1. Say that we flip a coin, and we get 4 heads in a row. If there is a 1/2 chance that a head will come up at the first trial, then there is 2^5 (1 in 32) chance of flipping another heads correct? But my question is this, what is the probability in this trial for instance if we assume that we have a 'master trial' of this sequence of events:

Tails...

If the percentage of chance of flipping a tails on the 5th flip is 96.875% as part of the trial of 5 coin tosses (1 in 32 chance of tossing a Heads on the 5th flip (3.125%).

Then what are the odds in percentage of Heads occuring AFTER a tails as part of the total trail of 6 tosses:

Tails...

And what are the odds in percentage of Tails occuring AFTER a tails as part of the total trail of 6 tosses:

Tails...
TAILS

If we take each outcome of 6 tosses a ONE trial, not indepentently. I do not want to simplify this in terms as 50/50 each time a coin toss is made because that is redundant, yes its true for 1 trial, but not the course of a 6 trial outcomes.

Am I right to assume that AFTER a tails after 4 consecutive heads, that tails is more likely? Or heads? I would assume tails seeing that heads has already appeared more times. Again, the law of large numbers can apply, my point is that it isn't 50/50 at the 6th coin flip toss, we know it is, but that we predict the sequence of events from the start of the 6 coin flip trial what the 6th result will be, heads or tails if that sequence of events happens?

2.

3. No, the odds are always 50/50 for each upcoming flip. Whether you like it or not.

Sorry to simplify, but it is indeed simple.

The odds of a long string of consecutive "heads" may be higher, but each flip is still 50/50.

4. That's too simplified for an answer I'm afraid it wasn't what I was looking for. Please grasp my concept.

At each point there is a 50/50 chance of heads or tails at each part of the trial.
At the start of the trail of 6 independant trials, the likelihood of those heads coming as many as they did as part of 6 trials wasn't likely. What is the probability of the two different outcomes I described as sequences occuring in each set of trials. If you don't know please say so.

I will elaborate. Accept this concept. Six trials of coin tosses are ONE conglomerate trial.

Heads 1/2 at stage 1 of 6 stage trial (50% at this point for tails as new individual trial)
Heads 1/4 at stage 2 of 6 stage trial (50% at this point for tails as new individual trial)
Heads 1/8 at stage 3 of 6 stage trial (50% at this point for tails as new individual trial)
Heads 1/16 at stage 4 of 6 stage trial (50% at this point for tails as new individual trial)
Tails 31/32 at stage 5 of 6 stage trial (Heads 1/32) - (50% at this point for tails as new individual trial)
Tails .... at stage 6 of 6 stage trial (50% at this point for tails as new individual trial)

I hope this cleared up the concept. What is the probability of tails coming at the 6th stage of this 6 stage trial. (I am aware that as a new trial it would be 50/50 (1/2) chance of tails. That isn't my point, my point is what are the odds of tails at this stage from starting at stage 1 and doing probability at that point to this point (Please don't argue 50/50 because during this sequence as a 6 trial sequence this sequence of events isn't 50/50, and tails at the 6th stage of 6 from stage 1 is not 50/50)

5. Your question is very unclear.

6. Originally Posted by mathman
Your question is very unclear.
What are the odds of the 6th occurance being tails after 4 heads and 1 tails in that order.

If the odds are 1 in 64 of getting 6 heads in a row (2^6)
If the odds are 1 in 64 in getting 6 tails in a row (2^6)

The odds of getting 4 heads in a row: 1 in 16 (2^4)
The odds of getting a tails after these 4 heads: 31 in 32 (2^5)

THE QUESTION
The odds of getting a tails after 4 heads and 1 tails. If the odds of a tails was 31/32 on the 5th flip and that occured. What are the odds of getting another tails after considering there were 4 heads before and 1 tails. ASSUMING we are talking about being at the first flip. NOT at the 6th.

7. Originally Posted by Quantime
I hope this cleared up the concept. What is the probability of tails coming at the 6th stage of this 6 stage trial. (I am aware that as a new trial it would be 50/50 (1/2) chance of tails. That isn't my point, my point is what are the odds of tails at this stage from starting at stage 1 and doing probability at that point to this point (Please don't argue 50/50 because during this sequence as a 6 trial sequence this sequence of events isn't 50/50, and tails at the 6th stage of 6 from stage 1 is not 50/50)
It is 50:50.
It is always 50:50.
It doesn't matter what "stage", it is 50:50.
It doesn't matter if you have had 6 or 1 million previous heads or tails, it is still 50:50.
It doesn't matter how you phrase the question, it is 50:50.

