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October 29th, 2018, 09:27 AM  #1 
Newbie Joined: Oct 2018 From: UK Posts: 5 Thanks: 0  Prob. Of getting Heads over Tails 60% or more of the time after N number of flips
I'm new to this site and mathematics is not my forte, so forgive me if I should have posted this in the High School Math Forum or I'm using incorrect terminology. I've always been fascinated with probability and problems like the following. Let's say the Probability of getting heads or tails is exactly 50/50 What is the probability of a coin landing on heads 60% of the time (or more) after N number of Flips? For specific examples, what is the probability that the coin will land on Heads >= 60% of the time after: a) 10 flips b) 1,000 flips c) 1,000,000 flips So that I can apply this to other situations (for example, dice), what is the formula for working this out assuming: N = NUMBER of events (in this case coin flips) S = Number of SUCCESSFUL outcomes (In this case Heads) O = Possible OUTCOMES/OPTIONS (Obviously in the case of a coin it should just be 2 – heads or tails but with a dice it would be 6) I know that c) should involve a much lower probability than b) or c) but I’m still not sure about the answers and formula I would apply to other situations when: N = NUMBER of events S = Number of SUCCESSFUL outcomes O = All Possible OUTCOMES/OPTIONS Thanks. Last edited by skipjack; October 29th, 2018 at 11:25 AM. 
October 29th, 2018, 12:42 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,704 Thanks: 670 
The distribution is binomial. Calculation is messy for a large number of trials, so normal approximation could be used to estimate solution. The probability of k successes after n flips is $\binom{n}{k}p^k(1p)^{nk}$, where $p$ is the probability of success. For coin flips, $p=\frac{1}{2}$. What you want is the sum over $k$ for the condition you set.
Last edited by skipjack; October 29th, 2018 at 02:22 PM. 
October 29th, 2018, 12:52 PM  #3  
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 550  Quote:
Let's start with 5 flips. The probability of getting exactly 60% heads, meaning 3 heads and 2 tails is $\dbinom{5}{3} * \left (\dfrac{1}{2} \right )^3 * \left ( 1  \dfrac{1}{2} \right )^2 = \dfrac{5!}{3! * (5  3)!} * \dfrac{1}{2^5} = \\ \dfrac{5 * 4}{2} * \dfrac{1}{32} = \dfrac{5}{16} = 31.25\%$ Not too bad a chance. The probability of getting at least 60% heads, meaning 3, 4, or 5 heads, is $\dbinom{5}{3} * \left (\dfrac{1}{2} \right )^3 * \left ( 1  \dfrac{1}{2} \right )^2 + \dbinom{5}{4} * \left (\dfrac{1}{2} \right )^4 * \left ( 1  \dfrac{1}{2} \right )^1 + \dbinom{5}{3} * \left (\dfrac{1}{2} \right )^5 * \left ( 1  \dfrac{1}{2} \right )^0 = \\ \dfrac{5!}{3! * (5  3)!} * \dfrac{1}{2^5} + \dfrac{5!}{4! * (5  4)!} * \dfrac{1}{2^5} + \dfrac{5!}{3! * (5  3)!} * \dfrac{1}{2^5} = \\ \dfrac{1}{32} * \left ( \dfrac{5 * 4}{2} + \dfrac{5}{1} + 1 \right ) = \dfrac{1}{32} * (10 + 5 + 1) = \dfrac{1}{2} = 50\%.$ These probabilities drop fairly quickly as the number of flips increases. At 10 flips we have, probability of exactly 6 heads is $\dbinom{10}{6} * \left (\dfrac{1}{2} \right )^{10} = \dfrac{10!}{6! * 4!} * \dfrac{1}{1024} = \dfrac{10 * 9 * 8 * 7}{4 * 3 * 2 * 1024} =\\ \dfrac{5040}{24576} \approx 20.51\%.$ The probability of at least 6 heads is $\dfrac{1}{1024} * \left ( \dfrac{10!}{6! * 4!} + \dfrac{10!}{7! * 3!} + \dfrac{10!}{8! * 2!} + \dfrac{10!}{9! * 1!} + \dfrac{10!}{10! * 0!} \right ) = \\ \dfrac{1}{1024} * (210 + 120 + 45 + 10 + 1) \approx 37.70\%.$ When we get to 100 flips, doing this by hand becomes tedious (although there are approximation tools and tables). Using a computer, I came up with: the probability of getting exactly 60 is approximately 1.08%, and the probability of getting at least 60 is approximately 2.84%. When you get to 1000 flips, the probability of getting 600 or more is going to drop almost to zero.  
October 29th, 2018, 02:00 PM  #4  
Newbie Joined: Oct 2018 From: UK Posts: 5 Thanks: 0 
Thank you for your answers. I didn't realise the calculation would be so complicated and involve so many iterations (for a large number of flips anyway). I assumed that a single formula would do the trick. Quote:
May I ask what computer program you use to calculate the result? Is it a computer program you made yourself or mathematical software I can download. It would be great If I could get a hold of some software that would calculate the results for me.  
October 29th, 2018, 02:27 PM  #5 
Global Moderator Joined: Dec 2006 Posts: 20,373 Thanks: 2010 
10! and 10 factorial are the same. Note that 0! = 1.

October 29th, 2018, 03:19 PM  #6 
Newbie Joined: Oct 2018 From: UK Posts: 5 Thanks: 0 
I think I understand now or kind of: The ((1/2)^10) is the probability of a head (1/2) to the power of the total number of flips (which is 10) or 1/1024 This is then multiplied by the sum of each possible combination. With each of the combinations you have 10! meaning 10 total flips divided by the chance of 6!, 7!, 8!, 9! or 10! heads multiplied by the chance of the inverse 4!, 3!, 2!, or 1! tail(s) I know that's a crude explanation but is that sort of correct? 
