a link to a Wiki about it: http://en.wikipedia.org/wiki/Sleeping_Beauty_problem

What beats me is that the article doesn’t mention (unless I missed it) the point I made in post 5 – which coin is being discussed? More specifically, what exactly is the question that we are trying to answer? The last time I heard that question, the answer was 42.

The article section (thirdier position) has a strange equation and assertion:

P(Tails and Tuesday) = P(Tails and Monday) = P(Heads and Monday)

It then says these are mutually exclusive.

Neither statement is true. P(Heads and Monday) = 1/2, and then if you flipped heads on Monday then P(Tails and Tuesday) = 0. P(Tails and Monday) = 1/2 and if you actually flipped tails on Monday then P(Tails and Tuesday) = 1/2. I’m taking the face mentioned to be flipped on the day mentioned. I cannot see how to make sense of the thirdier position.

Thanks for the problem Karl, definitely worthy of ToM. I shall think further on it, as this post was made hurriedly.

June 24th, 2011 at 8:29 am

Chris – as Jordan says in post 6 – there is only one coin toss.

The question, as I see it, appears to be when you are awoken – there are three possibilities:-

1. Its Monday – this is the first time you have been awoken

2. It’s Monday – you have been awoken, but you will be given an amnestic

3. Its Tuesday and this is your second awakening, but you don’t know it.

All three are possible when you are awoken – you just won’t know whether it was heads or tails, and whether you are in Scenario 1,2 or 3.

Does this mean that the coin toss was 1/2, scenarios are 1/3 – combined it means a probability of 1/6? (No! )

Trying to argue this reminds me of the scene in the Princess Bride with the Dread Pirate Robert facing off against Vissini over a poisoned goblet of wine…

June 24th, 2011 at 8:48 am

The only sensible way I can see it is that on average over 4 trials we will get:

MT

H-

H-

TH

TT

Each of those 4 possibilities is equally likely. Altogether we will have flipped 3 heads and 3 tails. If Chris was in fact asked on Monday, then the answer is 1 in 2. If it were in fact Tuesday, then on Tuesday that days flip would also be 1 in 2, and Mondays flip would be 1 in 1 and on average (for Tuesday) 1 head and 3 tails would have been flipped.

—

Hi Karl, I’ve just seen your post 14. Cases 1 and 2 are equally likely with probability 1/2. Case 3 is case 2, just later in time.

June 24th, 2011 at 8:52 am

… post 6 is wrong. The probability of there be one coin flip = the probability of there being two coin flips = 1/2.

All this assumes that the experimenters don’t partake and forget they’ve already finished the experiment.

June 24th, 2011 at 8:58 am

Chris – for post 6, I was agreeing with the number of coin tosses – there is only one on the Sunday night – there are no other coin tosses described in the question.

What follows with the three scenarios is based on the only coin toss on the Sunday night.

June 24th, 2011 at 10:10 am

Hi Karl. I see that I’ve not being paying proper attention to that detail (only one coin toss) despite SP’s comment on that point.

In that case, on Monday the correct answer would be 1/2. On Tuesday, the correct answer would be 0/1. Whichever answer he gave, Chris would only be right half of the time. Any other answer e.g. 1/3, would never be right.

Re-examining the Wiki:

P(Tails and Tuesday) = P(Tails and Monday) = P(Heads and Monday)

It then says these are mutually exclusive and then implies that each = 1/3.

I now realise that P(T and T) refers to the possibility of a second (hypothetical) coin toss and cannot see why it is being treated in the way that it is being treated. Further the “and” should be “on” e.g. P(T on T).

I accept that the question is ill-defined and so cannot be answered in the way that seems to be being suggested.

My dad wants me to go out with him, so I can’t finish up now. I shall ponder further.

June 24th, 2011 at 11:00 am

The precise question and the definition of answering it “correctly” is absolutely key.

If the question is “what is the probability that heads was flipped?” then the answer is “1/2″. It doesn’t matter what day it is. If it happens that it was Tuesday (and so tails was actually flipped), the probability (rather than the actual outcome) assessment is still 1/2 (I think, gulp).

If the goal was to say “heads” or “tails” and be right as often as possible, then you’d have to say “tails”. On Monday if heads was flipped, you’d be wrong, but you’d be right if tails was flipped, you’d also be right on the Tuesday. So by that standard, you’d be right twice as often as you’d be wrong, but you aren’t implying that the probability of throwing a heads isn’t 1/2.

July 1st, 2011 at 11:31 pm

The question does not make sense but if trying to figure out what 1 in 2 or 1 in 3 means will he answer heads in 2 or 3 days I believe he would answer Heads on either day since he was preprogramed to think that as he was awoken at 6am he was to think it as heads an that not until night time would be be drugged. Therefore he would spend the entire day believing that it was heads through out the day regardless if he was druged or not. The way the question is phrased it is directed to have the patient (victim) believe that if heads comes up he will wake at 6; this was a subliminal suggestion from the start.

July 10th, 2011 at 11:20 am

I agree with SP on post 1. Since “Chris” is aware if the details of the experiment, and wants to answer correctly, then that (first figuring out the day, then the coin toss) is the chain of thoughts he’ll have.

July 10th, 2011 at 12:05 pm

Hi Ankit, The big problem is, what exactly is the question that we’re being asked?