Guess what the answer is.

8. Originally Posted by Quantime
If the odds of a tails was 31/32 on the 5th flip and that occured
It wasn't 31/32, it was 50/50.

9. Originally Posted by Quantime
Originally Posted by mathman
Your question is very unclear.
What are the odds of the 6th occurance being tails after 4 heads and 1 tails in that order.

If the odds are 1 in 64 of getting 6 heads in a row (2^6)
If the odds are 1 in 64 in getting 6 tails in a row (2^6)

The odds of getting 4 heads in a row: 1 in 16 (2^4)
The odds of getting a tails after these 4 heads: 31 in 32 (2^5)

THE QUESTION
The odds of getting a tails after 4 heads and 1 tails. If the odds of a tails was 31/32 on the 5th flip and that occured. What are the odds of getting another tails after considering there were 4 heads before and 1 tails. ASSUMING we are talking about being at the first flip. NOT at the 6th.
Coins don't remember. So, the past has no bearing at all on the future. None. Every toss therefore has the same 50-50 probability of a head or tail (assuming a fair coin).

10. At each individual trial, yes, the chances of getting heads or tails are 50/50. But this irrelevant of the overall outcome which is determined by the number of individual outcomes and the number of consecutive trials.

In the outcome map below (split into 2 pictures, sorry), all the possible outcome paths are shown for a heads-or-tails experiment of 4 trials. At each individual trial, there is a 50/50 chance of getting heads or tails. But getting a certain combination through 4 trials is 1/16. There are 16 different combinations from the 4 consecutive trials. Getting heads 4 times in a row is a 1 in 16 chance, not a 1 in 2, because I'm not talking about immediate/general probability.

If I'm understanding the OP's question correctly, then this is simply a matter of middle school math, no offense intended. He's not talking about the chances of getting something at the immediate outcome, but the consecutive outcome (in other words, "in a row"). The chances of getting heads 4 times in a row is 1/16. Nothing non-trivial. Nothing non-standard.

So immediately, it's 50/50. ("immediate/general" probability)
Consecutively, it's variable. ("consecutive/path" probability)

But since this is elementary math, I could be misinterpreting the situation. Is this what you're talking about, Quantime?

11. Originally Posted by Quantime
THE QUESTION
The odds of getting a tails after 4 heads and 1 tails. If the odds of a tails was 31/32 on the 5th flip and that occured. What are the odds of getting another tails after considering there were 4 heads before and 1 tails. ASSUMING we are talking about being at the first flip. NOT at the 6th.
The way this question is worded does suggest we are talking about the 6th trial, of flipping the coin, and since the outcome of all trials is independent of the outcome of the other trials the probability (p) of tails on any flip of the coin is 0.5.

If however you rephrased the question to, "what is the probability of the following outcome of 6 trials in a coin tossing experiment, HHHHTT?" Then the probability, of this outcome, will be 0.0156. Of course any outcome of 6 independent trials will have the same probability, as the probability of heads or tails is the same, so the probability of the outcome HHHHTH will be 0.0156. However this probabilistic interpretation only applies if the trials have not yet been conducted, if we were to conduct the experiment and get HHHHT then the probability of HHHHTT becomes 0.5.

12. Thanks everybody for the insight, especially wallaby you understood the concept I was trying to explain. Brody, thanks for the informative post as well. Yes it is what I meant; the diagram perfectly represented what I was thinking.

13. Originally Posted by Quantime
Say that we flip a coin, and we get 4 heads in a row. If there is a 1/2 chance that a head will come up at the first trial, then there is 2^5 (1 in 32) chance of flipping another heads correct? But my question is this, what is the probability in this trial for instance if we assume that we have a 'master trial' of this sequence of events:

Tails...

If the percentage of chance of flipping a tails on the 5th flip is 96.875% as part of the trial of 5 coin tosses (1 in 32 chance of tossing a Heads on the 5th flip (3.125%).

