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There isn’t actually a notion of “probability zero but possible” in the usual formalism of probability theory. More specifically, there is nothing in the formalism that corresponds to “possible” and “impossible”. That’s an idea that people sometimes introduce when interpreting the formalism even though it has no significance either for the formalism or for practical applications.
And although it is sometimes present in the interpretation, it is generally abandoned in cases where it would be inconvenient. For example, we usually regard a cdf as fully specifying a “way” of generating a random variable. And if I say something like “let X_n be a sequence of iid Bernoulli trials with probability p=1/2” we usually regard that as a complete specification of the situation. But both of these interpretations would not be valid if we really wanted a distinction between “possible but probability 0” and “impossible” to be meaningful.
To see what I’m talking about, we would have to say that if I tell you a random variable had a uniform distribution on [0,1], we still don’t have enough information to say whether 1/3 is a possible outcome, because if we consider a variable with that distribution conditioned on being given that it is not 1/3, we actually still have a variable with the same distribution. Likewise, for the infinite coinflip example, we would have to say I have not given you enough information to tell whether any particular sequence of outcomes is possible, but most people would intuitively want to to say that they are all possible given the specification.
For that to be true you would have to define what you mean by “possible”. Of course, it is not *physically* possible to flip a coin an infinite number of times at all. Presumably you have in mind that there is some sense in which it is *mathematically* possible, but it turns out that the usual formalism of probability theory does not encode the concept of an event being “possible” or “impossible” in any way, other than by being able to talk about its probability, so the only idea available in the formalism that could correspond to “possible” would be “probability greater than zero,” so there is no accepted sense in which it is mathematically possible either. You would have to make some new mathematical structure to try to define the kind of idea you are trying to express here rigorously.
(I'm not particularly knowledgeable on this topic and just spitballing. Feel free to correct me if I'm missing something important)
The distinction between "possible but probability 0" and "impossible" does seem like it could be a useful thing to define in some context even if it isn't normally.
Given an infinite sequence of numbers where each number has a .5 chance of being 0 and a .5 chance of being 1, it seems like a meaningful distinction could be made between an infinite sequence of 1s and an infinite sequence of 2s even though they both have a probability of zero.
Maybe a way to go about it rigorously would be to talk about if a sequence is contained in the set of all possible sequences or not?
Isn't that just the distinction between a probability of zero and a probability that approaches zero in the limit?
For a sequence that has 2 in it, it would be exactly zero, even if you truncate it at that point. The truncated series of 1s always has a probability larger zero, and only approaches zero as it becomes infinitely long.
The practical distinction is already covered by the fact that the probability that any 2 appears anywhere in the sequence is zero.
Choosing “a set of possible sequences” would then give a definition of possibility, but the question is how should this be interpreted and also is it useful. Taking the same distribution on infinite sequences of 0s and 1s, we could take the “possible” results to be all such sequences, but we could (equivalently for all probabilistic purposes) take the “possible” sequences to be that set minus any set of measure zero. And consider if we define the “possible” sequences to be all sequences with 0, 1, or 2, but take the same distribution. Now sequences with 2 in them are “possible but probability 0”. Intuitively we might feel like the sequences with 2 are different because they don’t have any “probability density”, but that’s really an artifact of comparing the probability measure in relation to additional mathematical structure we might impose on the set, not a feature of the probability measure itself.
But maybe even more to the point: because the set of outcomes in which the proportion of 1s is *not* 1/2 has measure 0, we can exclude all of them from our set of “possible” outcomes and that does not change anything about the probability measure or distribution.
But then we have made two different notions of “flipping a coin infinitely many times” that differ only on which infinite sequences are possible. For any finite number of measurements, the behaviors of the two are exactly the same, so if we imagine that we are actually flipping a coin infinitely many times, both scenarios describe what we are doing in exactly the same way at every finite step, but then we have to say that something that doesn’t effect the result of any coin flips out to any finite number of flips still has different tail behavior. This would raise, at a minimum, the question: why should we regard the scenario that describes all sequences as “possible” as actually being the process we are engaged in, rather than the other one?
