Conjurer Posted September 2, 2019 Share Posted September 2, 2019 (edited) I ran into a coin flip problem where flipping 4 coins has a 6/16 or 3/8 probability of landing 2 heads and 2 tails. I expected this value to be 1/2, because you have a 50% chance of getting heads or tails. Then that is only 6 of the possible 16 outcomes, instead of 8. Then I realized that the number of possible outcomes where there is an even number of heads and tails actually decreases, compared to the total possible outcomes, the more times you flip the coin. I was wondering if someone could, please, troll me about the reason why this doesn't disprove the mathematics of probabilities, since the outcome of flipping more and more coins in a row approaches closer and closer to half of the coins being heads or tails. Edited September 2, 2019 by Conjurer Link to comment Share on other sites More sharing options...
Ghideon Posted September 3, 2019 Share Posted September 3, 2019 Short note on the specific case in the question: when tossing 4 coins you calculated the probability of getting exactly 2 heads and 2 tails. When tossing same coins a finite number of times you did not ask for the probability that the outcome is exactly 1/2 heads and 1/2 tails. As you said it will be approaching 50% heads and 50% tails in the long run. The situations are different and requires different formulas. 7 hours ago, Conjurer said: I was wondering if someone could, please, troll me about the reason why this doesn't disprove the mathematics of probabilities, since the outcome of flipping more and more coins in a row approaches closer and closer to half of the coins being heads or tails "Mathematics of probabilities" is a rather broad subject. Before attempting a longer explanation; are you questioning a more specific part of probabilities? For instance law of large numbers is applicable when discussing your question. Link to comment Share on other sites More sharing options...
Prometheus Posted September 3, 2019 Share Posted September 3, 2019 (edited) 10 hours ago, Conjurer said: I ran into a coin flip problem where flipping 4 coins has a 6/16 or 3/8 probability of landing 2 heads and 2 tails. I expected this value to be 1/2, because you have a 50% chance of getting heads or tails. Then that is only 6 of the possible 16 outcomes, instead of 8. This is correct. 10 hours ago, Conjurer said: Then I realized that the number of possible outcomes where there is an even number of heads and tails actually decreases, compared to the total possible outcomes, the more times you flip the coin. Not entirely sure what you mean. I generated a fewf binomial probability mass functions, with odd or even numbers of trials. With an even number of trials the most likely outcome is an odd number, while for an odd number of trials the most likely outcome is equal between an odd and an even number. We will expect an even number of heads or tails every time we have an odd number of flips. This is a consequence of the combinatorics of the binomial PMF (the n choose k bit). Does this answer your question? Edited September 3, 2019 by Prometheus Because Matlab and mornings don't mix 1 Link to comment Share on other sites More sharing options...
Ghideon Posted September 3, 2019 Share Posted September 3, 2019 (bold by me) 10 hours ago, Conjurer said: Then I realized that the number of possible outcomes where there is an even number of heads and tails actually decreases, compared to the total possible outcomes, the more times you flip the coin. I have (mis)interpreted the bold part to mean an identical number of heads and tails. So @Prometheus answer therefore may be more applicable. Link to comment Share on other sites More sharing options...
Conjurer Posted September 7, 2019 Author Share Posted September 7, 2019 On 9/3/2019 at 2:20 AM, Ghideon said: (bold by me) I have (mis)interpreted the bold part to mean an identical number of heads and tails. So @Prometheus answer therefore may be more applicable. 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. On 9/3/2019 at 1:35 AM, Prometheus said: Not entirely sure what you mean. I generated a fewf binomial probability mass functions, with odd or even numbers of trials. With an even number of trials the most likely outcome is an odd number, while for an odd number of trials the most likely outcome is equal between an odd and an even number. We will expect an even number of heads or tails every time we have an odd number of flips. This is a consequence of the combinatorics of the binomial PMF (the n choose k bit). Does this answer your question? 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. Link to comment Share on other sites More sharing options...
Ghideon Posted September 7, 2019 Share Posted September 7, 2019 1 hour ago, Conjurer said: I actually meant an identical number of heads and tails. Thanks for clarification. 1 hour ago, Conjurer said: 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. You seem to argue that the probability of exactly 1/2 of many and approximately 50% of many should be the same? It is not. For a large number of coin tosses there is actually a very small chance that there is exactly 1/2 heads and 1/2 tails. The only time it is probable to throw identical number of heads and tails is when the process is not fair or random. If I did a lot of tries and got 1/2 heads each time I would be suspicious. Example: Throw the coin 1000000000 times. What is the probability to throw exactly 1/2 of 1000000000 heads? Close to zero. What is the probability of throwing about 50% of them heads? Close to 1. And what is the probability of throwing heads the next time? It's 1/2 1 Link to comment Share on other sites More sharing options...
