No problem... thanks.

I thought I was pretty nice to you in my post, and you reply with this.

Can you read some of the things you posted again, please, like this:

and tell me you really think this post is in the spirit of open-minded discussion? It seems you're stubbornly insisting your point of view is correct without taking any account of the dozens of previous posts that disagree with you.

Thanks for chiming in. Yes it was getting a bit tilting / frustrating... but I'm sure I have been on the other end of being sure I was right when I was wrong.

Sigh. This problem is still baffling to me somewhat.

Suppose we want to answer the following question. Play the game one time. When the player sees his number he switches. What is the probability that his new number is lower.

It seems that there are 2 ways I can write a computer program to test this.

METHOD 1
Repeat a million times

Generate a sequence of coin flips
Generate the two numbers A, B
Choose one at random and call that N
Switch
Count how often switching increases or decreases our number[/i]

Answer will be 1/2 increase and 1/2 decrease

METHOD 2
Play the game 1 time. When the player looks at his slip his number is X and he wishes to know the probability that switching will increase / decrease. So he does the following experiment...
Repeat a million times

Generate a sequence of coin flips
Generate the two numbers A, B
Choose one at random and call that N
If N = X then we are interested in what happens so

Switch
Count how often switching increased or decreased our number

else

ignore... we are not interested in this case right now

end if

Answer will be 2/3 decrease and 1/3 increase


In other words, if we pick a number in advance N=X and test what happens when N=X we will get 1/3 and 2/3. And that is true for every value of X except 1 which will always increase. But if we don't choose N=X in advance and we simply count how often switching helps us in general we will get 1/2 and 1/2.

Still making my head spin. Like it calls into question the exact meaning of "probability".

100 Streaks, Using the Overall Probability Is:
H = 50 times
TH = 25 times
TTH = 12.5 times
etc.

^ I completely get that.

Here's what I don't get:
Maybe Bobf will be cool enough to explain it to me?

If we 'ignore' longer streaks, how can we not 'ignore' what should be a corresponding 'shorter' streak at the same time to get the probability for a 0 or 1, 1 or 2 or any other independent streak?

It seems like what you're saying we know is the longer streak is not possible, so we ignore it, but we factor in what should be the longer streak's corresponding 0 streak to calculate the probability of an either/or for a single 1T streak in isolation.

So, if the options are flipped exactly 0T or 1T on a given streak you're saying we should 'ignore' the 2T+ streaks (25), but not the corresponding 0T streak (25) we should not have without those longer streaks when we calculate the probability of a 0T or 1T for an isolated streak and that's what I'm not getting...

To me, it seems like for each 'streak' of 1T or more we should have a corresponding 0T streak based on the overall probability, but if we don't 'count' the longer streaks in the probability of getting exactly a 0T or 1T streak in isolation, how can we count the 0T streak that should only be added to the overall streaks when we run enough trials to have those longer streaks.

What I'm saying is:
H = 50 times (A)
TH = 25 times
T+ = 25 times (B)

We 'ignore' B for the calculation, but shouldn't there be a corresponding 25 streaks in A that we 'ignore' for a 0T or 1T toss in isolation?

IOW: We should have one 0T streak for a T streak of any length.
We should have a 1T streak for a streak of any length greater than 1T.

So...
We should have one 0T streak for each 1T streak we have.
We should have one 0T streak for each 2T streak we have.

So, overall, we should have two 0T streaks for a 1T streak and a 2T streak, but we should not have two 0T streaks per 1T streak if we're looking at a 1T streak in isolation, should we?

Where did the 2nd 0T streak come from when we're looking at a 1T streak in isolation?

Using 0T or 1T as an example, if we are looking only at an isolated streak, it seems we should only have a 1T streak and corresponding 0T streak as options, to me anyway.

I'm glad I'm not the only one! LOL

I really do appreciate all of your explanations and comments, even if you're baffled.

To uDrewAtThat

Try looking at it this way.