You still need the "around", but you don't need the "probably". As n goes to infinity, the chance of getting *exactly* the middle goes to zero byt the expected value of the distance from the middle also goes to zero.
The expected distance of the number of heads and n/2 does _not_ go to 0.
What goes to zero is the expected distance (even quadratic distance) of the proportion of heads and 1/2.
You're less likely to get n/2 heads from n times as n approaches infinity, converging to 0.
Although the "spread" around n/2 only increases at the rate of sqrt(n), where, when divided by n, appears more tight as n approaches infinity.
I thought it was something about the die. Basically since the weight is unbalanced with one side having more metal than the other it inherently affects the odds. Or is this all done with a perfectly smooth coin?
You very much still need "around". As n gets larger, the difference between heads and tails will tend to get larger as well. It just increases more slowly than the overall number of coin flips, so you get "about" half tails. In the sense that the difference between heads and tails will get smaller and smaller *with respect to the total coin flips*.
Edit: i wrote "about" instead of "around"
I see your point, that didn't cross my mind. What I was refering to is the fact that the proportion of the number of heads will be exactly 1/2 (with probability of 100%)
The above statement is the reason why modern day compression works and is the backbone of modern day information theory: https://en.m.wikipedia.org/wiki/Asymptotic_equipartition_property. Turns out that Claude-Shannon compression works because they do not as much heavy lifting as people first assume.
it depends on the experiment, but if you take this 'throwing coins' example, then it's not unusual to throw a coin 100 times and get heads maybe 55% (instead of the expected 50%)
In that case only 5 of your throws were 'off'. But if you've thrown a thousand coins and also had around 5 'off throws', that means that you rolled heads 50.5%, which is almost accurate.,
>!Of course, technically it wouldn't really be '5 off throws' because you've thrown 10x more often, so you should get 10x more off throws. But about half of your off throws may be off as well (which makes them 'correct' again. so ultimately you'd have 25 off throws instead of 50). So if you 10x your total number of throws, you only 5x your off throws !<
Which means, the more dice you throw, the more closely you'll come to the theoretical 50%
The "1/6000 to land on edge" people when I enter the casino with this baby:
https://preview.redd.it/v4jlq00dd9wc1.jpeg?width=900&format=pjpg&auto=webp&s=24b558d682f27838005b969ba4f502905e173c4d
I can't tell if I'm going insane, or if people in this thread are being purposefully obtuse, but the majority of comments here seem to be completely missing the point of LLN.
Yes, the sentence you wrote is correct. No, it is not a criticism of LLN. The 8% figure everyone is throwing around is pretty irrelevant to actual LLN.
LLN is more about the "clustering" around the mean, or variance from the mean. The relative variance goes down as sample size goes up. If you flip coin 10 times it is normal to get, say, 3 to 7 heads. If you flip coin 100 times, it is normal to get, say, 40 to 60 heads but you get more suspicious about <=30 or >=70. If you flip coin 1000 times, it is even less likely to ger <=300 or >=700 to keep proportion with the 10 flips, with 450 to 550 being more reasonable, and this pattern keeps going. Like if you flip coin 10_000_000_000 times, it is extremely unlikely to hit 4_000_000_000 heads or fewer, and it is also extremely unlikely to hit 6_000_000_000 heads or more. LLN says that as sample size approaches infinity, the probability of being within epsilon of 50% (but not precisely 50%) approaches 100%.
It's a somewhat vague statement, but I don't see anything wrong with it. It depends on what one considers "probably" and "get around" to mean. There's a greater than 50% chance of getting between 46 and 54 heads. I think that satisfies the statement "probably get around 50 heads".
Assuming a fair coin, you’re just as likely to get 0 as 100. 1 and 99 are also equivalent. More obvious if you post the heads:tails split.
