Conjurer

The number of possible outcomes with the expected value, actually, decreases compared to the total number of possible outcomes as the number of outcomes gets larger.

How does the average of the possible outcomes approach the expected value, when the probability of getting the same number of heads and tails becomes less likely, as you increase the number of coin flips?

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

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.

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?

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.

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.

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.

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!)

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

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 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.

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. Yes.

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

How do you find the probability of an event occurring? It seems like that would be the most basic fundamental rule of it.

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?

You got it backwards; that is the type of solution that I am looking for. If you had a box of 50 coins in it and dumped them out, you could count the number of heads or tails. You would find that it is close to 1/2 of them being heads or tails. You could keep dumping the box out and take the average of all the dumps, and you would find that the number of heads or tails approaches 1/2 the more times you dumped the box out. Then if you add the total number of times it is possible to get the same number of heads and tails by the total number of possible outcomes, you find that the probability decreases for doing that each time you generate more flips. Then it doesn't make sense to me how you could get P((nCr(H or T))/2^n) = 1/2 as n->infinity.

That is demonstrably false. You could tell a computer program to randomly pick a number 1 or 0 an increasing number of times. It will pick 1 or 0 half of the time, if you make it run enough trials.

I thought that reason could possibly be that there just wouldn't be a good chance that flipping heads or tails enough times would make you follow a path down the tree diagram that is close to the top or bottom of it. It could be more likely that you follow a path down the middle of the diagram, but that would seem to be able to fall prey to the gamblers fallacy. I was using the equations for combinations where n=#flips and r=#of heads or tails. Then I was taking that value and dividing it by 2^n in order to get the total number of times you would get the same number of heads and tails out of all possible outcomes from flipping it that number of times.

I seems like that would be fairly self explanatory from what you quoted me on, while asking this. I don't see how this statement or a corresponding statement could register correctly in my brain. I was teaching them the tree diagram. I cannot make a connection to what happens in the tree diagram to being more likely to getting the same number of heads and tails more frequently, since the probability of getting a heads or tails is 1/2.

How could a calculation even show that there is a probability of 1 of consecutive flips showing 49.999...% heads or tails as the number of flips approached infinity? So then the limit of the probability of getting heads or tails is 1/2? I am asking for the proof of the mathematics of probabilities. That is the problem you didn't understand. I heard that was what that problem was supposed to be, which was never solved as far as I know. Basically, I am just asking how you would show mathematically how it could be shown that you get the same result of what happens in reality if you consecutively flip a coin an increasing number of times. If someone sat there and did that, then they would find that the more times they flipped a coin, the closer and closer they would get to the same amount of them being heads or tails. That would occur every time they increased the number of consecutive flip in any instance they performed the experiment. They could count the number of heads or tails and determine that it is a fair coin, because half of the flips landed heads or tails. If it wasn't a fair coin, they could get some other probability. I just want to know how you could calculate the process of a fair coin showing this final result. I don't know how to solve this problem, and I am not aware of it ever being solved.

I heard that probabilities were not officially a branch of mathematics, because of this problem when I was in school. Then I never learned it growing up. Then it is now a course I am teaching as a student teacher. It really makes me wonder what changed to allow this to happen? I still have no proof that probabilities calculated this way is even accurate. How could a calculation even show that there is a probability of 1 of consecutive flips showing 49.999...% heads or tails as the number of flips approached infinity? So then the limit of the probability of getting heads or tails is 1/2? The question still remains. How do you mathematically show that consecutive coin flips can approach 1/2 probability of an event occurring? Other than just finding the final result and rounding 49.999...% of them being heads or tails? Mathematics is supposed to work forwards and backwards, but it still only seems to work backwards in this case. I don't know where or when this type of mathematics was proven or if it actually was.

I actually meant an identical number of heads and tails. The point I was trying to make is that the identical number of heads and tails decreases compared to the total number of possible outcomes as the number of trials increases. I don't see how this happens when the identical number of heads and tails approaches 1/2 if you were to just consider the total number of heads and tails and discover that half of them were heads or tails after a lot of trials. It makes it seems like there should be a different outcome, rather than what actually happens if you are not trying to just find the probability of heads or tails from the final result. I will give an example, getting the same number of heads and tails is 6/16 if you flip the coin 4 times. Now say I didn't actually see the flips being made, and I just came across a group of tables with only 6 out of 16 having the same number of heads and tails on them. I would then conclude that there wasn't a 1/2 probability of getting heads or tails, since there were not the same number of heads or tails on a majority of the tables, and only a minority of the tables had the same number of heads or tails on them. Say you did 8 flips of the coin, there would only be 70/256 instances where half of the flips resulted in heads or tails. 70/256 < 6/16 This happens each time you add more flips, you find that the total number of trials where the same number of heads and tails exist, it becomes a smaller number compared to the total number of possibilities. There are actually more times where there isn't the same number of heads and tails in a trial as the number of flips increases. I just don't understand how you could calculate getting to 1/2 probability from working the problem this way. The only way I know of where you could calculate this is just by adding the total number of flips together by half of the flips being heads or tails. Then the calculations do not seem to favor this result to where you could get there mathematically considering each consecutive flip.

You would have 1/10^24 chance of picking the right number. The events would be independent, so the previous draw would not affect the next draw. You would just still have the same chance of picking the correct winner. It would take 1/10^48 to pick the correct number twice in a row.

I just don't see why it would be so difficult to prove, or why it would be a complete waste of time. The photon should be the easiest particle to detect with modern technology. It doesn't make much sense in electrical engineering, and most experts in quantum mechanics will say what you said. Then they have no idea how this could be applied to EE. In EE, the electric field is only used to increase/decrease a voltage or create a current. Then this only occurs in the presence of another electromagnetic field or coil. The rest of the circuit carrying electrons will remain unchanged. Then any electron should be capable of absorbing any photon.

The photon is the force carrier of the magnetic field in quantum mechanics. Photons don't have a measurable amount of mass, but one could argue that they have energy, therefore they have mass. I am not sure it has actually ever been proven, directly by experiment, that photons are the force carrier of the magnetic field, either...