Basic statistics - Calculate distribution of winning I have a 100 sided fair dice with each side labelled 1 thru 100.  I win if the number rolled is 49 or higher (1% advantage).
1.  What is the probability of me winning exactly 500 rolls if the dice is rolled 1000 times?


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*What is the general formula for calculating the probability of winning exactly W rolls if:
P=probability of winning (52% if the above example)
N=total number of rolls

 A: When you roll the die once, the probability of a loss is $\frac{48}{100}=0.48$, and the probability of a win is $\frac{52}{100}=0.52$, not $0.51$. Therefore the probability of any particular sequence of $500$ wins and $500$ losses is $\left(\frac{48}{100}\right)^{500}\left(\frac{52}{100}\right)^{500}=0.48^{500}\cdot0.52^{500}=0.2496^{500}$. There are $\binom{1000}{500}$ ways to choose which $500$ rolls are wins, so there are $\binom{1000}{500}$ different sequences of $500$ wins and $500$ losses. The overall probability of getting one of these sequences is therefore 
$$\binom{1000}{500}\left(\frac{48}{100}\right)^{500}\left(\frac{52}{100}\right)^{500}=\binom{1000}{500}\cdot0.2496^{500}\approx0.005665\;.$$
If you want the probability of winning to be $0.51$, you need to set the lower limit for a win at $50$, not $49$.
As can be seen from the reasoning above, the general formula for the probability of winning exactly $W$ of $N$ rolls when the probability of a winning roll is $p$ is
$$\binom{N}Wp^W(1-p)^{N-W}\;.$$
A: The probability of winning a given game is $\dfrac{52}{100}$. Let this value be called $p$.
Thus, the probability of winning exactly 500 rolls out of 1000 rolls is:
$$\binom{1000}{500}p^{500}(1-p)^{500}$$
This is a Binomial Distribution. General formula given by:
$$\binom{N}{W}p^{W}(1-p)^{N-W}$$
You can use a Normal Approximation to solve this problem but I don't think that's the motivation of the question.
A: To use the normal approximation, we need the mean and variance of the number of trials you will win. The mean is $\mu = (0.52)(1000)=520,$ and the variance, using the binomial distribution, is $\sigma^2 = (0.52)(0.48)(1000) = 249.6$ 
Now to apply the normal approximation, we find the cdf of a normal distribution (with the above mean and variance) at the point $500.5$ and subtract off the cdf at $499.5$ 
To 5 decimals, these calculations yield a value of $0.01133$ This agrees with the binomial calculation to 5 decimals. In a spreadsheet I find the difference between the normal approximation value and the binomial value to be $5.78e-06$. I am unable to reproduce the value of $0.005665$ given in another answer. 