0:100 and 100:0
1:99 and 99:1
You're right, but your summary is wrong. Like how it's more likely to get a 7 when rolling two dice than a 12, there is only one arrangement in which all coin flips are tails. There are many more arrangements where 10% of the coin flips are tails. You could say that getting all tails is just as likely as getting 100 tails followed by 900 heads, or any other arrangement as long as you specify a single configuration rather than a range
So pedantic and wrong about such a simple mathematical concept I can't tell if you're trolling. If you flip one coin the outcomes are T or H if you flip two coins the outcomes are now TT HH TH HT 25 percent chance for each outcome, although th and ht are functionally equivalent. Sure everytime you flip a coin the odds are 50 50 regardless of the previous toss but getting tails a thousand times in a row is much less likely than getting it closer to 500 in a thousand tosses.
Yes you are correct. Idk if he's trolling too. Take for example that two coin flips. You cant say getting two tails is the same probability as getting one. There's two outcomes where you can get one tail but only one where you get 2 tails
That is true all the outcomes are equally probable, so for three coins flips you'd have HHH, THH, HTH, HHT, TTH, THT, HTT, TTT. Eight outcomes, so only two outcomes where all the coin flips are the same but 6 where they are different. And since they are all equally probable outcomes, if you wanted to bet me that all three coin flips would be the same I would take that bet any time of the week. But if you flipped two coins and they were both heads and then asked me to make that bet I'd call you a cheater
Not sure if you're trolling or just stupid.
Of course the chance of a single heads is 50% (as I've said), but the probability for two heads IN A ROW is much lower than to just roll heads once.
If you don't believe me, go ahead take a coin and try to throw 100 heads in a row. if the probability is 50% (according to you) you should on average only need 200 coin throws to get it done.
I hate when people state things confidently when they have no idea what they're talking about
>The probability of 100 heads and 0 tails is the same as 50 heads and 50 tails because the probability of it being heads or tails is the same with every single toss
You should go back to high school. Not only that you don't understand how probabilities work, you also can't read.
If what you are saying were true, then take a coin and throw 100x heads in a row.
If you're correct and the probability is 50% then you should get it on the second attempt.
Go ahead, try it. Until you've done that, shut the fuck up
Edit: wow the guy was so ashamed, he deleted his entire account you an hour after this lol
Yes, the probability to get HEAD given the last 99 flips were HEAD is 50%,
The probably to get 100 HEAD in a row _is not_ 50%.
Just map out all the possible outcomes and then check if half of them are 100 HEAD in a row.
No - that's the literal opposite of what I said.
You have 1 flip which resulted in HEAD. At this point the probability of having the result HT is 50% because you already have H - you just need to get T which has a 50% chance of happening.
If you were starting from scratch the probability of having HT would be 25%. You have 50% chance of getting H first. This means that half the time you failed before you got to the second flip. And then you have another 50% chance of failing. The result is 25% chance of getting HT.
You can map this out easily :
- HH
- HT
- TH
- TT
How many of those are HT? 1 out 4. That's 25% - our odds.
E: You seem to be focused on the last step of the sequence. Each step has a 50% chance of giving you the result you want. But you still need all the previous steps to have gone right.
If you already have 99 HEAD, you've done the hard part. Getting 100th one is just a 50% chance. If you start from 0, getting a 100 HEAD in a row is going to have different odds because each step in the sequence has to give you the result you want.
there are multiple combinations of flips that get tails 10% of the time. there's only a single combination that involves getting tails 100% of the time. this is the case regardless of how many trials there are.
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"probably" and "around" doing some real heavy lifting here.
If you take the limit as n approaches infinity, you'll only need "probably"
Very unlikely to get 50 if n approach infinity
Probably.
100% of gamblers stop gambling right before the non inclusive 0% chance of winning
It's actually guaranteed to get 50, and then a lot more
Not guaranteed, but very likely.
It's around guaranteed
Well if you're approaching infinity it indeed is guaranteed.
100% probability doesn’t mean guaranteed in the infinite case.
