Is probability saved somewhere and somehow ?

[danov]

Is probability "saved somewhere and somehow"?

sorry if stupid question...still at school...only know school-math...

We have following example:

We have a coin (with two sides 1 and 0) and we let it fall to the ground.

The probability for each side to be on top is (1/2) or 50%.

Now if we would throw the coin 2 times and get the side 1 both times.
The probability for this is (1/2)*(1/2) = (1/4) or 25%.
Thus the probability to get side 1 a third time is (1/2)^3 = (1/8) or 12,5%.
Therefore the probability to get side 0 in third throw is 1-(1/8) = 87,5%.
Right?


If we throw the coin 10 times and all the times we get the side 1, the
probability for this is (1/2)^10 = (1/1024) or ~0,098%.
Thus the probability to get side 1 again in 11th throw is (1/2)^11 = (1/2048) or 0,049%.
Therefore the probability to get side 0 in 11th throw is 99,95%.

Now the important point is:
It becomes more and more probable to get side 0.

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And this is my question:

Lets say PersonA threw the coin 10 times and got side 1 all the time.
Now as stated before the probability to get side 0 in 11th throw is 99,95%.

But PersonA stops after 10th throw and put the coin in a safe place.
One year later PersonA takes this coin out again and throws it:

Now is the probability to get side 0 = 99,95% or 50% ?

 

[nicksauce]

The answer to your question is 50%. [Edit] Also note that it doesn't matter whether or not you wait a year or a millisecond between tosses, all your logic is wrong.[/Edit]

The probability that I will flip two coins in a row and get heads both times is 25%. But on the second flip the chance of getting heads is 50% regardless of what happened on the first flip.

This might be helpful: Gambler's Fallacy

 

[arildno]

Hi, danov!

It is no wonder you are puzzled by this, because there is an underlying fallacy in the way of thinking here:

1. When we say that there is (1/2)^10 chance of getting 1 in 10 throws, THAT probability is what we assign PRIOR TO HAVING THROWN THE COIN AT ALL!
That is BEFORE you throw the coin for the first time, the chance to get 1 10 times in a row is extremely unlikely.

2. However, a totally different situation is the following:
You happen to have thrown 1 9 times ALREADY, and you ask yourself:
What is the chance of getting 1 on the one remaining throw?

Since the coin doesn't remember its own history, the probability for getting 1 on your last throw is just the probability of getting 1 on anyone throw, i.e, 1/2.

 

[danov]

The answer to your question is 50%. [Edit] Also note that it doesn't matter whether or not you wait a year or a millisecond between tosses, all your logic is wrong.[/Edit]

The probability that I will flip two coins in a row and get heads both times is 25%. But on the second flip the chance of getting heads is 50% regardless of what happened on the first flip.

This might be helpful: Gambler's Fallacy

As I said: Sorry if stupid question - I have no math skills yet but only my interest in understanding the world..

But I have a question to your answer:

If this would be so than Stochastics and staticstics wouldn't be possible or not?
I mean: The stochastic prognose for getting side 0 in next thow with 99.95% probability, after having thrown side 1 ten times.
But as you said the probability still should be 50%.

That would mean it is absolutely unimportant what we have thrown before and no prognose possible.
Not?

 

[arildno]

"The stochastic prognose for getting side 0 in next thow with 99.95% probability"

Not at all. You are completely misunderstanding this.

 

[WarPhalange]

There is a difference between the chance of getting X number of heads in a row, and getting heads in the next run. You are confusing the two, which are unrelated.

 

[HallsofIvy]

As I said: Sorry if stupid question - I have no math skills yet but only my interest in understanding the world..

But I have a question to your answer:

If this would be so than Stochastics and staticstics wouldn't be possible or not?
I mean: The stochastic prognose for getting side 0 in next thow with 99.95% probability, after having thrown side 1 ten times.
But as you said the probability still should be 50%.

That would mean it is absolutely unimportant what we have thrown before and no prognose possible.
Not?

Absolutely not. If you throw a fair coin 10 times and it comes up 1 every time, the probabililty it will come up 0 the next time is still 50%, not 99.95%. That is exactly what "stochastics" will tell you.

Now, "statistics" might tell you, after 10 side 1 in a row that the coin is not likely to be fair! But it would also include a probability that that prediction is wrong.

 

[ephedyn]

If it would help you to understand the concept, let's simplify the problem.

A denotes tossing the coin and getting side 0,
B denotes tossing the coin and getting side 1.

Can you draw a tree diagram for 2 consecutive tosses and label, and figure out the values of:
- P(A)
- P(B)
- P(A|A)
- P(B|A)
- P(A|B)
- P(B|B)
- P(A intersect A)
- P(A intersect B)
- P(B intersect A)
- P(B intersect B)

And for that matter, do you see the difference between the set of events:
- (A|B)
- (B intersect B)'
Now, ask yourself, from your question, are you interested in (A|B) or (B intersect B)'?

SPOILER:
P(A) = 1/2
P(B) = 1/2
P(A|A) = 0.5
P(B|A) = 0.5
P(A|B) = 0.5
P(B|B) = 0.5
P(A intersect A) = 0.25
P(A intersect B) = 0.25
P(B intersect A) = 0.25
P(B intersect B) = 0.25

- (A|B) = 0.5
- (B intersect B)' = P(A intersect A) + P(A intersect B) + P(B intersect A) = 0.75
Your question is better phrased as, "What is the probability of A occurring GIVEN that B has already occurred consecutively for 10 times?"