Then what are the odds in percentage of Heads occuring AFTER a tails as part of the total trail of 6 tosses:

Tails...

And what are the odds in percentage of Tails occuring AFTER a tails as part of the total trail of 6 tosses:

Tails...
TAILS

If we take each outcome of 6 tosses a ONE trial, not indepentently. I do not want to simplify this in terms as 50/50 each time a coin toss is made because that is redundant, yes its true for 1 trial, but not the course of a 6 trial outcomes.

Am I right to assume that AFTER a tails after 4 consecutive heads, that tails is more likely? Or heads? I would assume tails seeing that heads has already appeared more times. Again, the law of large numbers can apply, my point is that it isn't 50/50 at the 6th coin flip toss, we know it is, but that we predict the sequence of events from the start of the 6 coin flip trial what the 6th result will be, heads or tails if that sequence of events happens?

You're asking an audience "so much smaller" than the greater game, right?

What's you're achievement here?

Acceptance?

Or proving these people know nothing?

14. Originally Posted by Quantime
Thanks everybody for the insight, especially wallaby you understood the concept I was trying to explain. Brody, thanks for the informative post as well. Yes it is what I meant; the diagram perfectly represented what I was thinking.
Glad to help. Just got a little frustrated that people started getting a little too die-hard on the 50/50 thing. And by the way, you should be careful dealing with path probabilities. From your posts the information seems to be a little mixed up, see these two points:

1) When you're writing probabilities, make sure not to use 1/32 and 50/50 in the same context or sentence. I made the mistake myself, but then realized. You either say 1/2 vs 1/32 ... or 50/50 vs 0.033125/99.96875. As the first kind (1/2) is the proper fraction form, and the second (50/50) represents the contrasting odds out of 100. As you can see, the former is much easier in most cases.

2) Look back on your 6 trials. The chance of getting HHHHHH is 1/64. Getting heads on the 6th trial consecutively considering the previous outcomes is 1/64. But that does not mean getting tails on the 6th trial is 63/64 (without you consciously aware of the previous outcomes). Getting HHHHHH and HHHHHT are both 1/64 chances, because there are 64 unique pathways. In other words, getting a different pathway than HHHHHH is 63/64 (and not just getting tails in general), since there are 63 other possible pathways besides that specific set of outcomes. It's a little mind-jumbling, but do you understand?

15. Originally Posted by brody
Just got a little frustrated that people started getting a little too die-hard on the 50/50 thing.
Yeah, sorry about that

I didn't (and still don't really) "get" the question being asked. I liked mathman's wallaby's fuller answer (much better than mine, I'm not ashamed to say). And yours obviously helped Quantime. I'll have to take another look at it...

16. Originally Posted by brody
[Getting HHHHHH and HHHHHT are both 1/64 chances, because there are 64 unique pathways.
And, because the chances are equal, it is ... 50:50. (Sorry, couldn't resist )

17. Originally Posted by davros
You're asking an audience "so much smaller" than the greater game, right?

What's you're achievement here?

Acceptance?

Or proving these people know nothing?
What is your purpose here? To troll? My question has been clearly answered thanks to the help of these guys, so now what is your purpose in being provacative?

Regards, QT

18. Originally Posted by Strange
Originally Posted by brody
Just got a little frustrated that people started getting a little too die-hard on the 50/50 thing.
Yeah, sorry about that
No biggie It's cool.

I didn't (and still don't really) "get" the question being asked.
Don't worry. This thread just really over-complicated a simple concept. It's basically asking about probability concerning outcomes in a row. That's it, in a row. The OP, and me too I'll admit, turned this primordial blob of a lesson into a 3-d maze. Quantime called it a "single conglomerate trial" and in the sense "from the 1st flip" and I said "consecutively". It's just looking at the overall experiment, all the trials together.

And, because the chances are equal, it is ... 50:50. (Sorry, couldn't resist )
Haha :P I can't ... resist ... either, but ... It's actually 0.015625 : 0.015625 : 0.015625 ... (64 times) since each is a 64th of a 100 probable and distributed equally.