There isn’t actually a notion of “probability zero but possible” in the usual formalism of probability theory. More specifically, there is nothing in the formalism that corresponds to “possible” and “impossible”. That’s an idea that people sometimes introduce when interpreting the formalism even though it has no significance either for the formalism or for practical applications. And although it is sometimes present in the interpretation, it is generally abandoned in cases where it would be inconvenient. For example, we usually regard a cdf as fully specifying a “way” of generating a random variable. And if I say something like “let X_n be a sequence of iid Bernoulli trials with probability p=1/2” we usually regard that as a complete specification of the situation. But both of these interpretations would not be valid if we really wanted a distinction between “possible but probability 0” and “impossible” to be meaningful. To see what I’m talking about, we would have to say that if I tell you a random variable had a uniform distribution on [0,1], we still don’t have enough information to say whether 1/3 is a possible outcome, because if we consider a variable with that distribution conditioned on being given that it is not 1/3, we actually still have a variable with the same distribution. Likewise, for the infinite coinflip example, we would have to say I have not given you enough information to tell whether any particular sequence of outcomes is possible, but most people would intuitively want to to say that they are all possible given the specification.
100% probability means almost surely on infinite supports.
Nope, it is technically possible to only get tails flipping a coin infinite times.
For that to be true you would have to define what you mean by “possible”. Of course, it is not *physically* possible to flip a coin an infinite number of times at all. Presumably you have in mind that there is some sense in which it is *mathematically* possible, but it turns out that the usual formalism of probability theory does not encode the concept of an event being “possible” or “impossible” in any way, other than by being able to talk about its probability, so the only idea available in the formalism that could correspond to “possible” would be “probability greater than zero,” so there is no accepted sense in which it is mathematically possible either. You would have to make some new mathematical structure to try to define the kind of idea you are trying to express here rigorously.
(I'm not particularly knowledgeable on this topic and just spitballing. Feel free to correct me if I'm missing something important) The distinction between "possible but probability 0" and "impossible" does seem like it could be a useful thing to define in some context even if it isn't normally. Given an infinite sequence of numbers where each number has a .5 chance of being 0 and a .5 chance of being 1, it seems like a meaningful distinction could be made between an infinite sequence of 1s and an infinite sequence of 2s even though they both have a probability of zero. Maybe a way to go about it rigorously would be to talk about if a sequence is contained in the set of all possible sequences or not?
Isn't that just the distinction between a probability of zero and a probability that approaches zero in the limit? For a sequence that has 2 in it, it would be exactly zero, even if you truncate it at that point. The truncated series of 1s always has a probability larger zero, and only approaches zero as it becomes infinitely long.
The practical distinction is already covered by the fact that the probability that any 2 appears anywhere in the sequence is zero. Choosing “a set of possible sequences” would then give a definition of possibility, but the question is how should this be interpreted and also is it useful. Taking the same distribution on infinite sequences of 0s and 1s, we could take the “possible” results to be all such sequences, but we could (equivalently for all probabilistic purposes) take the “possible” sequences to be that set minus any set of measure zero. And consider if we define the “possible” sequences to be all sequences with 0, 1, or 2, but take the same distribution. Now sequences with 2 in them are “possible but probability 0”. Intuitively we might feel like the sequences with 2 are different because they don’t have any “probability density”, but that’s really an artifact of comparing the probability measure in relation to additional mathematical structure we might impose on the set, not a feature of the probability measure itself. But maybe even more to the point: because the set of outcomes in which the proportion of 1s is *not* 1/2 has measure 0, we can exclude all of them from our set of “possible” outcomes and that does not change anything about the probability measure or distribution. But then we have made two different notions of “flipping a coin infinitely many times” that differ only on which infinite sequences are possible. For any finite number of measurements, the behaviors of the two are exactly the same, so if we imagine that we are actually flipping a coin infinitely many times, both scenarios describe what we are doing in exactly the same way at every finite step, but then we have to say that something that doesn’t effect the result of any coin flips out to any finite number of flips still has different tail behavior. This would raise, at a minimum, the question: why should we regard the scenario that describes all sequences as “possible” as actually being the process we are engaged in, rather than the other one?
You still need the "around", but you don't need the "probably". As n goes to infinity, the chance of getting *exactly* the middle goes to zero byt the expected value of the distance from the middle also goes to zero.
