For a fair lottery game where the odds of $1$ ticket winning are $1$ in $p$, where you can spend a total of $K$ dollars, and where you will spread your ticket purchases equally among $n$ draws, prove that the odds of winning at least once decrease as $n$ increases.
An online "discussion" I'm having about the relative merits (such as they are) of different strategies of playing the lottery. With a fixed amount of money to spend on lottery tickets, he claims that it's better to spread your ticket purchases out over multiple draws, while I say it's better to spend all the money on unique tickets for one draw.
I can show that I'm correct using absolute numbers, but I'm trying to prove a general case.
$$p \ge K \ge n \ge 1$$
The odds of losing every time (never winning) are given by: $${\left(p-\frac K n\over p\right)}^n$$
I can throw "actual" numbers in there for p, K, and n and show that playing \$100 all at once is better than playing \$10 ten times or \$1 one hundred times.
I want to show the general case. That is, I want to show that for any given p and K, the odds of losing every time: $${\left(p-\frac K n\over p\right)}^n$$ increase as n increases from 1 to K. I started with the presumption: $${\left(p-\frac K {n+1}\right)^{n+1} \over {p^{n+1}}} > { \left(p-\frac K {n}\right)^{n} \over {p^n}}$$ but college math was a long time ago, and I'm stuck at this point (I've probably gone off in the wrong direction): $${\left(p-\frac K {n+1}\right)^{n+1} \over \left(p-\frac K {n}\right)^{n}} > p$$
Any pointers on how I can "simplify" that expression or otherwise continue with a proof?
When you say it is better to play all your money in one draw, that is correct if by better you mean a better chance of winning something. Each ticket has the same expected value. The improvement from playing the same draw comes from the fact that you cannot win more than once. Let's contrast throwing a die. Playing one number on two separate rolls gives you $\frac {11}{36}$ chance of winning at all, $\frac {10}{36}$ that you win once and $\frac 1{36}$ that you win twice. The total winnings are $\frac {12}{36}=\frac 26=\frac 13$. If you play two numbers on one roll, you can only win once, but your chance of winning is $\frac 26=\frac 13$ Yes, you have a higher chance of winning once, but the same total winnings. The other side of the "discussion" may be focusing on expected value instead of the chance of total loss.
What you are trying to prove is that $(1-\frac an)^n$ is monotonically increasing with $n$ (where $a=\frac Kp$) It starts at $1-a$ and approaches a limit of $\exp(-a)$ as $n \to \infty$ If you expand the binomial, it starts $1-a+\frac {n(n-1)}{2n^2}a^2-\dots$ When $a \lt 1$ the terms are decreasing and the second order term increases with $n$