Approaching 1/2 Probability

[studiot]

16 minutes ago, Conjurer said:

Then it doesn't make sense to me how you could get (nCr(H or T))/2^n = 1/2 as n->infinity.

This is why I asked you what your definition of probability is.

18 minutes ago, Conjurer said:

You got it backwards; that is the type of solution that I am looking for

I don't understand this.

Let me work it forwards a bit more fore you

Let P(n) be the probability that n trials will contain an equal number (n/2) of heads and tails.

p(0) = 1 since there are 0 heads and zero tails

P(1) = 0 since there is either one head or one tail

P(2) = 0.5 since there are 4 possibilities, two of which have an equal number of heads and tails

P(3) = 0 since there are no possibilities with an equal number of heads and tails

over to you

[Ghideon]

16 minutes ago, Conjurer said:

If you wanted to determine if a coin was fair or not, you would run enough trials to get a 5 sigma over and over where you were able to count the number of heads or tails, and that ended up being close to half of the flips.  The closer that was to 1/2, then the "fairer" the coin would be.  How else would you determine if a coin was fair or not?

Did you understand the difference between the two equations I posted? Which one of them do you claim to be incorrect?

[Conjurer]

3 minutes ago, Ghideon said:

Did you understand the difference between the two equations I posted? Which one of them do you claim to be incorrect?

The one that says the absolute difference between the number of heads and tails approaches infinity with an increasing number of flips.

Edited September 8, 2019 by Conjurer

[uncool]

4 minutes ago, Conjurer said:

The one that says the absolute difference between the number of heads and tails approaches infinity with an increasing number of flips.

Do you understand what "absolute difference" means?

[Conjurer]

14 minutes ago, studiot said:

This is why I asked you what your definition of probability is.

I don't understand this.

Let me work it forwards a bit more fore you

Let P(n) be the probability that n trials will contain an equal number (n/2) of heads and tails.

p(0) = 1 since there are 0 heads and zero tails

P(1) = 0 since there is either one head or one tail

P(2) = 0.5 since there are 4 possibilities, two of which have an equal number of heads and tails

P(3) = 0 since there are no possibilities with an equal number of heads and tails

over to you

To find the probability of a series of events occurring you have to find the total amount of combinations it occurs divided by the total number of possible outcomes. 

https://www.calculator.net/permutation-and-combination-calculator.html?cnv=4&crv=2&x=85&y=24

You can put in 4 as n, because that is the number of flips.  We want half of the outcomes to be heads or tails, so then r=2.

Then you find that out of 4 flips, there are 6 possible combinations you can get an even number of heads or tails.

Then to find the total number of outcomes, you would take 2^n where n=4, which comes out to 16.

Then you have a 6/16 or 3/8 chance of getting the same number of heads and tails with 4 coin flips.

3 minutes ago, uncool said:

Do you understand what "absolute difference" means?

Yes.

Edited September 8, 2019 by Conjurer

[studiot]

7 minutes ago, Conjurer said:

To find the probability of a series of events occurring you have to find the total amount of combinations it occurs divided by the total number of possible outcomes. 

https://www.calculator.net/permutation-and-combination-calculator.html?cnv=4&crv=2&x=85&y=24

You can put in 4 as n, because that is the number of flips.  We want half of the outcomes to be heads or tails, so then r=2.

Then you find that out of 4 flips, there are 6 possible combinations you can get an even number of heads or tails.

Then to find the total number of outcomes, you would take 2^n where n=4, which comes out to 16.

Then you have a 6/16 or 3/8 chance of getting the same number of heads and tails with 4 coin flips

Well I make it that there are 12 possible outcomes for n = 4.

I also make it that 6 of these result in an equal number of heads and tails.

But then I don't rely on dodgy information from the net.

Have you considered Strange's tree diagram?

Edited September 8, 2019 by studiot

[Conjurer]

4 minutes ago, studiot said:

Well I make it that there are 12 possible outcomes for n = 4.

There are actually 16.  There are 2 possible outcomes for the first flip, H or T.  There are two possible outcomes for the second flip, H or T.  Then you multiply 2x2 to get 4.  You can continue this process to 2^4=16.  

12 is the permutations or outcomes without replacement.  Coin flipping has replacement of the starting conditions, so it is nPr=n^r, like how it describes in the middle of the page.

https://www.calculator.net/permutation-and-combination-calculator.html?cnv=4&crv=2&x=85&y=24

8 minutes ago, studiot said:

Have you considered Strange's tree diagram?

Draw the diagram yourself, you will get 16 ends on the end of the tree drawing it out.  I have already done that.