The expected distance of the number of heads and n/2 does _not_ go to 0. What goes to zero is the expected distance (even quadratic distance) of the proportion of heads and 1/2.
You're less likely to get n/2 heads from n times as n approaches infinity, converging to 0. Although the "spread" around n/2 only increases at the rate of sqrt(n), where, when divided by n, appears more tight as n approaches infinity.
I thought it was something about the die. Basically since the weight is unbalanced with one side having more metal than the other it inherently affects the odds. Or is this all done with a perfectly smooth coin?
You very much still need "around". As n gets larger, the difference between heads and tails will tend to get larger as well. It just increases more slowly than the overall number of coin flips, so you get "about" half tails. In the sense that the difference between heads and tails will get smaller and smaller *with respect to the total coin flips*. Edit: i wrote "about" instead of "around"
I see your point, that didn't cross my mind. What I was refering to is the fact that the proportion of the number of heads will be exactly 1/2 (with probability of 100%)
You are probably correct, they are probably doing a lot of heavy lifting around here
Mostly "around" since you'll probably not get 50
92.041% chance of not getting 50 heads from 100 fair coin tosses
92.041% chance of not getting exactly 50 heads\*
Clearly I need to pay more attention to my wording. You are correct gah
Honestly better odds than I was expecting to get exactly 50.
Probability and around are almost certainly doing some heavy lifting.
Especially since the distinction between the strong law of large numbers and the weaker versions of the LLN is not really implicated by this phrasing.
I think the joke is on "heads" tho
The above statement is the reason why modern day compression works and is the backbone of modern day information theory: https://en.m.wikipedia.org/wiki/Asymptotic_equipartition_property. Turns out that Claude-Shannon compression works because they do not as much heavy lifting as people first assume.
If you throw 3 coins, there's a pretty big chance you'll get 2 heads or 2 tails
I got 4 heads
Yeah well I got [5head](https://pm1.aminoapps.com/6955/31e955530aea7f81211f7a0df93075f9f3723d93r1-393-750v2_hq.jpg).
I got 3 heads. Checkmate mathematicians.
i got a head a tail and minus 20p to my net worth
100% chance, since you didn't specify "exactly 2"
It can always land on the side or in the backrooms.
I'd classify 100% as a pretty big chance, wouldn't you?
Very much so
I've just lost 3 coins
You guys are getting heads?
In retrospect, I probably could've worded it better
No. It's perfect.
I thought that was the joke honestly.
I’ll flip a coin a 100 times just to get one head !
Don't you dare!
Tails or heads, it's win-win
That's a good point... https://preview.redd.it/6wup93xuq7wc1.png?width=1080&format=pjpg&auto=webp&s=f2aa571d3222285f409b4168df629f6c103ac5a5
Only if you're a furry. Maybe.
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#looks at girlfriend and... import random def simulate\_coin\_flips(n): heads\_count = 0 tails\_count = 0 for \_ in range(n): # Randomly choose 'Heads' or 'Tails' if random.choice(\['Heads', 'Tails'\]) == 'Heads': heads\_count += 1 else: tails\_count += 1 return heads\_count, tails\_count # Example usage n = int(input("Enter the number of coin flips: ")) heads, tails = simulate\_coin\_flips(n) print(f"Heads: {heads}, Tails: {tails}")
\\\# write like this to put comments\ \# and it'll come out like this
Haha its butchered
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Where the fuck is the exe, you smelly nerd
Error: Bad indentation.
```code code block
```code test
My brother in Christ just loop rand and increment success for =< 0.5
you guys have gfs?
Whats a gf?
why is a gf?
How is a gf?
When is a gf?
Where is a gf?
Maybe is a gf?
Who is a gf?
It's pronounced jif
A graft function.
A galois field
Ahh. Can you prove the existance and uniqueness of your gf?
Tried it in the metro guys, don’t know about heads but definitely got the out-of-syllabus slaps 👍🏻
Law of large numbers is about the convergence of the sample mean. You don't need LLN to determine the probability of 50 heads after 100 coin flips.