19. Originally Posted by brody
Don't worry. This thread just really over-complicated a simple concept. It's basically asking about probability concerning outcomes in a row. That's it, in a row. The OP, and me too I'll admit, turned this primordial blob of a lesson into a 3-d maze. Quantime called it a "single conglomerate trial" and in the sense "from the 1st flip" and I said "consecutively". It's just looking at the overall experiment, all the trials together.
To simplify further (even though it's not needed any more) i'd say that this is simply a question about Binomially distributed random variables,

20. Originally Posted by wallaby
To simplify further (even though it's not needed any more) i'd say that this is simply a question about Binomially distributed random variables,
Ah, much easier to understand now. Don't forget that the simple exaction function of the discrete distribution for a general Bernoulli trial can help. Remember we're trying to make this easy.

Originally Posted by Strange
I liked mathman's wallaby's fuller answer
Lol. I thought you were being bitterly sarcastic with mathman.

21. Originally Posted by Quantime
Originally Posted by mathman
Your question is very unclear.
What are the odds of the 6th occurance being tails after 4 heads and 1 tails in that order.

.
Sorry, the answer is still 50/50

22. yeah well wait for what im preparing:

express a a number as the place where it apears first in random sucesion

the complete works of shakespeare appear on position 10 trillionth of pi

23. Well, that makes as much sense as most of your posts....

24.

25. Originally Posted by luxtpm
thanx for the self portrait.

26. yeah therefore with my rambling im more likely to produce shakespeares work, given enough time, than you whos stuck on nonsensenonsensenonsensenonsensenonsensenonsenseno nsensenonsensenonsense

27. Since that's what you post, how else can I respond? That's a problem on your end, not mine.

28. Yes well, we can safely ignore luxtpm's contributions here.

However, let's fix up our notation a bit. By convention, probabilities are expressed as a fraction of unity. So, for a single coin flip, the probability of H or T is 0.5. It is acceptable (but not usual) to write this as 50%. The notation 50:50 is meaningful, but refers to ODDS, not to probability, a different thing, whereas 50/50 has no meaning in either context.

So. Flip a (fair) coin. The probability of either H or T is 0.5. For what follows, let's write this as . Let's call this our "base case"

Now flip this same coin twice and count the number of possible outcomes. There are 4, right? So let's write the probability of each possible outcome as .

I assume as an hypothesis that for any flips the probability of each possible outcome is and set to prove that the probability of the -th flip being a certain chosen one is

But this is just which implies from my base case that, for any arbitrary , each subsequent flip has 0.5 probability

This is a poor example of "proof by induction"

29. Originally Posted by Guitarist
However, let's fix up our notation a bit. By convention, probabilities are expressed as a fraction of unity. So, for a single coin flip, the probability of H or T is 0.5. It is acceptable (but not usual) to write this as 50%. The notation 50:50 is meaningful, but refers to ODDS, not to probability, a different thing, whereas 50/50 has no meaning in either context.
Well, you could see 50/50 in the fractional form of probability (being in correct notation), where the cardinality of the sample space is 50 and the outcomes are inevitable. Although, of course, it would be better to write 1 or (as what would be better in that case) 100%. But I wouldn't say 50/50 has no statistical meaning at all.

And by the way, sorry for the messy notation in my previous posts!

But this is just which implies from my base case that, for any arbitrary , each subsequent flip has 0.5 probability
So are you saying that the probability is still 50-50? If yes, I would disagree. Sorry I was a little confused by your proof.

30. Originally Posted by luxtpm
yeah well wait for what im preparing:

express a a number as the place where it apears first in random sucesion

the complete works of shakespeare appear on position 10 trillionth of pi
Ah, so you wanna prove Pi's normal, eh? Are you serious about this? Because I would be exceedingly interested to see a proof. Otherwise, this is totally irrelevant to probability.

31. This is the same type of deal than trying to work out the odds of a certain lottery number coming up. If you have 7 numbers with a possibility of being any number from 1 to 49, with each number only coming up once of course, how do you work that out? As far as I understand, any sequence has the same chance as any other, even 1234567?

