Weighted Coin

[fairyrak]

I have a coin and it is unfair because it has just 43.035048% of producing tails

if i throw it N times:

What's the probability that in N there were actually MORE THAN 44.0182% of tails?

What's the formula to get it if i replace 43.035048% with let's say z%?

[jim360]

Any particular reason the numbers used are so awkward?

Best to answer the final question, and then you are free to insert whatever the heck numbers you like. The answer is going to be based on the "binomial distribution", which gives that if an event either happens (with probability z) or does not (with probability 1-z), then the probability of k occurences of the event in N trials is:

{N!/[k! (N-k)!]} *z^k*(1-z)^(N-k)

where N! means "N factorial" and, say, 4! = 4*3*2*1 = 24; and p^k is p to the power of k. To adapt to the case you are interested in when you want, say, q% successes or greater out of the N times you throw it, then you would replace k in the formula above by k=(q*N/100), and then sum between this value of k and k = N.

It ain't pretty, I'm afraid. To get a handle on what you might expect, though, if we take N = 100 trials, z = 43% and q = 44%, then the probability that you got more than 44/100 tails is about 45%.

[fairyrak]

Because i usually play something with a lot of players and the players have range from % 35 to 50% of doing a certaing thing and the lots of decimals are regrettably important for us

[fairyrak]

anyways if it's not easy to find the real %, then now way, let's go with the estimation for integer numbers

[fairyrak]

We are doing something and N is going to be increasing in time! but as N increases the %s will change. And the tendency will be hopefully more and more towards a greater % of not being MORE THAN 44.0182% because of so many tries imply that the 43.035048% is something good.

and for me it's fun to know my progress

[jim360]

I'm not generally a fan of numbers in a problem being anything other than relatively round integer numbers (eg 1, 5, 10, 25 etc) because there is not really much additional insight gained from using other numbers -- unless by some accident things cancel when you use round numbers, I suppose, and don't if you use more arbitrary ones.

But anyway, the formula given above will give you the specific case you wanted under the replacement z = 043035048, and q = 44.0182, and while the formula given isn't exactly the most elegant ever it is hopefully clear what's going on. If not then a brief explanation:

if you throw the coin N times and want k tails then the probability of getting exactly k tails in succession would be just z^k, but you also have

[jim360]

hmm, hit enter too soon...

if you throw the coin N times and want k tails then you also need N-k heads, so you first produce a probability which is the probability of k tails in a row followed by N-k heads in a row, giving the z^k*(1-z)^(N-k) factor. The bit at the front counts the number of distinct orderings in which this can happen.

[fairyrak]

N is going to be very big, impossible to calculate like 200! with my pc, I better will try with statistics , although in excel I'm afraid I will need a very large amount of columns if I want to have a good number

[jim360]

If you are wanting to do the same thing for a particularly large value of N then there's a lovely result known as the Central Limit Theorem that allows you to use a normal distribution instead: see https://en.wikipedia.org/wiki/Normal_distribution for the new probability distribution, and further down it will tell you how to calculate the cumulative distribution, that you'd need to estimate the probability if you wanted more than such-and-such a number of tails. Not sure excel will be any good at integration, but there are places online that will do the work for you such as this one:

http://www.danielsoper.com/statcalc3/calc.aspx?id=53

[fairyrak]

thanks