If a cricketer has a $37$% chance of hitting a six, what is the probability of him hitting a six in 2 deliveries? 
If a cricketer has a $37$% chance of hitting a six, what is the
probability of him hitting a six in 2 deliveries?

Our intuition says the probability will increase. So it will be $2(37/100)$. But it doesn't, seem right. Also, it won't decrease like $(37/100)^2$.
What is the correct answer? My friend says he calculated the answer to be $55.5$%. Is that correct?
 A: You have to add (considering you want to compute the probability of him hitting a six at least one time in two deliveries):

*

*The chances of him hitting a six first and then NOT hitting a six.

*The chances of him NOT hitting a six first and then hitting a six.

*The chances of him hitting a six in both deliveries.

Knowing that the chances of him hitting a six are $\frac{37}{100}$, we conclude that the chances of him not hitting a six are $\frac{63}{100}$, so this gives us the following result:
$$\frac{37}{100}\cdot\frac{63}{100} + \frac{63}{100}\cdot\frac{37}{100}+\frac{37}{100}\cdot\frac{37}{100}=\boxed{\frac{6031}{10000}},$$
so we can conclude that the probability of him hitting at least a six is $\frac{6031}{10000}$, to be said $60.31$%.
Another way faster to do this is to just calculate the probability  of him NOT hitting a six both times (since it's the only case where the cricketer does not hit any six), and subtract that number to $1$. This would give us the same result:
$$1-\frac{63}{100}\cdot\frac{63}{100}=\frac{6031}{10000}.$$
A: Following your logic, the probability would be $3\cdot(37/100)>1$ in three deliveries.
It's true that the chances of hitting a six in three deliveries are very high, but the probability cannot exceed $1$.
In this kind of problems it's easier if you compute the probability of not hitting a six in $n$ deliveries, because these are independent events and the combined probability is obtained by multiplying the probabilities of the single events.
Thus the probability the cricketer doesn't hit a six in $n$ deliveries is $(1-p)^n$, where $p$ is the probability of hitting a six.
In order to find the probability you want, just compute the complement:
$$
1-(1-p)^2=1-\dfrac{67^2}{100^2}=\frac{6031}{10000}=60.31\%
$$
In three deliveries
$$
1-(1-p)^3\approx75\%
$$