Prometheus Posted September 7, 2019 Share Posted September 7, 2019 26 minutes ago, Ghideon said: Example: Throw the coin 1000000000 times. What is the probability to throw exactly 1/2 of 1000000000 heads? Close to zero. Just to emphasise this point: the probability of exactly half is close to zero, even though the expectation is 500000000 (which is half heads). With so many possible outcomes the probability of even the most likely outcome has only a tiny chance of occuring, as you can see from the PMF of 90000 flips below. Which is what you're seeing, Conjurer, as you perform more trials, the probability of getting exactly half heads is always decreasing, though it always remains the most likely of all possible outcomes. That should be clear on the other graph which is just zoomed in around the expected value. Link to comment Share on other sites More sharing options...
Conjurer Posted September 8, 2019 Author Share Posted September 8, 2019 (edited) 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. Edited September 8, 2019 by Conjurer Link to comment Share on other sites More sharing options...
uncool Posted September 8, 2019 Share Posted September 8, 2019 (edited) 39 minutes ago, Conjurer said: I heard that probabilities were not officially a branch of mathematics, because of this problem when I was in school. What "problem"? Probability theory and statistics have been a branch of mathematics since the 1600s, if you start counting with Pascal and Fermat. 39 minutes ago, Conjurer said: 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 have no idea what you are trying to ask here. Edited September 8, 2019 by uncool Link to comment Share on other sites More sharing options...
Conjurer Posted September 8, 2019 Author Share Posted September 8, 2019 4 minutes ago, uncool said: What "problem"? 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? 4 minutes ago, uncool said: I have no idea what you are trying to ask here. 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. Link to comment Share on other sites More sharing options...
uncool Posted September 8, 2019 Share Posted September 8, 2019 (edited) It sounds like you're asking about the law of large numbers, a theorem known for binary variables since 1713. 14 minutes ago, Conjurer said: I am asking for the proof of the mathematics of probabilities. This is too vague to even mean something. Proof is for specific statements. What is the specific statement that you are asking for proof of? 14 minutes ago, Conjurer said: 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. This isn't true, actually. They would find that the ratio of heads would approach 1/2, but that the number of excess heads would diverge from 0. Edited September 8, 2019 by uncool Link to comment Share on other sites More sharing options...
studiot Posted September 8, 2019 Share Posted September 8, 2019 13 minutes ago, Conjurer said: 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. If you are teaching probability, how would you show this assertion to be true? Remember that every second toss represents an odd number of tosses so for every second toss as Ghideon says. As a matter of interest what definition of 'probability' are you teaching your students since you mention the correspondence to reality? Link to comment Share on other sites More sharing options...
Conjurer Posted September 8, 2019 Author Share Posted September 8, 2019 (edited) 17 minutes ago, studiot said: If you are teaching probability, how would you show this assertion to be true? I seems like that would be fairly self explanatory from what you quoted me on, while asking this. 17 minutes ago, studiot said: Remember that every second toss represents an odd number of tosses so for every second toss as Ghideon says. I don't see how this statement or a corresponding statement could register correctly in my brain. 17 minutes ago, studiot said: As a matter of interest what definition of 'probability' are you teaching your students since you mention the correspondence to reality? 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. Edited September 8, 2019 by Conjurer Link to comment Share on other sites More sharing options...
Strange Posted September 8, 2019 Share Posted September 8, 2019 34 minutes ago, Conjurer said: 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. Quote 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 would start with combinatorics: https://en.wikipedia.org/wiki/Combinatorics You can easily write down all the combinations of 4 heads and tails. If you do this for different numbers of tosses, you will observe that the number is always 2N. It would not be hard to derive this mathematically. Similarly, if you count how many of those combinations have the same number of heads and tails, you can work out the formula for that. Form those two things, you can calculate the probability. Probability theory developed from these sort of observations. I have read that Pascal first started looking at it because a friend (Guy de Maupassant?) asked him why he kept losing when gambling. By analysing how many combinations of playing cards in a hand, rolled fo the dice, etc you can calculate the odds. (The odds that you will lose, generally.) Link to comment Share on other sites More sharing options...