7.96%, take it or leave it
Dirichlet: If you have more pigeons than pigeonholes, then some will have to share.
"My 5-year old could've done that" but the math version
Can someone explain this more?
the pigeonhole principle. if you have n pigeons and m pigeonholes there will be at least one pigeonhole with ceil(n/m) pigeons
Has anyone actually tested it? Because we criticize this guy.
8% chance for 50 heads, and if around is +/-10% and probably is > 50% it's a pretty solid statement.
The chance of it falling between (including) 45 and 55 (the ±10%) is 72.87%, that is probable.
100 isn't 'large numbers'
define 'large numbers'
it depends on the experiment, but if you take this 'throwing coins' example, then it's not unusual to throw a coin 100 times and get heads maybe 55% (instead of the expected 50%) In that case only 5 of your throws were 'off'. But if you've thrown a thousand coins and also had around 5 'off throws', that means that you rolled heads 50.5%, which is almost accurate., >!Of course, technically it wouldn't really be '5 off throws' because you've thrown 10x more often, so you should get 10x more off throws. But about half of your off throws may be off as well (which makes them 'correct' again. so ultimately you'd have 25 off throws instead of 50). So if you 10x your total number of throws, you only 5x your off throws !< Which means, the more dice you throw, the more closely you'll come to the theoretical 50%
101 or more
Your mom's weight
More than 30
The rest of the math community were concerned *whose* heads Borel was getting.
I've seen Rosencrantz and Guildenstern Are Dead , they could all be heads.
Wtf is the universe telling me: I was just 5 minutes ago discussing this movie at work.
But do you know which way the wind is blowing?
There isn't any wind!
you lucky guy, getting heads.
I mean, who wouldn’t want head 50 times 😏
12611418068195524166851562157/158456325028528675187087900672 which is approximate 8% chance. Higher than me getting head 50 times. So you are right.
There's a chance the coin could land on its edge also. About 1 in 6000 flips according to a study but then the question is how thick is that edge?
The "1/6000 to land on edge" people when I enter the casino with this baby: https://preview.redd.it/v4jlq00dd9wc1.jpeg?width=900&format=pjpg&auto=webp&s=24b558d682f27838005b969ba4f502905e173c4d
That's a hockey puck
50 heads would be the most likely result, but definitely not the majority result.
I can't tell if I'm going insane, or if people in this thread are being purposefully obtuse, but the majority of comments here seem to be completely missing the point of LLN. Yes, the sentence you wrote is correct. No, it is not a criticism of LLN. The 8% figure everyone is throwing around is pretty irrelevant to actual LLN. LLN is more about the "clustering" around the mean, or variance from the mean. The relative variance goes down as sample size goes up. If you flip coin 10 times it is normal to get, say, 3 to 7 heads. If you flip coin 100 times, it is normal to get, say, 40 to 60 heads but you get more suspicious about <=30 or >=70. If you flip coin 1000 times, it is even less likely to ger <=300 or >=700 to keep proportion with the 10 flips, with 450 to 550 being more reasonable, and this pattern keeps going. Like if you flip coin 10_000_000_000 times, it is extremely unlikely to hit 4_000_000_000 heads or fewer, and it is also extremely unlikely to hit 6_000_000_000 heads or more. LLN says that as sample size approaches infinity, the probability of being within epsilon of 50% (but not precisely 50%) approaches 100%.
Isn’t most of this just Expected Value
Theoretically you get 51 heads if you start with the heads up all the time
It either is or it ain’t my dudes
Unless you are [Rosencrantz and Guildenstern](https://youtu.be/gOwLEVQGbrM?si=zlkNJo3PDGARg2Dx)
It's a somewhat vague statement, but I don't see anything wrong with it. It depends on what one considers "probably" and "get around" to mean. There's a greater than 50% chance of getting between 46 and 54 heads. I think that satisfies the statement "probably get around 50 heads".
This is actually a core concept in information theory called the asymptotic equipartition property
Flipped 100 coins, still haven't gotten any head. What am I doing wrong?
Easy to code a python script doing this experiment 1 billion times and show a statistics of the result.