Say if you flip the coin 6 times, there would be a 20/64 chance or 5/16 chance to get the same number of heads and tails.  This comes out to be 0.3125, but 4 flips had a 3/8 or 0.375 chance.  Then the probability of getting the same number of heads and tails decreases as you increase the number of flips.

In reality, it would produce closer to an equal number of heads or tails with more flips, so one could determine that the odds of getting heads or tails is 1/2 with a fair coin. 

[studiot]

You are right, there are 16.

Well done +1

So how does this progress finding a formula to take the limit of?

[Ghideon]

25 minutes ago, Conjurer said:

The one that says the absolute difference between the number of heads and tails approaches infinity with an increasing number of flips.

Can you provide a mathematic reason why you believe that? You seem to claim that gamblers ruin theorem is not correct. Another way of stating what I said is:

Quote

Let two players each have a finite number of pennies (say,  for player one and  for player two). Now, flip one of the pennies (from either player), with each player having 50% probability of winning, and transfer a penny from the loser to the winner. Now repeat the process until one player has all the pennies.

If the process is repeated indefinitely, the probability that one of the two player will eventually lose all his pennies must be 100%. 

 http://mathworld.wolfram.com/GamblersRuin.html

That means that no matter how many pennies the players may start with, eventually one of the players will have all the pennies. 

 

 

 

 

[Conjurer]

23 minutes ago, studiot said:

So how does this progress finding a formula to take the limit of?

That's what I am wondering.

How could you ever show that 

n!/(n^r(r!(n-r)!)) -> P(1/2), when n->infinity, and r=n/2, while n is an even integer

Edited September 8, 2019 by Conjurer

[studiot]

5 minutes ago, Ghideon said:

If the process is repeated indefinitely,

This is the key to the definition of probability.

Probability is defined as a limit, which means the sequence we are generating with my p(n) has to converge if there is a limit.

Can you see the problem with convergence of the sequence that is emerging, and that ghideon is pointing to?

Do you know what a limit is?

[Conjurer]

2 minutes ago, Ghideon said:

Can you provide a mathematic reason why you believe that? You seem to claim that gamblers ruin theorem is not correct. Another way of stating what I said is:

 http://mathworld.wolfram.com/GamblersRuin.html

That means that no matter how many pennies the players may start with, eventually one of the players will have all the pennies. 

 

That mathematical reasoning behind it is that mathematics is based on reality, and reality isn't based on mathematics.

The reason why one of the players ends up with all the pennies is because with an infinite number of flips, they could eventually reach the number of flips in a row to where one player loses them all.  Then if they continued to play the game after they lost, it could still balance out to where each player lost and won the same number of times.

I tried using wolfram alpha and it seems to be wanting to do factorials today, somewhat.  I don't see why this doesn't come anywhere near 1/2...

 

lim_(n->∞) integral(n!)/(n^r (r! (n - r)!)) dn = ∞/(r!)

[uncool]

1 hour ago, Conjurer said:

The reason why one of the players ends up with all the pennies is because with an infinite number of flips, they could eventually reach the number of flips in a row to where one player loses them all. 

...that's pretty much what Ghideon said.

[Conjurer]

14 minutes ago, uncool said:

...that's pretty much what Ghideon said.

I don't think it either confirms or denies the scenario I was talking about.  It is a different situation.  It would be more like the situation I was describing if the game didn't end.  If the game didn't end, then it should still come out the same amount of winners and losers, just like you would expect the heads or tails to turn out to.  You would just be combining multiple flips into an entire game in that case.

[Ghideon]

When tossing a fair coin n times: the difference between number of heads and number of tails approaches infinity:
[math] \lim_{n \rightarrow  \infty } |h-t| \rightarrow  \infty [/math]

You claim the above to be incorrect, can you provide your alternate version? Please use math symbols.

5 hours ago, Conjurer said:

I don't think it either confirms or denies the scenario I was talking about.  It is a different situation.  It would be more like the situation I was describing if the game didn't end.  If the game didn't end, then it should still come out the same amount of winners and losers, just like you would expect the heads or tails to turn out to.  You would just be combining multiple flips into an entire game in that case.

Your description is vague, can you formulate your claims in a more formal way?  The math of gamblers ruin says theorem states that the game always ends when players start with a limited amount of coins. The game cannot go on forever with limited amount of pennies. But if you want the game to have the possibility to go on forever; change the game so players start with an unlimited amount of pennies. The equation above still tells you what happens. 

My interpretation is that you claim:
[math] \lim_{n \rightarrow  \infty } |h-t| =0 [/math] where n=number of tosses of a fair coin, h=number of heads, t= number of tails
But the above equation is not true. The amount of heads and tails will not balance. It will deviate as I pointed out earlier. 