32. if you search here for number 11 or 22 or 33 or 44.... it appears 10% less than number 12.23,34,45,56...

verify it searching with your browser searcher

you take a sucesion of pi big enough the whole works of shakespeare would appear

therefore you compress the whole works of shakespeare into the position it appears in pi

(better do it a matrix like the secret message of the bible which was a square matrix of letters and read in the 4 senses of the surface of letters)

33. Originally Posted by luxtpm
you take a sucesion of pi big enough the whole works of shakespeare would appear
Prove it.

therefore you compress the whole works of shakespeare into the position it appears in pi
What if the number of bits needed to express that position (if it exists) is greater than the number of bits needed to represent all of Shakespeare's work plus video recordings of all performances? Then it wouldn't be much use.

(better do it a matrix like the secret message of the bible which was a square matrix of letters and read in the 4 senses of the surface of letters)
Oh dear, he's off on another random tangent.

What is the probability that luxtpm can make write two coherent sentences?

34. Originally Posted by KALSTER
As far as I understand, any sequence has the same chance as any other, even 1234567?
Absolutely. And of course you are less likely to share the prize because so many people will think, "that could never happen."

35. if you search with a browser searcher in a random secuence or pi you will find number 1111111 10% less times than 1234567

36. "What if the number of bits needed to express that position (if it exists) is greater than the number of bits needed to represent all of Shakespeare's work plus video recordings of all performances? "

this is the only that made sense of what you wrote the rest is trolling

if in the number PI the message is longer you express it with square root of two or with any irrational number in which you save space

so you express the complete works of shakespeare as a position in CERTAIN irrational number

37. Originally Posted by luxtpm
this is the only that made sense of what you wrote the rest is trolling
Actually, the really important bit was this:
Originally Posted by luxtpm
you take a sucesion of pi big enough the whole works of shakespeare would appear
Prove it.

38. Originally Posted by luxtpm
if you search with a browser searcher in a random secuence or pi you will find number 1111111 10% less times than 1234567
So? Use a longer random sequence and the difference will decrease. Not all sequences will appear exactly the same number of times (it would indicate the sequence wasn't random if that happened).

39. Originally Posted by Strange
Originally Posted by luxtpm
this is the only that made sense of what you wrote the rest is trolling
Actually, the really important bit was this:
Originally Posted by luxtpm
you take a sucesion of pi big enough the whole works of shakespeare would appear
Prove it.
on 10^2 digits of pi it will randomly appear a two digit string from shakespeare on 10^1000 string of digits will appear a 1000 digit string from shakespeare

so if the whole works of shalespeare take a trillion digits avrage it will appear on pi after 10^1 trillion digits

see how you were trolling?

40. Originally Posted by luxtpm
on 10^2 digits of pi it will randomly appear a two digit string from shakespeare on 10^1000 string of digits will appear a 1000 digit string from shakespeare

so if the whole works of shalespeare take a trillion digits avrage it will appear on pi after 10^1 trillion digits
This is only true if Pi is normal: Normal number - Wikipedia, the free encyclopedia (which is unproven).

41. haha youre funny

check yourself if you need farther proof :

http://stuff.mit.edu/afs/sipb/contrib/pi/

i
s like you belive in the good nature of the universe and at the same time you belive the horror which MEDIA portraits which is contradictory and you dont belive personal experience into pi being pseudorandom

then tell me a winning betting system for pi you can use starting in any staring point of the secuence

that is prove me is not normal

man prove even im real

edit:

besides i sayd average, of course pi is not normal only berautifull info is contained in it as the real universe not the FAKE media

42. Why does your posting style "decay" over time? You start being vaguely sensible and then get more and more irrational with every post.

43. i use both hemispheres in syncronicity some of the time the right creative and left logic

but the problem is that i still have a strong hemispheric dominance and at day have the creative hemisphere mostly working and at night the logic one

i can tell by which nostril i breath: at day left nostril at night right nostril

so what you liked was yesterday night

i bet most of you bretahe permanently only throught the right nostril, thats bad for the brain and spirit

edit:

well but i apologize i highjacked the thread ill open a new one

44. Originally Posted by luxtpm
but the problem is that i still have a strong hemispheric dominance and at day have the creative hemisphere mostly working and at night the logic one
That whole left brain/right brain thing is a complete myth, you know.

i can tell by which nostril i breath: at day left nostril at night right nostril
Oh good grief.

