Binomial computation: Probability of winning at least 175 games out of 900 trials

Question

We roll a die, if it stops on 1, we win, if not we lose.

So we have $\frac{1}{6}$ chance of winning. What is the probability of winnign at least 175 times out of 900 rolls ? We can see this as a repetition of independent games so we can use Binomial distribution. I want to find $$P(\text{Win} \geq 175)$$ but the binomial formula will only give me $P(\text{Win} = 175)$ so I will need to do that for $176,..900$. Or I can also compute $1-P(\text{Win} < 175)$ but I still need to do 175 calculations..
How can I do it ?

Answer 1

Two methods are easier than computing with the binomial PDF by hand:

(1) Use software. Suppose the binomial random variable is $W \sim \mathsf{Binom}(n=1000, p = 1/6).$ You seek $P(W \ge 175) = 1 - P(X \le 174) = 0.2515.$ Using R statistical software where pbinom is a binomial CDF and dbinom is a binomial PDF, we get the exact answer.

 1 - pbinom(174, 1000, 1/6)
 ## 0.2514526
 k = 175:1000;  sum(dbinom(k, 1000, 1/6))
 ## 0.2514526

Use a normal approximation: $\mu = E(W) = np = 1000(1/6) = 166.6667$ and $\sigma = SD(W) = \sqrt{np(1-p)} = \sqrt{1000(1/6)(5/6)} = 11.7851.$ Then $$P(X \ge 175) = P(X > 174.5) = P\left(\frac{W - \mu}{\sigma} > \frac{174.5 - 166.68}{11.7851}\right)\\ \approx P(Z > 0.6647) \approx 0.2531,$$ where $Z$ is standard normal and the approximate probability can be obtained using printed standard normal tables. (You will probably not get exactly the same result because without interpolation you will enter the printed table with 0.66 instead of 0.6647. In any case one should not expect better than two place accuracy using a normal approximation to a binomial distribution.)

The figure below shows the exact binomial distribution (vertical black bars) along with the 'best fitting' normal distribution. The relevant probability is to the right of the dotted red line.