Ghideon Posted September 8, 2019 Share Posted September 8, 2019 (edited) 1 hour ago, Conjurer said: 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. I do not get the above. "The closer and closer they would get to the same amount of them being heads or tails." That sounds like the opposite of what is known about games and probabilities*. The more times you toss a fair coin the larger the probability that there will be a large difference between number of heads and tails. Here is an attempt at using mathematic symbols** for my statement. When tossing n times: the limit for number of heads divided by number of tails is 1. [math] \lim_{n \rightarrow \infty } \big(\frac{h}{t}\big) = 1 [/math] When tossing n times: the difference between number of heads and number of tails approaches infinity is unbounded. (edit, got some help from Strange) [math] \lim_{n \rightarrow \infty } | h-t | \rightarrow \infty [/math] I remember this being taught in connection to Gamblers Ruin Theorem https://en.wikipedia.org/wiki/Gambler's_ruin#Fair_coin_flipping. *) First occurring in "On Reasoning in Games of Chance", 1657? **) Not exactly sure if "equals infinity" is the correct way to write what I intend to say. Edited September 8, 2019 by Ghideon rightarrow instead of equals, grammar 2 Link to comment Share on other sites More sharing options...
Conjurer Posted September 8, 2019 Author Share Posted September 8, 2019 12 minutes ago, Strange said: I would start with combinatorics: https://en.wikipedia.org/wiki/Combinatorics 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. 14 minutes ago, Strange said: You can easily write down all the combinations of 4 heads and tails. If you do this for different numbers of tosses, you will observe that the number is always 2N. It would not be hard to derive this mathematically. Similarly, if you count how many of those combinations have the same number of heads and tails, you can work out the formula for that. Form those two things, you can calculate the probability. 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. Link to comment Share on other sites More sharing options...
studiot Posted September 8, 2019 Share Posted September 8, 2019 (edited) 24 minutes ago, Conjurer said: 1)I seems like that would be fairly self explanatory from what you quoted me on, while asking this. 2)I don't see how this statement or a corresponding statement could register correctly in my brain. 3) 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. 2) I apologise, I didn't finish the second statement. The comment is linked to the first, however. 1) I asked for a mathematical proof of your assertion. Something anyone (especially a student) is entitled to ask of a maths teacher. self explanator is not good enough. I would expect something along the lines given the probability of n coin tosses P(n) = ?, where The probability of P(n+1) is (by induction) some smaller value You would need to start with n = 0 - equal probability P(0) = 1 p(1) = zero, yes 0. p(2) = ? And show that as n increases the mathematical function you have proposed does indeed tend to 0.5 Edited September 8, 2019 by studiot spelling Link to comment Share on other sites More sharing options...
Conjurer Posted September 8, 2019 Author Share Posted September 8, 2019 5 minutes ago, Ghideon said: The more times you toss a fair coin the larger the probability that there will be a large difference between number of heads and tails. 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. Link to comment Share on other sites More sharing options...
uncool Posted September 8, 2019 Share Posted September 8, 2019 1 minute ago, Conjurer said: 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. He is correct; he is talking about the absolute difference between the number of heads and tails. Only the ratio will approach 1/2. Link to comment Share on other sites More sharing options...
studiot Posted September 8, 2019 Share Posted September 8, 2019 So how are you getting on with my sequence? Link to comment Share on other sites More sharing options...
Conjurer Posted September 8, 2019 Author Share Posted September 8, 2019 (edited) 33 minutes ago, studiot said: 2) I apologise, I didn't finish the second statement. The comment is linked to the first, however. 1) I asked for a mathematical proof of your assertion. Something anyone (especially a student) is entitled to ask of a maths teacher. self explanator is not good enough. I would expect something along the lines given the probability of n coin tosses P(n) = 0.5, where The probability of P(n+1) is (by induction) some smaller value You would need to start with n = 0 - equal probability P(0) = 1 p(1) = zero, yes 0. p(2) = ? And show that as n increase the mathematical function you have proposed does indeed tend to 0.5 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. Edited September 8, 2019 by Conjurer Link to comment Share on other sites More sharing options...
Ghideon Posted September 8, 2019 Share Posted September 8, 2019 19 minutes ago, Conjurer said: 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. Show how it is false then. This could be part of the misunderstanding? Link to comment Share on other sites More sharing options...
Conjurer Posted September 8, 2019 Author Share Posted September 8, 2019 2 minutes ago, Ghideon said: Show how it is false then. This could be part of the misunderstanding? 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? Link to comment Share on other sites More sharing options...
uncool Posted September 8, 2019 Share Posted September 8, 2019 That's not an answer to the very specific statement Ghideon made. Link to comment Share on other sites More sharing options...
Conjurer Posted September 8, 2019 Author Share Posted September 8, 2019 6 minutes ago, uncool said: That's not an answer to the very specific statement Ghideon made. How do you find the probability of an event occurring? It seems like that would be the most basic fundamental rule of it. Link to comment Share on other sites More sharing options...
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