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I'm writing a story and I'm trying to figure out the probability of a coin toss scenario.

What is the probability of getting 51% correct guesses out of 100?
Group answer rounded is: 0.078
How would you say this for non math people like myself using the X out of X statement? The chances of this happening are X times out of X.

What is the probability of getting 51% correct guesses out of 1000?
Group answer rounded is: .0207
What is the probability of getting 51% correct guesses out of 10000?
Group answer rounded is .00108
What is the probability of getting 51% correct guesses out of 100000?

Thank you.

I understand I am being downvoted because I did not show my research for this question. I'm not a student or a math professional and I did not think you all wanted to see the list of google searches and all the pages I've read to try to figure out this question.

Yes, I did google it. This is my third day trying to find info on this.
Yes, I did try Wolfram Alpha. I had no idea how to form the question.
Yes, I did think this was the place to go for a novice to ask silly stupid questions.

I apologize for the intrusion. Many thanks to those who did answer.

Clarification.

Jan has a random number generator that is 0 and 1, like a coin flip. Jan has another random number generator that is 0 and 1, guessing head or tail.

The probability that Jan’s second generator guessing correctly 51% of the time out of 100 guesses would be [ number here ] out of [ number here ]. {I’m assuming this would be very reasonable like 49 times out of 100 Jan could get 51% correct guesses out of 100.}

The probability that Jan’s second generator guessing correctly 51% of the time out of 1,000 guesses would be [ number here ] out of [ number here ]. {This would be harder, for Jan to get 51% correct out of 1,000 it would be 1 out of 100?}

The probability that Jan’s second generator guessing correctly 51% of the time out of 10,000 guesses would be [ number here ] out of [ number here ]. {This would be up there like 1 in 7 billion?}

Then this one is way out there but thought it would be interesting to know, The probability that Jan’s second generator guessing correctly 51% of the time out of 100,000 guesses would be [ number here ] out of [ number here ].

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    Do you mean exactly $51\%$ of correct quesses or at least $51\%$ of correct quesses. – 2017-02-03
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    I don't get the downvotes. Someone who does not know the subject asks a question that is easy when you know probability and very hard if you don't. That is fine by me. Upvoted. – 2017-02-03

3 Answers 3

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You can ask Wolfram Alpha this sort of question:

Probability that X=51 where X has a binomial distribution with n=100 and p=.5

Probability that X=51 where X has a binomial distribution with n=100 and p=.5
answer : 0.078

Probability that X=510 where X has a binomial distribution with n=1000 and p=.5
.0207

Probability that X=5100 where X has a binomial distribution with n=10000 and p=.5
.00108

If you meant at least 51% just change X=51 to X>=51. The numbers become $.46$, $.27$ and $.023$.

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    I spent 30+ mins unsuccessfully on Wolfram Alpha trying to figure out how to find the formula. Thank you. – 2017-02-03
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    @YanceyG Yes, Alpha is a bit *beta* on some interpretations (!). Note that you can do that on Excel if you know the Binomial probability functions too. – 2017-02-03
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For the first one it's 

100!/(51! * 49! * 2^100)

which Google search says is 

0.0780286641

or about 3/4 of 1%.

The second one is 

1000!/(510! * 490! * 2^1000)

which is more than Google can handle. It will be very tiny.

Other folks may provide more answers.

Are you sure the question is the one you want to ask? Your story might work better if you asked for "between 50% and 51%" or "between 50% and 52%" or "between 50.5% and 51.5%". Those would be easier to calculate, too.

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    The value of the second one is $0.0206564$. The value of the third one is $0.00107986$. The value of the fourth one is $5.19466\cdot10^{-12}$. – 2017-02-03
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    @Jan Feel free to edit those into the answer. – 2017-02-03
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    Thank you, I would prefer not to do that (there is an odd feeling about it, what if I say something completely nonsensical, others will think its you). – 2017-02-03
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    @Jan If you edited nonsense into my answer I'd roll back the edit. But these are good additions. Your adding them makes this site look like the cooperative venture it is. But I'll do it later if you don't. – 2017-02-03
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The random variable for number of right guesses is binomial distributed as

$X\sim Bin(100, 0.5)$. Therefore $$P(X=51)=\binom{100}{51}\cdot 0.5^{51}\cdot 0.5^{49}=7.8\%$$

For large $n$ the distribution of $X$ can be approximated by the Normal distribution:

$$P(X=51)\approx \Phi\left(\frac{51+0.5-50}{\sqrt{25}}\right)-\Phi\left(\frac{51-0.5-50}{\sqrt{25}}\right)$$ $$=\Phi(0.3)-\Phi(0.1)=0.6179-0.5398=7.8\%$$

Even for $n=100$ the approximation is good enough for the most purposes. The larger $n$ is the better is the approximation.