The coins have been tossed for 800 sequences.

There were about:
400 H
200 TH
100 TTH
50 TTTH
etc

Now say you get a paper with 3.
You know the sequence was either H or TH.

Can you now see that the probability (conditionally on the sequence being H or TH) is 2/3 H and 1/3 TH?

If we know that the sequence was H or TH then we know H is twice as likely to have occurred.

Rob

I know I took them into account and now it seems I might not be quite as far off, nutty, incorrect or loony as it would appear if you only read the other posts and don't consider the point I tried to make repeatedly in this thread.

Look at the 2 different ways Bobf said he can write the program and the two different answers he got... The different results are exactly what I said was the difference in the Conditional and Overall Probability... Go figure.

Sorry for not sounding overly friendly, but the first time you posted anything in a thread I've been in was too tell me how wrong and tilting I was when it's starting to appear I might not be that wrong after all...

Kind of tilting to have someone hop into a thread and tell me I'm wrong and tilting, when I'm pretty sure I was not wrong.

Check Bobf's previous post about the two different ways he can write the program and the two different answers he got... The answers he came up with are what I said were the difference between the Conditional and Overall probability.

I still don't think I was wrong.
(In fact, now I'm pretty sure I was right.)

I just didn't want to argue about it any more, because no one wanted to listen, but apparently Bobf is pretty cool and decided to test...

Why is there a difference in his tests?

That's what I tried to figure out / point out in my question to him... See 4 posts previous to this one.

No. I'm still sure that if you flip a coin until H comes and tell me that either 1T came or 2T came, the probability that 1T came is 2/3.

And to try to briefly answer your other post. I am not ignoring only longer streaks. I'm ignoring any streak that did not occur.

If you tell me that 1T or 2T come i am crossing off both shorter and longer streaks that I know did not occur.
H
TTTH
TTTTH
TTTTTH
etc.

and I am counting everything that could have occurred
TH 25%
TTH 12.5%

I don't know what you mean by "corresponding streaks" in that post either.

How about this...

I understand how you came up with up and down 50% of the time each in one of your computer simulations and it has to do with what I've been saying, but I'm tired of posting on this topic and trying to explain where I'm coming from and what I'm looking at.

The difference in your simulations makes sense to me, but if they don't to anyone else, then I'm not going to try and explain it except to say the short answer is:

In this case, the conditional probability for a specific streak is 1/2 (up/down) and the overall probability of the streaks when looked at as a whole is 1/3 (up/down).

Ok, but I don't agree. My simulation still had two slips of paper being chosen at random. That's where the 50/50 comes from in Method 1 of my simulation. The 50/50 does not come from TH and TTH being equally likely given that 1T or 2T occurred.

But I am tired and can't think so clearly now.

Thanks again for explaining... I'm tired too.

I understand what I was saying and why... Maybe no one else will ever get it and maybe I really am nutty and wrong, but I've tried my best to explain what I'm seeing and why I think there's a difference and we're not getting anywhere, so no big deal, it's just 'one of those things' I can understand what I'm thinking and why, but I guess I really can't explain to someone else.

Sorry to anyone I've been 'tilty' with...

Maybe we'll meet again in a different thread and get along better

Cliffs for My Position:

If I was given the chance to play the game once and switch, I would switch any number. If I was given the chance to play 10+ times I would only switch if the number was 1...

The reason why is in my posts.

Method 2:
1/3 chance of increase when switching for any number of tails>1.


Method 1:
1/4 chance of getting 0 tails and a 1 on the sheet (100% increase)
3/4 chance of getting >1 on the sheet (1/3 chance of increase for each of these)
Total chance of increase with switch:
1/4 + (3/4 * 1/3) = 1/4 + 1/4 = 1/2


For both Methods the chance of an increase when switching is 1/3 when you have a number greater than 1 on the sheet.

The 50% overall increase in Method 1 comes from the abundance of sheets with 1 on them.

Rob