Probability Roulette Problem Suppose the Roulette table has 37 numbers (European Roulette table). During 37 spins, I always do the same bet: 35 numbers straight (35 chips in 35 different numbers). 
Then:


*

*the probability of winning the 37 consecutive spins is $(\frac{35}{37})^{37}\approx 0.1279$,

*the probability of losing at least once during the 37 spins is $1-(\frac{35}{37})^{37}\approx 0.8720$,

*the probability of losing exactly once during the 37 spins is $\binom{37}{1}(\frac{2}{37})(\frac{35}{37})^{36}\approx 0.2705$

*the probability of losing exactly twice during the 37 spins is $\binom{37}{2}(\frac{2}{37})^2(\frac{35}{37})^{35}\approx 0.2782$

*the probability of losing exactly 3 times during the 37 spins is $\binom{37}{3}(\frac{2}{37})^3(\frac{35}{37})^{34}\approx 0.1855$ 

*the probability of losing exactly $m$ times during the 37 spins is $\binom{37}{m}(\frac{2}{37})^m(\frac{35}{37})^{37-m}$, being decreasing when $2\leq m\leq 37$, reaching the lowest value $(\frac{2}{35})^{37}$ (losing the 37 spins).


Why the probability of losing exactly twice during the 37 spins is greater than the probability of losing exactly once during the 37 spins? I expected to get a decreasing sequence of probabilities starting from $m=1$, not from $m=2$.
This tells us that 37 is a number of spins big enough to lose exactly twice being more probably than losing exactly once, but 37 isn't a number of spins big enough  to lose exactly more than 3 times.
For which numbers $n$ of spins, we have a decreasing sequence of probabilities
$\binom{n}{m}(\frac{2}{37})^m(\frac{35}{37})^{n-m}$ ($1\leq m\leq n$)? How to find such $n$? What is special about these $n$'s?
 A: This is, as you probably know, a sequence of Bernoulli trials. The expected number of successes is $p \cdot n$. Thus, the expected number of failures is $(1-p) \cdot n$. In this case, that means we should expect to lose $\frac{2 n}{37}$ times. In order to get a decreasing sequence, we need to get that expectation below 2. That is, take $n \le 36$ (or $n \lt 36$) if you want strictly decreasing probabilities.
A: Let $P_n(k)$ be the probability of losing exactly $k$ times. We examine the ratio $$\frac{P_n(k+1)}{P_n(k)}.$$ This is equal to
$$\frac{\binom{n}{k+1} \left(\frac{2}{37}\right)^{k+1}\left(\frac{35}{37}\right)^{n-k-1}}{ \binom{n}{k} \left(\frac{2}{37}\right)^{k}\left(\frac{35}{37}\right)^{n-k}               }, $$
which simplifies to
$$\frac{n-k}{k+1}\cdot \frac{2}{35}.\tag{1}$$
If probabilities are to decrease strictly from $k=1$, we want $\frac{n-1}{2}\cdot \frac{2}{35}$ to be $\lt 1$. This gives $n\lt 36$. 
Remark: Note that the fraction $\frac{n-k}{k+1}$ is decreasing, From (1) we can find for any given $k$ the smallest $n$ such  that $P_n(k)$ is non-decreasing  up  to $k$, and then decreasing. 
A: Second question:
It has to be $Bin(n,\frac{2}{37}|1) > Bin(n,\frac{2}{37}|2)$.
${n \choose 1} \left( \frac{2}{37} \right)^1\cdot \left( \frac{35}{37} \right)^{n-1} >{n \choose 2} \left( \frac{2}{37} \right)^2\cdot \left( \frac{35}{37} \right)^{n-2}$
$n\cdot \frac{2}{37} \cdot \frac{37}{35} \cdot \left( \frac{35}{37} \right)^n > \frac{n\cdot (n-1)}{2} \cdot \left( \frac{2}{37} \right) ^2\cdot \left( \frac{37}{35} \right) ^2\cdot \left( \frac{35}{37} \right)^{n}$
After dividing the equation by $\left( \frac{35}{37} \right)^{n},n, \frac{37}{35} $ and $\frac{2}{37}$ and cancelling out 37, we get
$1 >\frac{n-1}{2} \cdot  \frac{2}{35} \Rightarrow 35 > n-1 \Rightarrow 36 > n $
$\Rightarrow 1 \leq n \leq 35 $
