Anyway to solve this problem without calculation? The problem is 


  Airlines find that each passenger who reserves a seat fails to turn up with probability 0.1 independently from other passengers. So Teeny Weeny Airlines always sell 10 tickets for their 9 seat aero plane while Blockbuster Airways always sell 20 tickets for their 18 seat airplane. Which one is more often overbooked?


The brutal way to solve this is trivial.The probability for Teeny Weeny Airlines to be overbooked is $p_1=0.9^{10}≈0.3487$, and the probability for Blockbuster Airways to be overbooked is $p_2=0.9^{20}+C_{20}^{19}×0.9^{19}×0.1≈0.3917$. So Blockbuster Airways should be more often overbooked.

My question is -- is there any more graceful solution without really calculating the probability? Thank you!
1. Qualitative approach: let the number of passengers not showing up for airline $i$ ($=1,2$) be $$N_i\sim Binom(i\cdot n,\frac 1 {n})$$ with $n=10$ in our case. Note that $N_i$ is centered around $i$ - its mean, mode and median are equal to $i$. When you double the number of passengers, you squish and spread out distribution of $N_1$ into that of $N_2$ - hence the probability of getting the mean/median number of missing passengers exactly goes down (i.e. $P(N_1=1)&gt;P(N_2=2)$). This probability must get distributed both to the right and to the left of this mean/median value - hence $$p_1=P(N_1&lt;1)&lt;P(N_2&lt;2)=p_2$$

2. Poisson approximation to the binomial: $N_i\approx Pois(i)$:
$$p_1=P(N_1=0)\approx e^{-1}$$
$$p_2= P(N_2=0,1)\approx e^{-2}(1+\frac {2^1} {1!})&gt;e^{-1}$$

3. Exact analysis:

$$p_1=(1-\frac 1 n)^n=e^{n\log(1-\frac 1 n)}&gt;e^{-1-\frac 1 {2(n-1)}}&gt;e^{-1}(1-\frac 1 {2(n-1)})$$
$$p_2=p_1p_1+2np_1p_1\frac {1/n}{1-1/n}=p_1p_1(1+\frac 2 {1-1/n})&gt;p_1e^{-1}(1-\frac 1{2(n-1)})(3+\frac 2 {n-1})=p_1e^{-1}(3+\frac 1 {2(n-1)}-\frac 1 {(n-1)^2})\ge\frac 3 ep_1$$
 for all $n\geq 3$.

Check $n=2$ separately to get $$\frac {p_2} {p_1}&gt;\frac 3 e &gt; 1.1$$ for all $n\ge 2$.
For the same amount of tickets sold, odds are in favor of WA.

Consider two successive TWA flights vs. a single WA flight. Given the possible numbers of passengers turning up, the numbers of overbooked flights per $20$ tickets are

$$\begin{align}
10, 10\to 2, 1\\
10, 9\to 2, 1\\
9, 10\to 2, 1\\
10, &lt;9\to 1, 0\\
&lt;9, 10 \to 1, 0\\
&lt;9, &lt;9\to 0, 0\end{align}$$
never in defavor of WA.