/sci/ - Science & Math

It's not possible for the odds to become 2/3 or greater. Due to the fact that the coin is flipped an infinite number of times, it is not possible to be able to calculate. It is impossible to determine just how many heads are in an infinite number of coin flips.

infinite number of flips = 100% chance as the coin will eventually occupy every square foot in existence an infinite number of times.

/thread

>>2911027

I have a hard time believing that you get 2/3 a.s. There are infinitely many paths were you don't get to 2/3. Therefore, at t->inf you still have a positive probability of no 2/3? My measure theory sucks but it's not entirely clear to me that these infinitely many paths have measure zero rather than some positive probability.

>>2911658
No.

fuck you it can land on its side.

0%
infinite flips = infinite heads and infinite tails

there are no "paths," as that would be a quantification.

think about the wording of the questing.

On first toss, you get H or T. That is, there is a 50% chance that the ratio exceeds 2/3 (1/1 heads > 2/3 heads).

On second toss, if you got a tail on first toss, there is no chance that the ratio will exceed 2/3 (outcomes are TH and TT for first toss tails).

If first toss was heads, you already exceeded 2/3 on the first toss.

On third toss, if you got TH on first two, you can match 2/3 by getting THH (as opposed to THT). If first two tosses were TT, you cannot exceed 2/3. If first toss was H, you've already exceeded 2/3.


The question "what is the probability that, after an infinite number of flips, that the ratio of heads to tails is 2/3" IS MUCH DIFFERENT THAN "What are the odds that the overall (cumulative) proportion of heads (i.e. # heads / # flips) EVER reaches 2/3 or more?"

From a fairly crude empirical analysis, my gut says that the answer is 1/sqrt(2).

I don't think you quite understand the concept of infinity.

You agree that a random walk in one dimension has a 100% chance of returning to its starting point, right? That's what flipping a coin is. Heads is a step forward and tails is a step back.

>>2911767
I agree

>>2913328
my gut says the answer is 2/3

only because after 6 throws is is 21/32

Independent trials.
Binomial distribution.

I'm sure you can figure it out.

so many people in this thread don't understand the question.

and it is a fine question, though i'm having trouble answering it.

Let me get this straight: the possible events are infinite binary sequences where at each index the number is chosen randomly and uniformly (0 or 1) and independently of the preceding indices, right? And OP is now asking what the ratio a/b is where b is the "number" of all such sequences and a is the "number" of those sequence where the "ratio" heads/tails is at least 2/3, right? Well a=0 by the definition of a random experiment with two equally possible outcomes and thus a/b=0 which translates to: OP, the probability is 0%.

If at any point I was wrong in assuming OP's statement then OP should learn to express himself more clearly. Also, OP is a fag.

>>2913377
that is both tl;dr and wrong, it goes like this:

keep throwing a coin and recording the results

if at any point there is a 2/3 or greater H to T ratio of coins thrown so far then stop

what is the probability you will stop?

at first sight it seems fiendishly difficult

>>2913386
OP's post was tl;dr as well, your argument is invalid. The way you phrased it it's actually readable. As I said in my last paragraph, OP should apparently learn to express himself more clearly. As for the question, you have my attention. Gimme a sec.

http://en.wikipedia.org/wiki/Kolmogorov%27s_zero-one_law

It can only be 0 or 1. Since it can't be 0, it has to be 1.

>>2913397
what argument? your post was wrong and so long i didn't finish it.

>>2911027
wording of the question isn't ambiguous.

You simply need to find the percent of the paths where the heads/tails ratio is 2/3 or greater.

also, >>2913162

>>2913386
Won't the ratio be 1/0 on the first throw? So don't you have to stop then? Shouldn't there be some kind of buffer there?

>>2911684
it is 100% you dumbfuck. with an infinite number of flips it is bound to happen eventually.

due to the nature of the question the answer can only be 0% or 100%

>>2914360
This happens only with a probability of 50%. Thus we can already conclude that the probability OP is looking for will be higher than 50%.

Probability is a lie

100%. If the odds of something happening in a given amount of time is non-zero, and you allow an infinite amount of time, then that thing will eventually happen.

