Probability of Punctures for a group of cyclists The matter of the probability of punctures occurring cropped up during a ride yesterday with a friend. 
His view is this, (As we can't let a subject drop....  ;-) )
"Eric,
There must be more chance of a puncture occurring if there are more riders.
The way I see it is this:
If there is a 20% chance of a rider getting a puncture, then that means there is a 1 in 5 chance of a puncture for that rider. As a fraction that is 1/5. It follows that if there were 5 riders, the chances would be 5/5 for the group (i.e., one puncture). It follows that if there were 10 riders it would be 10/5 which means there would be 2 punctures. 
Therefore, the more riders there are, the more there is a chance of a puncture."
My view was that yes, 1/5 (20%) can be a figure to agree on, but that figure is stand alone, irrespective of the number of riders on the ride. 
I know probability is very counter intuitive to the human mind, so what seems to be the case is not always the case. 
Can anyone give a view, given that we're probably both high school grade mathematicians...if we're lucky!
Eric
 A: The word "chance" is very loosely used in the question. Let's examine the case for 5 riders.
If it is the expected value that is sought, E[x] = np, so for  5 riders, $5\cdot\frac15 = 1$
If it is the probability that is sought,
P(none of the 5 have a puncture) = $(\frac{4}{5})^5 = \frac{1024}{3125}$
Thus P(at least one puncture occurs) = 1 - $\frac{1024}{3125} = \frac{2101}{3125},\approx 0.67$
A: If there are $N$ riders, the probability of having no puncture is $$\Bbb{P}(\text{ no puncture }) = \bigg(\frac{4}{5}\bigg)^N $$
and that decreases with $N$. As the numbers of riders increase, the probability of occuring a puncture increases.
A: This can be interpreted as a Bernoulli process. The probability of any single rider getting a puncture is (say) $1/5$. The probabilities of the different riders getting punctures are all independent, so to find the probability of certain people getting punctures, multiply $0.2$ for all the people who get punctures and $0.8$ for all the people who don't. 
To have no punctures, multiply $0.8$ by itself for as many people as you have. It's less than $1$, so as you raise it to higher and higher powers, it gets smaller and smaller. That means as you have more and more people, your group has a smaller and smaller chance of escaping the trip unscathed.
You can find the exact formulas from the binomial distribution.