45. Originally Posted by KALSTER
This is the same type of deal than trying to work out the odds of a certain lottery number coming up. If you have 7 numbers with a possibility of being any number from 1 to 49, with each number only coming up once of course, how do you work that out? As far as I understand, any sequence has the same chance as any other, even 1234567?
Yes indeed, but here the stats get a bit different. I am not too sure how, as am one of those saddos who have a zero probability of winning i.e. I have never played

Let's see if I understand the game.

a) once a number is drawn it cannot be drawn again

b) order is unimportant

c) seven numbers are drawn

Is this correct? Anyway, assuming it is, the following seems to be true.

First assume you need to guess correctly all 49, then the probability of doing so is
But if only seven are allowed, we need to get a bit "fancy". There is an ubiquitous gadget in Math called the binomial coefficient which is written as which we read as "from n (objects) choose p (objects)".

It is a fact that which suggests itself as our solution, namely that there are ways to guess the lottery correctly, and so the probability of any single entry will simply be where the exponent denotes the multiplicative inverse

Anyone with a calculator (which doesn't include me!) can find a numerical solution. I suspect it is a rather small probability.......

46. Originally Posted by Guitarist
Anyone with a calculator (which doesn't include me!) can find a numerical solution. I suspect it is a rather small probability.......
If you enter (\frac{49!}{7! \times 42!})^{-1} in Wolfram Alpha, you get: 1/85900584 or about 1.16 x 10^-8

That is quite a small number.

47. Originally Posted by KALSTER
This is the same type of deal than trying to work out the odds of a certain lottery number coming up. If you have 7 numbers with a possibility of being any number from 1 to 49, with each number only coming up once of course, how do you work that out?
This is called conditional probability with dependent events. Since it involves non-replacement (same number can't show up twice or more in the sequence), it gets a little tricky.

49 x 48 x 47 x 46 x 45 x 44 x 43 = 432938943360. (mathematically, take the factorial of the individual sample space and reduce the operation of the largest terms to the number of trials).

So if you pick a number, any number (even 1-2-3-4-5-6-7), you'll have a chance of hitting jackpot. So "1 in a million" is sort of an understatement, a big understatement.

But truth is, that's a relatively easy lottery. Many lotteries use individual jumbling balls to make each number, allowing replacement.

(I'm not totally sure my method or calculation is right, so any correction is appreciated)

48. Originally Posted by Guitarist
It is a fact that which suggests itself as our solution, namely that there are ways to guess the lottery correctly, and so the probability of any single entry will simply be where the exponent denotes the multiplicative inverse
Looks like Guitarist beat me to it. You used a different method and got a different result. But you're probably much more knowledgeable on the subject than I am.

I thought you would take the factorial of the sample space, but minimizing its terms to the number of trials. I'm not sure why you would use the method above though.

49. Guitarist --> Hero

I guess my original question was the add up the odds, I mean say having 4 heads followed by 2 tails is one outcome but short of the 63 other ones along that path, therefore I thought that because the other 63 alternatives had gone, there was a larger probability there would be a tails instead of a heads.

I am now starting to think about if there is a term of quantum probability and the natural balance of baryoness, strangeness etc, seeing as the randomness of those particles at the quantum level still seem to add up to the same balance. The odds of heads or tails at that point is 50:50, but is that really an oversimplification?

Say to flip 18 heads in a row would be very unlikely. Flipping another after is very likely. Doesn't make sense but its true apparently. Just because we have already made it passed the previous 18 with heads odds of getting a 19th is 50:50 but if it made that 19 it would be 1 in 524,288 of getting 19 heads in a row true. Still 50:50

50. Deleted. Terribly sorry about that.

51. Nice work my friend. You got yourself a 7 day suspension for this very nice piece of mathematics.

Have a nice holiday

52. What the hell that was totally out of the blue on brody, what caused his overt aggression? Guitarist whats going on here, how come brody is quoting something you wrote, but didn't?

53. I don't believe I am unduly betraying confidences here, but there has been some discussion about this in the "Mod Corner".

There seem to be 2 schools of thought, namely

brody was drunk and forget himself

brody's account was hacked, and the abuse was not from HIM but from some hacker.I note that the abuse was probably from a mobile phone - note the "u" for "you" and "ur" for "you're", often referred to as "text-speak". I have no idea if this is relevant

Only brody himself can clarify this, and that would require me to lift the 7 day suspension which I believed at the time was lenient. For this, I would require a membership consensus. Personally I don't feel so inclined, but I am willing to be guided by the members

So your thoughts are?