The probability of A occurring AND 10 prior consecutive occurrences of B is (1/2)^11.
The probability of B occurring AND 10 prior consecutive occurrences of B is (1/2)^11.
There are 2 total possible outcomes, and only 1 pertains to A occurring after 10 consecutive occurrences of B. You asked, "[what] is the probability to get side 0 = 99,95% or 50%?"
= relevant outcome / total number of possible outcomes
= (relevant outcome / total number of possible outcomes) / (total number of possible outcomes / total number of possible outcomes)
= (relevant outcome / total number of possible outcomes) / [sum(1toTOTAL)(possible outcome/total number of possible outcomes)]
= probability of A occurring AND 10 prior consecutive occurrences of B / (probability of A occurring AND 10 prior consecutive occurrences of B + probability of B occurring AND 10 prior consecutive occurrences of B)
= (1/2)^11 / [(1/2)^11 + (1/2)^11]
= 0.5

Alternatively, what is the probability of A occurring GIVEN that B has already occurred consecutively for 10 times? 0.5.

P.S.: I haven't slept for 2 days, so I hope I got it right for yer.

 

[silkop]

sorry if stupid question...still at school...only know school-math...

We have following example:

We have a coin (with two sides 1 and 0) and we let it fall to the ground.

The probability for each side to be on top is (1/2) or 50%.

Now if we would throw the coin 2 times and get the side 1 both times.
The probability for this is (1/2)*(1/2) = (1/4) or 25%.
Thus the probability to get side 1 a third time is (1/2)^3 = (1/8) or 12,5%.

Wrong.

Let me explain probability to you from a Bayesian perspective. It may be more helpful than what you're taught at school. The gist of it is translating expressions like P(A|B) into English (it reads "probability of A given B" - but what does it mean - read on!).
A - some well-defined proposition, which can be true or false in reality
B - everything that you already know to be true
P(A|B) - given that you know B to be true, how probable is A? To answer this, you assume that the "world" can be in n possible states, in some of which A is true, in others it's false. The relative proportion of the A-true states to the total number of possible world states is what we refer to as probability.

Logical propositions like A and B can be combined. AB means "both A and B are true". A or B means pretty much what it says - A is true or B is true or both are true.

There are just two rules for combining probabilities. These rules can be derived from some desirable assumptions, you can ask if you are interested in them, but for now you will have to trust me they always work:
product (chain) rule: P(AB|C) = P(A|BC) P(B|C)
sum rule: P(A or B|C) = P(A|C) + P(B|C) - P(AB|C)

Now back to your coin example. Always start with defining the propositions that you wish to consider. It's very important to have clear propositions to work with:

X - all that we know about (idealized) coins
H1 - result of the first throw is heads (side 0)
T1 - result of the first throw is tails (side 1)
H2 - result of the second throw is heads
T2 - result of the second throw is tails
H3 - result of the third throw is heads
T3 - result of the third throw is tails

Let's state your assumptions about probabilities:

P(H1|X) = P(T1|X) = 0.5

This is equivalent to saying: "If all I was told is that an idealized coin was thrown, then I'd only know that one of two possibilities is true: the first result was heads or the first result was tails".

Furthermore, because it is an idealized coin, you could repeat the same statement regarding the second and third throw:

"If all I was told is that an idealized coin was thrown, then I'd only know that one of the two possibilities is true: the second [third] result was heads or the second [third] result was tails". In probability notation:

P(T3|X) = P(T2|X) = P(T1|X) = 0.5

The above expresses our assumption that each throw result is independent from all others. If your coin had memory, it would of course be a wrong assumption. But normal coins don't seem to have such memory and idealized coins most definitely don't (it is precisely what we mean by "idealized").

So, with all of the above, you are correct about
P(T1T2|X) = P(T2|T1X) P(T1|X) = P(T2|X) P(T1|X) = 0.5 * 0.5 = 0.25

(note how the product rule was applied to get from P(T1T2|X) to the already defined P(T2|X) and P(T1|X) and thus to actual numbers!)

But you are incorrect about
P(T3|X) = 0.125

which you confused with
P(T1T2T3|X) = P(T3T2|T1X) P(T1|X) = P(T3|T2T1X) P(T2|T1X) P(T1|X) = P(T3|X) P(T2|X) P(T1|X) = 0.5 * 0.5 * 0.5 = 0.125

Translated to natural language the above is "If all I was told was that a coin was thrown, then I'd only know that three-tails-in-a-row is one of the eight possible three-throws results. (You can list all the other seven ones if you wish.)"

Once again, note that the proposition T3, which means "third throw result is tails" is not the same as T1T2T3, which means "first, second, AND third throw result is tails".

 

[Hurkyl]

Lets say PersonA threw the coin 10 times and got side 1 all the time.

Okay. Given this information, what is the probability that all ten of those coin flips got side 1?

 

[silkop]

Lets say PersonA threw the coin 10 times and got side 1 all the time.

Okay. Given this information, what is the probability that all ten of those coin flips got side 1?

P(A|A) = 1, unless A is impossible, in which case P(A|A) is undefined.