October 29th, 2018, 05:20 PM  #7  
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 550  Quote:
The binomial distribution involves independent yes/no decisions. For example, rolling a die and getting a 3 or not a 3. There are only two possibilities, success and failure. And what you get on one roll does not depend on what you got on a previous roll and foes not affect what you get on a later roll. If p = the probability of success, then the probability of failure = 1  p. With me so far? In the special case, when p = 1/2, then (1  p) = 1/2. So the basic formula for k successes in n trials is $\dbinom{n}{k} * p^k * (1  p)^{(nk)}, \text { where } \dbinom{n}{k} = \dfrac{n!}{k! * (nk)!}.$ $p^k$ is sort of obvious: it is the probability of k successes on specific trials. $(1  p)^{(nk)}$ is the probability of (n  k) failures on specific trials. But what is that binomial coeffecient? It tells us how many specific trials of the type that we are interested in are possible, Let's consider 3 successes out of five trials. We could have success on the first trial, second trial, and third trial. (1) We could have success on the first trial, second trial, and fourth trial. (2) We could have success on the first trial, second trial, and fifth trial. (3) We could have success on the first trial, third trial, and fourth trial. (4) We could have success on the first trial, third trial, and fifth trial. (5) We could have success on the first trial, fourth trial, and fifth trial. (6) We could have success on the second trial, third trial, and fourth trial. (7) We could have success on the second trial, third trial, and fifth trial. (8 ) We could have success on the second trial, fourth trial, and fifth trial. (9) We could have success on the third trial, fourth trial, and fifth trial. (10) The general formula for how many ways that you can have k successes out of n trials is $\dbinom{n}{k} = \dfrac{n!}{k! * (n  k)!}.$ Because we have actually counted the number for 3 successes out of 5 trials and found the answer to be 10, let's see what the formula says. $\dbinom{5}{3} = \dfrac{5!}{3! * (5  3)!} = \dfrac{5 * 4 * 3 * 2 * 1}{3 * 2 * 1 * 2!} = \dfrac{5 * 4}{2 * 1} = 10.$ As for my calculation when n = 100, I just used an excel spreadsheet, nothing fancy.  
October 29th, 2018, 07:12 PM  #8  
Newbie Joined: Oct 2018 From: UK Posts: 5 Thanks: 0  Probability and Telepathy Experiments Quote:
However it's impossible to use excel to work out 200+ rolls of a dice or coin flips as the numbers are too high. For example 170! is >7.26E+306 . Calculators just don't go that high. Is there any way of obtaining an approximation for factorial's in the thousands range? Why is this important? According to Dr Dean Radin, who is a parapsychology researcher, there now exists hard evidence for the existence of extrasensory perception (ESP) such as Telepathy. However, the effect is so subtle that a lot of trials have to be carried out before the odds against chance become over a billion to one. One of the experiments involves the ganzfeld experiment under highly controlled conditions where the sender and receiver are completely separated. The ganzfeld experiments are among the most recent in parapsychology for testing telepathy. One person is assigned the role of sender and another the receiver. A computer randomly generates 4 images from a pool of thousands of images. Then the computer randomly selects 1 of those images leaving 3 decoy images The sender has then got to mentally send the correct image out to the receiver. The receiver is shown the correct image plus the 3 decoy images. There should be a 1 in 4 chance or a 25% chance of the receiver selecting the correct image assuming Telepathy does NOT exist. However we are seeing results in the range of around 34%. That's a 9% higher average than what should be expected. For some special populations, like siblings, twins or motherdaughter pairs, the hit rates are even higher. This wouldn't be interesting if the sample size was just a few hundred trials. However I've heard it stated that the odds against chance of getting 34% rather than 25% after 2,500 trials is over a billion to one. I was trying to figure out a formula to calculate whether the billion to one estimate was accurate assuming 2,500 trials with a success rate of 34%. Last edited by skipjack; October 29th, 2018 at 10:24 PM.  
October 29th, 2018, 10:37 PM  #9 
Global Moderator Joined: Dec 2006 Posts: 20,373 Thanks: 2010  
October 30th, 2018, 05:22 AM  #10  
Newbie Joined: Oct 2018 From: UK Posts: 5 Thanks: 0  Quote:
For example, a replication of Dean Radin’s double slit consciousness experiment. What is more remarkable is that trained mediators perform better at these experiments than the untrained indicating that the cause was indeed a mindmatter relationship. Their findings, across the work series, claim that the observed effects: a) globally support the mindmatter interaction hypothesis, ie. the causal effect of the participant’s intention in the optical system; b) cannot be explained as procedural or analytical artifacts, as the control sessions (without participants present) resulted in no significant differences between the intentionpresent and intentionabsent epochs; c) are stronger for participants with contemplative practices training, e.g. meditation; Here is the abstract to the Scientific paper in question: Motivated by a series of reported experiments and their controversial results, the present work investigated if volunteers could causally affect an optical doubleslit system by mental efforts alone. The participants' task in the experimental sessions alternated between intending an increase in the (realtime feedbackinformed) amount of light diffracted through a specific single slit versus relaxing their intentional effort. In total, 240 sessions contributed by 171 volunteers were recorded. The first 160 sessions were collected in an exploratory mode, and those data revealed statistically significant differences between the intention and relax conditions. Here is a link to the scientific paper: https://osf.io/zsgwp/  

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60%, flips, heads, number, prob, tails, time 
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