If I flip a coin forever, is it possible that the coin will always land on heads? Or will it eventually land on tails?
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Yes, it’s theoretically possible that head always comes.
The probability for this is very low (everything equal to 0), but it is possible.
[And… no, this is not a contradiction that the probability is 0 and the event is still possible. Such events are called “almost impossible“ events. Note that ‘almost impossible” not the same is as “impossible”.]
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On the other hand, one can also say that there will almost certainly be a number at some point.
[Also here the difference between ‘almost safe and secure. There won’t be a number at some point. It will only come almost certainly sometime number.]
Thanks again.
Could one say, as a conclusion, that such an event in Aleph-0 does not occur “safely” attempts, but in Aleph-1 attempts has an actual entrance probability greater than 0 but smaller than Aleph-1?
The problem is that you can’t dice until all eternity. For every finite time horizon, there is a positive though so small probability that never comes. The mathematical lime against infinite then supplies a zero. You can interpret it in such a way that there is a number at some point. But it is a mathematical interpretation far from our reality.
Hello.
The law of the great numbers states that each event will occur in its frequency according to its probability.
So if you throw the coin up to all eternity, the probability that you always get your head is so low that you can count it with 0. This means that the probability for at least one number increases to 1.
Actually, you will get infinitely often head and infinitely often number.
I even had a few times in a row on edge. Very unbelievable.
This mathematically corresponds to a limit value formation.
The likelihood that we do not have a “head” is to
Now we will increase n as desired to “Grenzwert infinite” and get
Limes( 1 / 2^n ) for n -> infinite = 0
The limit value does not come close to zero but IST 0.
The probability of the event “never head” is therefore NULL for infinitely many litters.
EDIT:
What I described above applies to a single attempt with infinite litters.
As the question is asked, however, there are apparently two infinity.
One could imagine that one would have infinitely many attempts to achieve a number of infinitely many “head” events.
Then the lime formation would be
Limes [ m * Limes( 1 / 2^n ] for n -> infinite, m -> infinite
The limit value would coincide here
Limes (k / 2^k ) for k -> infinite
And as 2^k grows exponentially faster than k he would be NULL again here.
Achill sends greetings.
Yes, for ever means that there is never a conclusion after so many attempts.
A similar consideration: No matter how high the likelihood of infinite possibilities is theoretical – what happens at the most attempt has happened 100%. 😉
When everything goes to right things, the number of head and number will be the same in the long term.
The fact that only one side comes up is very unlikely.
She’s going to change.
In theory, this can happen, but the probability is more than extremely low.
PS: I did this as a teen and could determine what’s coming
But the likelihood of this is halved with every throw.
It is very unlikely, but possible. The coins are independent.
P(always head, literally always)=1/2^infinite
I thought you can’t just use as a number
about something like this:
1/Infinite = 0
Formed:
1 = 0 times infinite
power
1 = 0
what is wrong of course
That’s why he didn’t paint a symbol of equality, but a –>.
Formal correct would have to stand there: lim (1/2)^x = 0 for x against infinity
Yeah, right, you have to write, go against infinite, but come out almost on the same
It’s very unlikely.
At least if you have infinitely many people who cast infinitely many coins.
To what extent it makes a difference how many people do it. ?
Theoretically, this is possible.