>>2914375
Continuing: we can in fact already conclude that the probability will be greater than 75%.

>>2914388
Going a step further I can already say that the probability OP is looking for is at least 13/16. Shit's getting quite big already, but looking at the was the probabilities develop from step to step I can't see yet if it will converge or not. Like OP I intuitively believe that it isn't 100%.

>>2914412
Now I've obtained that the probability is at least 7/8.

Wouldn't it be the sum of a number for fractions?

you'll have a 1/2 chance of it after the first throw (if it comes up heads). Add to this a 1/8 chance if it comes up THH, then whatever chance it is for 5 flips and so on.

I'm rather emasculated by this question, good job OP

>>2914386
The probability of the ratio never exceeding 2/3 is non-zero. Therefore it will never reach it.

Flawed logic is flawed

>>2914652

Can we create a summation that models this accurately. Is there a pattern to be found? I am only thinking this would work because I am doing statistical thermo and the partition function is a summation of probabilities, although it is not exactly the same.

You flip it an infinite amount of times.

There are no probabilities.

It lands on heads an infinite amount. It lands on tails an infinite amount. It lands on its edge an infinite amount.
1:1:1

Let's go binary! Let H be 1 and T be 0.
After n flips, we can represent our sequence of outcomes as an integer in binary with a range between 0 and 2^n - 1. Let S(n) be the number of "successful" sequences by OP's query after n flips. The desired probability after n flips is then P(n) = S(n) / 2^n

For example, after 1 flip:
H = 1 *
T = 0
S(1) = 1 => P(1) = 1/2

After 2 flips:
HH = 11*
HT = 10*
TH = 01
TT = 00
S(2) = 2 => P(2) = 1/2

After 3 flips:
HHH = 111*
HHT = 110*
HTH = 101*
HTT = 100*
THH = 011*
THT = 010
TTH = 001
TTT = 000
S(3) = 5 => P(3) = 5/8

The number of "successful" sequences is at the very least 2^(n - 1). Additional "successful" sequences can be found by examining sequences between 2^(n - 2) and 2^(n - 1) where the number of 1s in the binary representation is greater than the number of 0s. In the case n is even, however, sequences between 2^(n - 2) and 2^(n - 1) with equal 1s and 0s are also successful. For example, consider n = 4. The sequence THHT or 0110 = 6 is between 2^(4 - 2) = 4 and 2^(4 - 1) = 8.

Maybe more in a minute. It's butts time for now. OP, your question makes the endpoints of my facial lip region curl upward on both ends.

Datas:

P(1) = 1 / 2 = 0.5
P(2) = 2 / 4 = 0.5
P(3) = 5 / 8 = 0.625
P(4) = 10 / 16 = 0.625
P(5) = 20 / 32 = 0.625
P(6) = 42 / 64 = 0.65625
P(7) = 84 / 128 = 0.65625
P(8) = 168 / 256 = 0.65625
P(9) = 343 / 512 = 0.669921875
P(10) = 686 / 1024 = 0.669921875
P(11) = 1372 / 2048 = 0.669921875
P(12) = 2774 / 4096 = 0.67724609375
P(13) = 5548 / 8192 = 0.67724609375
P(14) = 11096 / 16384 = 0.67724609375
P(15) = 22335 / 32768 = 0.68161010742188
P(16) = 44670 / 65536 = 0.68161010742188
P(17) = 89340 / 131072 = 0.68161010742188
P(18) = 179408 / 262144 = 0.68438720703125
P(19) = 358816 / 524288 = 0.68438720703125

>>2915009
P(20) = 717632 / 1048576 = 0.68438720703125
P(21) = 1439140 / 2097152 = 0.68623542785645
P(22) = 2878280 / 4194304 = 0.68623542785645
P(23) = 5756560 / 8388608 = 0.68623542785645
P(24) = 11534438 / 16777216 = 0.68750607967377

Anyone else have something else towards this?

Finding a pattern has to be based on some factorial, continued fraction, or perhaps even prime numbers as the way the "valid" sequences seem to appear at higher amounts of flips (e.g. TTHHHH on 6 flips and TTTTHHHHHHHTHHH on 15 flips)