 

 

[studiot]

 

working

 

Edited September 9, 2019 by studiot

[Conjurer]

On 9/9/2019 at 12:05 AM, Ghideon said:

When tossing a fair coin n times: the difference between number of heads and number of tails approaches infinity:
limn→∞|h−t|→∞

You claim the above to be incorrect, can you provide your alternate version? Please use math symbols. 

 

The probability of getting heads or tails is 1/2, so you should get either heads or tails, half of the time...

You are claiming that as a result of flipping a coin over and over, you should end up only getting either heads or tails most of the time, after flipping it a lot.  What you are saying just doesn't work out with observational reality!  

I am the one asking for the proof to this problem!  I am the one that doesn't understand how probability theory has become accepted without such a proof.  The point of this post is me asking you for a proof of why the difference in the total amount of heads and tails doesn't approach infinity.  In reality, you will never encounter a situation where you just flip so many heads, and then a coin just starts flipping tails a predominantly larger amount forever.

[uncool]

18 minutes ago, Conjurer said:

The point of this post is me asking you for a proof of why the difference in the total amount of heads and tails doesn't approach infinity. 

And Ghideon is telling you that you are asking for a proof of something false. 

As you flip an increasing number of times, the absolute difference between the number of heads and tails will tend to grow, but the ratio of heads will tend to 1/2. 

Edited September 12, 2019 by uncool

[Conjurer]

3 minutes ago, uncool said:

And Ghideon is telling you that you are asking for a proof of something false. 

As you flip an increasing number of times, the absolute difference between the number of heads and tails will tend to grow, but the ratio of heads will tend to 1/2. 

He is wrong for making that equation based on the example he gave.  The example doesn't actually go to infinity.  That is the key difference, and the source of his mistake.  Given more chances, yes, you could tend to a larger count of something in a row that is larger than another something, but something would have to halt that process for it to work.  

I agree that a limited number of flips would eventually provide a winner of all the coins, and it would become more often to get the same result of heads or tails if the limited number of times it was flipped was increased.  

He is just saying I don't need a proof, because his theory is false.

[uncool]

Can you make a precise statement that you think is false?

[Conjurer]

Mathematically, there is no way to prove that the rules of probability could provide an answer (as far as we know) to a repeated event , increased repeatedly more times, approaches the same probability of that event occurring a single time, if it continued on forever, considering all the possible outcomes.

Like I said, flipping a coin should get you heads or tails half the time.  Therefore, the probability of flipping a coin and getting heads or tails is 1/2, since it landed either heads or tails approximately half the time.  Could you then prove that the rules probability can show that you should get heads or tails half the time if you flipped it an infinite amount of times from the possible outcomes?

How could we know that a probability, based on the total number of possible outcomes, could even be true if it doesn't tend towards the probability of the event occurring one single time, when it is increased a greater amount of times?

Edited September 12, 2019 by Conjurer

[uncool]

12 minutes ago, Conjurer said:

Like I said, flipping a coin should get you heads or tails half the time.  Therefore, the probability of flipping a coin and getting heads or tails is 1/2, since it landed either heads or tails approximately half the time.  Could you then prove that the rules probability can show that you should get heads or tails half the time if you flipped it an infinite amount of times from the possible outcomes?

As I said earlier:

On 9/8/2019 at 2:36 PM, uncool said:

It sounds like you're asking about the law of large numbers, a theorem known for binary variables since 1713.

If you don't mean the law of large numbers, then I still can't discern a precise statement you think is false.

[Conjurer]

8 minutes ago, uncool said:

As I said earlier:

If you don't mean the law of large numbers, then I still can't discern a precise statement you think is false.

So, what?  I am guessing you are expecting me to have some kind of eureka moment here.  I don't get what point you are trying to make by just saying, law of large numbers.  

[uncool]

I expect that you should be able to look something up when explicitly told its name.

https://en.wikipedia.org/wiki/Law_of_large_numbers

"In probability theory, the law of large numbers (LLN) is a theorem that describes the result of performing the same experiment a large number of times. According to the law, the average of the results obtained from a large number of trials should be close to the expected value, and will tend to become closer as more trials are performed."

[Conjurer]

6 minutes ago, uncool said:

I expect that you should be able to look something up when explicitly told its name.

https://en.wikipedia.org/wiki/Law_of_large_numbers

"In probability theory, the law of large numbers (LLN) is a theorem that describes the result of performing the same experiment a large number of times. According to the law, the average of the results obtained from a large number of trials should be close to the expected value, and will tend to become closer as more trials are performed."

That's like exactly like what I have been trying to explain to you!  You get that part now?