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06-25-2012, 12:45 AM   #46
grinder

Join Date: Feb 2008
Posts: 423

Quote:
 Originally Posted by uDrewAtThat? I have been... Thanks to Bobf for taking the time to post.
No problem... thanks.

06-25-2012, 12:49 AM   #47
old hand

Join Date: Aug 2011
Posts: 1,690

Quote:
 Originally Posted by uDrewAtThat? I do wonder how people learn if not by discussing though? Maybe you should just not read threads that tilt you... I really didn't know this thread was required reading. Sorry for the inconvenience.
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:

Quote:
 Originally Posted by uDrewAtThat? The Conditional Probability and the Overall Probability differ in this case. You keep calculating and telling me about the Overall Probability, which I get and pointed out in my previous post. You are Not Getting the Conditional Probability. The Overall Probability is Not the same as the Conditional Probability, and I am Not incorrect about the Conditional Probability of the Specific Streak ending at 2 or 3.
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.

06-25-2012, 01:08 AM   #48
grinder

Join Date: Feb 2008
Posts: 423

Quote:
 Originally Posted by kamikaze baby It's way too tilting to read all of this, but since bobf asked if others could chime in, I will: it's getting more than a bit frustrating to hear uDrew insist that everyone else use conditional probability here when that is precisely what everyone is already doing. bobf clearly understands conditional probability perfectly, as do many of the other posters in this thread. uDrew, does it not make you question your point of view at all when not one other person agrees with you, including several people who are obviously quite fluent in mathematics?
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.

 06-25-2012, 01:45 AM #49 grinder   Join Date: Feb 2008 Posts: 423 Re: Baffling Math Paradox 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 timesGenerate 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 timesGenerate 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 soSwitch Count how often switching increased or decreased our numberelseignore... we are not interested in this case right nowend ifAnswer 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".
06-25-2012, 01:55 AM   #50
centurion

Join Date: Jun 2012
Location: On the Button
Posts: 159

Quote:
 Originally Posted by bobf H = 50 times TH = 25 times TTH = 12.5 times etc. 50% of the time you got a second T but only 25% of the time you got exactly one more tail.

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?

Quote:
 Originally Posted by bobf In your experiment, suppose every time you got exactly 1T or exactly 2T you told someone else "I got 1T or 2T". How many times did you tell them that? 25 + 12.5 times or about 38 times. Now suppose every time you told that person "I got 1T or 2T" that person tried to predict whether it was 1T or 2T. Is it 50/50 for them? Were there 19 1T and 19 2T? No. There were twice as many 1T ad 2T. They should predict 1T if they want to be right more often than not.
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.

Last edited by uDrewAtThat?; 06-25-2012 at 02:02 AM.

06-25-2012, 01:59 AM   #51
centurion

Join Date: Jun 2012
Location: On the Button
Posts: 159

Quote:
 Originally Posted by bobf Sigh. This problem is still baffling to me somewhat.
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.

 06-25-2012, 02:30 AM #52 centurion     Join Date: Jul 2010 Location: Sheffield Posts: 105 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 Last edited by Scouse Rob; 06-25-2012 at 02:46 AM.
06-25-2012, 02:40 AM   #53
centurion

Join Date: Jun 2012
Location: On the Button
Posts: 159

Quote:
 Originally Posted by kamikaze baby 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.
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.

Last edited by uDrewAtThat?; 06-25-2012 at 02:53 AM.

06-25-2012, 02:43 AM   #54
centurion

Join Date: Jun 2012
Location: On the Button
Posts: 159

Quote:
 Originally Posted by Scouse Rob 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 the 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? Rob
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.

Last edited by uDrewAtThat?; 06-25-2012 at 02:49 AM.

06-25-2012, 02:50 AM   #55
grinder

Join Date: Feb 2008
Posts: 423

Quote:
 Originally Posted by uDrewAtThat? 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.
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.

06-25-2012, 02:59 AM   #56
centurion

Join Date: Jun 2012
Location: On the Button
Posts: 159

Quote:
 Originally Posted by bobf I don't know what you mean by "corresponding streaks" in that post either.

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).

06-25-2012, 03:07 AM   #57
grinder

Join Date: Feb 2008
Posts: 423

Quote:
 Originally Posted by uDrewAtThat? 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.

 06-25-2012, 03:15 AM #58 centurion     Join Date: Jun 2012 Location: On the Button Posts: 159 Re: Baffling Math Paradox 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 Last edited by uDrewAtThat?; 06-25-2012 at 03:28 AM.
 06-25-2012, 03:28 AM #59 centurion     Join Date: Jun 2012 Location: On the Button Posts: 159 Re: Baffling Math Paradox 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.
06-25-2012, 04:13 AM   #60
centurion

Join Date: Jul 2010
Location: Sheffield
Posts: 105

Quote:
 Originally Posted by bobf METHOD 1 Repeat a million timesGenerate 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 timesGenerate 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 soSwitch Count how often switching increased or decreased our numberelseignore... we are not interested in this case right nowend ifAnswer will be 2/3 decrease and 1/3 increase

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

Last edited by Scouse Rob; 06-25-2012 at 04:23 AM.

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