I don't really understand what you're saying, and I don't think any of the other posters above were assuming coins have memories or whatever you're suggesting. Find the probability you are given a 9. That can happen in one of two ways:

- the sequence of flips was T, H, which will happen 1/4 of the time. Then half the time we are given a '9'. So the probability is 1/8 that we get a '9' *and* that the other slip has a '3'.

- the sequence of flips was T, T, H, which will happen 1/8 of the time. Then half the time we are given a '9'. So the probability is 1/16 that we get a '9' *and* that the other slip has a '27'.

Notice that, when we are given a '9', it's twice as likely for the other slip to have a '3' than to have a '27', which is where the other posters got their 2/3 and 1/3 probabilities.

It doesn't matter what the probability we were give a 9 was, any more than it matters what the probability of being dealt a 9 or 99 is after you receive your hand in poker... What matters is the probable outcome relative to the fact we were 'dealt' a 9.

Being 'dealt' a 9 what we know about the sequence of the flips is it was at least T, so given the fact there is 1 T, what are the chances there was a 2nd T?

It's not 1/3!

Look at your example: T H or T T H

Either the next flip was a T or the next flip was an H and the odds of it being a T or an H are 50% each.

Let's look at the number 27...
We know there were at least 2 Ts in a row.

What odds would you give me knowing there have been 2Ts in a row if I bet there would be 3?

If you look at the overall probability and say 3 only happens 1/8 times, so you give me 7 to 1, I'd take that bet all day, because once we get to 2 the odds of there being 3 are 1/2, not 1/8.

Flip it around: If someone said I've flipped 2 Ts in a row, so I'll give you 4 to 1 if I get 3Ts, because 3 only happens 1 in 8 times, would you take the bet?

P(flipping T tails) = 1/2^(T+1)

P(seeing N = 3^M) = 1/2 * P(M tails) + 1/2 * P(M-1 tails)
= 1/2 * 1/2^(M+1) + 1/2 * 1/2^(M)
= 1/2^(M+1) * (1+1/2)
= 3/2^(M+2)

Thus when we see the paper with 3^M on we know T is either M or M-1. We can use Bayes' theorem:

P(T=M | see 3^M) = P(see 3^M | T=M) * P(T=M) / P(see 3^M)
= 1/2 * (1/2^(M+1)) / (3/2^M+2)
= 2^(-1 - (M+1) + (M+2)) / 3
= 1/3

and

P(T=M-1 | see 3^M) = P(see 3^M | T=M-1) * P(T=M-1) / P(see 3^M)
= 1/2 * (1/2^(M)) / (3/2^M+2)
= 2^(-1 - (M) + (M+2)) / 3
= 2/3

Thus the EV's are:

EV(switch) = P(T=M | see 3^M) * (3N) + P(T=M-1 | see 3^M) * (N/3)
= (1/3) * 3N + (2/3) * N/3
= (11/9) N

EV(stay) = N

When we stay we get N, when we switch we have an EV of (11/9)N. Thus we always switch.

Your logic is flawed. Even if we know that 1 tail came it does not mean that 2 tails is just as likely as 1 tail because half the time we get the second tail, we also get three or more tails. If I tell you that 1+ tails occurred it is still true that 1 tail occurs twice as often as two tails.

Let's count.

Let's assume you saw a 3 and switch.

1000 trials

500 times: 0 tails
250 times: 1 tails
250 times: 2+ tails but we can ignore these because 3 is impossible to see in this case.

agree so far?

assuming you see a 3 and switch...

500 times there are 0 tails and (1,3) is written and 250 of those you see a 3 and switch to 1.
250 times there are 1 tails and (3,9) is written and 125 of those you see a 3 and switch to 9.

so
250 times you see 3 and switch to 1
125 times you see 3 and switch to 9

I agree with all your calculations. But suppose you calculate the EV(switch | T = t) and EV(stay | T = t). In other words, calculate from the viewpoint of the person writing on the slips who knows T = t but does not know which slip you picked. In that case EV(switch) = EV(stay) = average of the two numbers.

I believe that if you ran a monte carlo simulation of this with one person calculating the EV(switch)-EV(stay) for each value of T it would convert to 0 for each T, and another person calculating EV(swtich)-EV(stay) for each value of N it would convert to 2/9N for each N. If a third person calculated EV(switch)-EV(stay) for the game, that would not converge.

50% of the time there will be a 3rd. But 50% of those times there will also be a 4th. So there will be twice as many exactly 2T's as exactly 3T's. So given that you know that there are either exactly 2 or exactly 3 tails it still is not 50/50 that there are exactly 2 tails.