54. Personally I smell hack, but we're not going to find out until after easter when brody shall rise again.

55. I don't think inebriation explains this, i'd have thought brody would have taken out his anger on someone he actually had a dislike of if this were the case. (i've never seen anything to indicate that Guitarist fits this category) I also thinks this is hacking.

56. A happy hacker ( or unhappy ) is at work.

57. Both his last two messages were from the same IP address. If it wasn't him, it must have been someone that had access to his computer.

58. Perhaps this might be a good time to introduce another probability problem, which may be of interest to the problem solvers.

The problem:
The two envelopes problem, also known as the exchange paradox, is a brain teaser, puzzle or paradox in logic, philosophy, probability and recreational mathematics, of special interest in decision theory and for the Bayesian interpretation of probability theory.
Specifically:
Let us say you are given two indistinguishable envelopes, each of which contains a positive sum of money. One envelope contains twice as much as the other. You may pick one envelope and keep whatever amount it contains. You pick one envelope at random but before you open it you are offered the possibility to take the other envelope instead.
The implications of switching:
The switching argument: Now suppose you reason as follows:
1. I denote by A the amount in my selected envelope.
2. The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2.
3. The other envelope may contain either 2A or A/2.
4. If A is the smaller amount the other envelope contains 2A.
5. If A is the larger amount the other envelope contains A/2.
6. Thus the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2.
7. So the expected value of the money in the other envelope is

8. This is greater than A, so I gain on average by swapping.
9. After the switch, I can denote that content by B and reason in exactly the same manner as above.
10. I will conclude that the most rational thing to do is to swap back again.
11. To be rational, I will thus end up swapping envelopes indefinitely.
12. As it seems more rational to open just any envelope than to swap indefinitely, we have a contradiction.

The puzzle: The puzzle is to find the flaw in the very compelling line of reasoning above.

59. Originally Posted by Write4U
7. So the expected value of the money in the other envelope is

The puzzle: The puzzle is to find the flaw in the very compelling line of reasoning above.[/COLOR]
Right there above.

Spoiler (white on white) follows NO PEAKING!!

You are not allowed to pre-multiply a "prize" by a probability. This is tantamount to the assertion that, all other things being equal, your chances of winning are proportional to the prize on offer.

Note also that in so doing the factor in the product on the RHS, the sum of all probabilities, is greater than 1 which makes no sense

60. Originally Posted by Guitarist

You are not allowed to pre-multiply a "prize" by a probability. This is tantamount to the assertion that, all other things being equal, your chances of winning are proportional to the prize on offer.

Note also that in so doing the factor in the product on the RHS, the sum of all probabilities, is greater than 1 which makes no sense
My answer is also in white... i hope.

The expected value is what is presented in the equation you quoted, this represents what we would be expected to win if we repeated the test a sufficiently large number of times. (there is no problem with this value surpassing one) However you are correct that the chances of winning are unaffected by the prize on offer, it's only the expected winnings of each contestant that changes. (in this particular case the expected winnings don't change either way) Also i'm pretty sure the expected winnings should be 3A/2, or 3A/4 depending on which envelope you define as the standard amount, and not 5A/4)

This seems to be a variant of the Monty Hall Problem, unlike in the traditional Monty Hall Problem though there is no advantage to be had by switching.

61. Originally Posted by wallaby
Originally Posted by Guitarist

You are not allowed to pre-multiply a "prize" by a probability. This is tantamount to the assertion that, all other things being equal, your chances of winning are proportional to the prize on offer.

Note also that in so doing the factor in the product on the RHS, the sum of all probabilities, is greater than 1 which makes no sense
My answer is also in white... i hope.

The expected value is what is presented in the equation you quoted, this represents what we would be expected to win if we repeated the test a sufficiently large number of times. (there is no problem with this value surpassing one) However you are correct that the chances of winning are unaffected by the prize on offer, it's only the expected winnings of each contestant that changes. (in this particular case the expected winnings don't change either way) Also i'm pretty sure the expected winnings should be 3A/2, or 3A/4 depending on which envelope you define as the standard amount, and not 5A/4)

This seems to be a variant of the Monty Hall Problem, unlike in the traditional Monty Hall Problem though there is no advantage to be had
Don't hide the answers, it limits discussion.