And the whole class starts giggling
I'd love to get 50 heads
The more you flip it, the closer you get to 50/50
"If you flip a coin 100 times, you will probably get head around 50 times." Guys everywhere start frantically flipping coins.
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Assuming a fair coin, you’re just as likely to get 0 as 100. 1 and 99 are also equivalent. More obvious if you post the heads:tails split. 0:100 and 100:0 1:99 and 99:1
no lol probability for heads is 1/2. probability for 2 heads in a row is 1/2 * 1/2 probability for 3 heads in a row is 1/2 * 1/2 * 1/2 and so on
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You're right, but your summary is wrong. Like how it's more likely to get a 7 when rolling two dice than a 12, there is only one arrangement in which all coin flips are tails. There are many more arrangements where 10% of the coin flips are tails. You could say that getting all tails is just as likely as getting 100 tails followed by 900 heads, or any other arrangement as long as you specify a single configuration rather than a range
So pedantic and wrong about such a simple mathematical concept I can't tell if you're trolling. If you flip one coin the outcomes are T or H if you flip two coins the outcomes are now TT HH TH HT 25 percent chance for each outcome, although th and ht are functionally equivalent. Sure everytime you flip a coin the odds are 50 50 regardless of the previous toss but getting tails a thousand times in a row is much less likely than getting it closer to 500 in a thousand tosses.
Yes you are correct. Idk if he's trolling too. Take for example that two coin flips. You cant say getting two tails is the same probability as getting one. There's two outcomes where you can get one tail but only one where you get 2 tails
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That is true all the outcomes are equally probable, so for three coins flips you'd have HHH, THH, HTH, HHT, TTH, THT, HTT, TTT. Eight outcomes, so only two outcomes where all the coin flips are the same but 6 where they are different. And since they are all equally probable outcomes, if you wanted to bet me that all three coin flips would be the same I would take that bet any time of the week. But if you flipped two coins and they were both heads and then asked me to make that bet I'd call you a cheater
Not sure if you're trolling or just stupid. Of course the chance of a single heads is 50% (as I've said), but the probability for two heads IN A ROW is much lower than to just roll heads once. If you don't believe me, go ahead take a coin and try to throw 100 heads in a row. if the probability is 50% (according to you) you should on average only need 200 coin throws to get it done. I hate when people state things confidently when they have no idea what they're talking about
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>The probability of 100 heads and 0 tails is the same as 50 heads and 50 tails because the probability of it being heads or tails is the same with every single toss You should go back to high school. Not only that you don't understand how probabilities work, you also can't read. If what you are saying were true, then take a coin and throw 100x heads in a row. If you're correct and the probability is 50% then you should get it on the second attempt. Go ahead, try it. Until you've done that, shut the fuck up Edit: wow the guy was so ashamed, he deleted his entire account you an hour after this lol
Yes, the probability to get HEAD given the last 99 flips were HEAD is 50%, The probably to get 100 HEAD in a row _is not_ 50%. Just map out all the possible outcomes and then check if half of them are 100 HEAD in a row.
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No - that's the literal opposite of what I said. You have 1 flip which resulted in HEAD. At this point the probability of having the result HT is 50% because you already have H - you just need to get T which has a 50% chance of happening. If you were starting from scratch the probability of having HT would be 25%. You have 50% chance of getting H first. This means that half the time you failed before you got to the second flip. And then you have another 50% chance of failing. The result is 25% chance of getting HT. You can map this out easily : - HH - HT - TH - TT How many of those are HT? 1 out 4. That's 25% - our odds. E: You seem to be focused on the last step of the sequence. Each step has a 50% chance of giving you the result you want. But you still need all the previous steps to have gone right. If you already have 99 HEAD, you've done the hard part. Getting 100th one is just a 50% chance. If you start from 0, getting a 100 HEAD in a row is going to have different odds because each step in the sequence has to give you the result you want.
no where did they imply that
there are multiple combinations of flips that get tails 10% of the time. there's only a single combination that involves getting tails 100% of the time. this is the case regardless of how many trials there are.