I think the 1/3 probability of tripling your money is right.

If you see 3^T on your paper then:

The first T-1 coins landed tails.
The possibilities for coins T and T+1 are [H,H], [H,T], [T,H] and [T,T].
[T,T] is impossible so the possibilities are reduced to [H,H], [H,T] and [T,H].
There is a 2/3 chance the other paper has 3^(T-1). [H,T], [H,H]
There is a 1/3 chance the other paper has 3^(T+1). [T,H]

It is twice as likely that (a head was tossed on coin T) than (a tail was tossed on coin T AND a head was tossed on coin T+1).

I think.

I agree always switch is the correct answer, but I think the correct probability depends looking at the specific situation rather than the 'overall series of flips', doesn't it?

What you are doing seems to be the same as looking at the 4 flush on the flop, a blank on the turn and thinking 'well, I know the flush hits 34.5% of the time from the flop to the river, so I'm going to use the same odds (34.5% = overall probability flop to river) in this isolated incident on the turn as I would from the flop to the river.', or, even more closely analogous: Looking at the 4 flush on the flop and thinking 'well, the overall probability of hitting a flush is roughly 11%, so that's my equity right now.'

We already know the series of flips in question ended at 1 or 0 tails when we see a 3, and if we look at 'the whole' (overall probability) it seems we will not have the right odds for this specific flip any more than we would have the correct odds of hitting the flush by the river when we have 4 on the flop if we use the overall probability (11%) of hitting the flush from pre-flop to the river.

What we know if there is a 3 on the paper:
There is either 1 or 0 tails in that series.
There is an equal chance in that specific series there was 1T or 0T.

If we say 'overall the series could keep going' and use that number for our calculation, we are doing essentially the same thing as we would be using the odds for the flush hitting from pre-flop to the river when we have 4 to a flush on the flop, aren't we?

If we're not, please explain how I'm incorrect in looking at the specific event (series) using the conditional probability we have rather than the overall probability everyone seems to think it correct?

If I should use the overall probability for the specific coin toss, should I also use the overall probability of hitting a flush (roughly 11%) when I have a 4 flush on the flop rather than the conditional probability of 34.5%?

Overall I believe that switching in not superior to staying. They are equal.

Suppose we randomly create 1 billion (or any large finite number of) flipping sequences, thus generating 1 billion number pairs. We then let two players Switch and Stay play the game repeatedly (like 1 million times) against each of those number pairs. The EV of the two players is equal.

I actually don't agree that switching is correct. I agree that the EV(switch | some N) > EV(stay | some N) but EV(switch | some T) = EV(stay | some T).

I think relating it to poker is only going to complicate matters.

yes, but it's twice as likely that it's 0.

no

128 flips
H 64
TH 32
TTH 16
TTTH 8
TTTTH 4
TTTTTH 2
etc.

when we see a 3 we know that either (H 64) or (TH 32) occurred, doesn't mean they are equally likely to have occured. There are twice as many H as TH.

put it another way. suppose i told you that either H occurred or TH occurred above. which is more likely to have occurred?

if i tell you that obama died yesterday and either he died of a heart attack or he was hit on the head by a meteor does that mean its 50/50 that he was hit by a meteor?

Yes, you got it. He's forgetting the bolded part... I think.

No. When you 'throw out' the longer series of flips, you also have to eliminate 1/2 of the 0T flips to get the right odds for that specific series. We know there was an end point < 2T.

64 - 16 - 8 - 4 - 2 = 32

If you were picking a random piece of paper out of a hat from all given pieces of paper (say your 1,000,000 test) series, you would be correct, but we're only concerned with 1 series and 2 pieces of paper involved, not the outcome of all series.

There's more than one thread in SMP on the 2 envelope paradox if anyone wants to search for them.

Don't do any math for a moment. Always switching without even having to look at the slip can not possibly be a +EV move. If you're picking one at random, switching to the other one shouldn't earn or lose you anything. Imagine giving one slip of paper to one person and the other to another person. Would it be +EV for both of them to switch slips with each other? Where is this EV coming from?

i'm not throwing them out. i'm counting what happens every time we see a three.

in 128 trials
H occurs 64 times and 1/2 of those or 32 times we see a three
TH occurs 32 times and 1/2 of those or 16 times we see a three
in all other cases we do not see a three

so we see a three 48 times.

of those 48 times that we see a three
32 times it was because it was H and the other number was 1
16 times it was because it was TH and the other number was 9