Say envelope A contains $10. Envelope B could contain 10/2 =$5, or 10x2 = $20. Thus by switching the possible gain would be greater than the possible loss. If I switch and I am wrong I will still have$5. If I am right I will have $20. I effect I would only risk$5 to gain $10. 62. There is another version of this: the game show contestant (or something), where a contestant has to choose between different doors and then also has an option to switch. I think it was on this forum a while back. 63. Originally Posted by KALSTER There is another version of this: the game show contestant (or something), where a contestant has to choose between different doors and then also has an option to switch. I think it was on this forum a while back. It is called "The Monte Hall Problem" Monty Hall problem - Wikipedia, the free encyclopedia 64. Originally Posted by Write4U Don't hide the answers, it limits discussion. Say envelope A contains$10. Envelope B could contain 10/2 = $5, or 10x2 =$20. Thus by switching the possible gain would be greater than the possible loss.

If I switch and I am wrong I will still have $5. If I am right I will have$20.

I effect I would only risk $5 to gain$10.
And this is why i hate Bayesian statistics problems, i miss the point every time.

65. Originally Posted by wallaby
Originally Posted by Write4U
Don't hide the answers, it limits discussion.

Say envelope A contains $10. Envelope B could contain 10/2 =$5, or 10x2 = $20. Thus by switching the possible gain would be greater than the possible loss. If I switch and I am wrong I will still have$5. If I am right I will have $20. I effect I would only risk$5 to gain \$10.
And this is why i hate Bayesian statistics problems, i miss the point every time.
But my little example does not solve the problem of switching. The paradox remains.

66. Look, you are conflating probability with game strategy - and as I tried to explain in my hidden post, this is less related to probability than to the stakes involved (also contestant psychology). If you want to move the discussion in that direction it's fine with me.

First note this "problem" as originally stated has very little to do with either Bayes or Monty Hall.

Simply this: the probability of guessing the envelope with the higher amount is 0.5, that of the lower amount is likewise 0.5. The probability of switching higher-to-lower is the same as the probability of switching lower-to-higher, both being 0.5.

The Monty Hall "problem" arises from the fact that there are 3 boxes, and the red-herring that Monty opens one to reveal a goat (or whatever). It's a red-herring because it tricks people into believing that this revelation somehow changes the probability of the original choice being "correct". It doesn't, it remains at 0.3333.....

But the second choice now has probability 0.5 of being correct, and so one has the ratio 0.333...:0.5 for the first and second choices respectively.

Better switch!!

67. Originally Posted by mathman
Originally Posted by KALSTER
There is another version of this: the game show contestant (or something), where a contestant has to choose between different doors and then also has an option to switch. I think it was on this forum a while back.
It is called "The Monte Hall Problem"
Monty Hall problem - Wikipedia, the free encyclopedia
oh of course if you can change a LIMITED times it doesnt matter

of course if you can change INFINITE TIMES is better

this is old:

68. Originally Posted by Write4U

But my little example does not solve the problem of switching. The paradox remains.
You're right. This question has been driving me mad so i've done some research and while i'm not sure if it resolves the paradox i think i've found where it goes wrong from the Bayesian point of view. (i guess someone can correct me if i'm wrong)

The way the problem has been phrased we have that after picking one envelope at random we must determine what we expect the other envelope to contain, without looking at the contents of the envelope that has already been chosen. So we have no information on the contents of either envelope and the probability that ours contains the greater amount is 0.5. The problem has then proposed that we assign a value to the contents of our envelope (A), then we consider two scenario's:
1. The other envelope contains the greater amount (2A),
2. The other envelope contains a lesser amount (),
However at step 6 we have assigned each of these scenario's a probability of 0.5, since it is equally likely that the other envelope contains an amount greater or less than the amount in our envelope. When we assign a value to our envelope the contents on the other envelope become conditional on 'A', . Here lies the problem.

By the definition of conditional probability, . Since we don't know what value our envelope has or what the sample space of 'A' is then we cannot assign a probability to or by extension .

It's some what of a weak argument, but it's all i can gather. (Asides from the fact that this paradox doesn't exist in frequentist probability theory)

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