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April 3rd, 2017, 09:34 PM   #31
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Originally Posted by Mariga View Post
Romsek that's incorrect. It is not under binomial. Take a coin and flip it 50 times. count the number of heads and tails. You will notice it is around 25 give or take a few heads and 25 give or take a few tails. eg maybe 22 heads and 28 tails. it even gets better for 100 flips.. way better for 1000 flips and so forth.
assuming a fair coin with independent flips it is absolutely correct.

Your understanding of probability seems a bit lacking.
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April 4th, 2017, 12:25 PM   #32
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What's even better? I think you are wrong anyway. But that needn't worry you as long as you understand why you are so likely to get something close to a 50/50 split. Hint: it's to do with the "probability space". The important thing about the probability space is that is includes plenty of sequences where only 10% or fewer of the flips come up heads.
Sorry, guys, I sometimes don't explain things quite well coz I assume people do get me. But I think you have explained it well there. That the more we increase the sample size, the closer it gets to the 50/50 split. Let's say if we have 5 flips only. It is very possible to have only 20% as heads.. i.e. one head and 4 tails. If we have 10 flips, the chances of having just one head is very little, and the way to calculate it is though binomial distribution as Romsek explained. Still, if we have a million flips, chances of getting a thousand heads only is even smaller. In fact, chances of even getting 400,000 heads and 600,000 tails is also quite small.
Also, the number of consecutive heads or tails doesn't matter as long as they would eventually cancel each other out. Which explains why some numbers, e.g. 19 or 27, would have long chains but still would eventually reduce to 1.

Last edited by skipjack; April 6th, 2017 at 08:29 AM.
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April 4th, 2017, 12:45 PM   #33
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I agree with Romsek. You don't really understand probability. In particular, you don't seem to comprehend the concept of the Probability Space.

It's also false to suggest that in Collatz' chains we are randomly selecting odd or even numbers. From the moment you pick your first number, the rest of the chain is deterministic. Probability doesn't come into it.
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April 5th, 2017, 12:24 AM   #34
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I also agree. I don't have much understanding of probability and hence my vague explanation. But I do hope you guys get my point. If anything is unclear in what I have so far written, ask me to clarify.
On the point of whether the outcomes are probabilistic or deterministic, now that's a tough one. I will hence pose two arguments.

1.The first number we take, n is random and hence it is undetermined, and though the outcomes are determined by the formula, still, the outcomes are undetermined since the first one wasn't. So each outcome has a probability of 0.5 of being odd or even. That holds.

2.This point is a little stronger. Let's say we pick the first number, say 23. We have already determined it is odd, but we can't determine the next number in the sequence is unless we calculate it using the sequence's formula. It relates quite well to this example; let's say there are two boxes in front of you and you are told there is a cat in one of the boxes. But you can easily determine which box has a cat by opening at least one box.
So, when we have number 23, we do not know whether the next outcome will be odd or even until we calculate it. So chances are 50/50 same as the way chances are 50/50 for the cat.

Last edited by skipjack; April 6th, 2017 at 08:31 AM.
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April 5th, 2017, 12:48 AM   #35
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,,, when we have number 23, we do not know whether the next outcome will be odd or even until we calculate it.
Of course that is not true. Given 23, the entire rest of the sequence is determined, not random. 23 is odd so we multiply by 3 to get 69 and add 1 to get 70. Since 23 is odd, 3 times an odd number is odd, and adding 1 makes it even.

Can you think of any odd number that when you multiply it by 3 and add 1 you don't get an even number? I'd love for you to give me 50-50 odds on that. I'll bet even every time and I'll keep winning every bet.

Now continuing from 70, every number can be written as some odd number times a power of 2. In this case it's 2 times 35. Now 35 is odd and the next number must be even.

If you ever hit a power of 2, you collapse down to 1 and you're done. But if not, you collapse down to some odd number (by dividing by 2 over and over till you can't divide by 2 any more) and then you have an odd number and the next number is even ...

It's not a problem subject to any randomness at all. Given any number, the eventual outcome is determined, as someone already noted. The only question is how to figure out what the outcome must be. It's a problem that already has an answer, we just can't figure out whether every sequence collapses to 1 or not.

Last edited by skipjack; April 6th, 2017 at 08:33 AM.
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April 5th, 2017, 01:21 AM   #36
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Also, you may probably want to ask whether it is possible to have a chain of let's say 100 consecutive odds.. or even an infinite number of consecutive odds so that the sequence doesn't reduce. Because of the probability of 0.5 as earlier stated. And why do I insist it it always 0.5 and is not possible for it to be 1?

The formula for getting the consecutive odds, 1.5n+0.5, if u replace n with the first 8 odd numbers, you get the following results;
2,5,8,11,14,17,20,23...
you notice that half of the odd numbers yield even numbers. So for any odd number, there is a 50/50 chance of the outcome being odd or even.
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April 5th, 2017, 01:29 AM   #37
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Maschke, it seems we are not on the same page. If you have read my proof, you will notice I don't use 3n+1 because it is obvious the next integer would be even. Instead, I use (3n+1)/2 and hence the next integer is not obviously either odd or even.

Last edited by skipjack; April 6th, 2017 at 08:34 AM.
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April 5th, 2017, 10:36 AM   #38
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Maschke,it seems we are not on the same page. If you have read my proof, you will notice I don't use 3n+1 because it is obvious the next integer would be even. Instead, I use (3n+1)/2 and hence the next integer is not obviously either odd or even.
Then that's your error. You're completely misunderstanding the problem. Whenever you encounter an odd number, the next step is even; then you pull out the largest power of 2 that you can. If your even number was a power of 2, you're done; else you have some odd number other than 1, and you multiply by 3 and add 1 to get a larger even number. You're just confusing yourself by invoking (3n+1)/2. That's a huge obfuscation that has nothing to do with the problem.

Last edited by Maschke; April 5th, 2017 at 10:39 AM.
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April 5th, 2017, 03:29 PM   #39
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I can't say that I agree that your way of looking at it is the only correct way.
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April 5th, 2017, 03:48 PM   #40
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Quote:
Originally Posted by Mariga View Post
So for any odd number, there is a 50/50 chance of the outcome being odd or even.
For any odd number, there is either a 100% chance that the outcome is odd or there is a 100% chance that it is even. It's not a probablistic system, it's deterministic.

Even if it were probablistic, the probability space would still contain an infinite sequence of odd numbers. (Actually there are more sequences that that one, an infinite number in fact, which would disprove the conjecture but one is sufficient to exhibit the flaw in your proof).

Think about how monumentally unlikely it is that human beings appeared in the universe or that Leicester won the English Premier League last season. If we only ever got results within a standard deviation of the mean, neither would have happened - and probability theory wouldn't make sense.

A one in a million chance has to occur roughly one time in a million or it isn't a one in a million chance. The statement that something randomly occurs 50% of the time and doesn't the other 50% gives rise to the inescapable and logical conclusion that (infinitely) long chains of each state are possible and no less likely than any other infinitely long chain of states.
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