1

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?

Ivo Terek
  • 77,665
user159407
  • 11
  • 1

1 Answers1

1

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 $n$

Ross Millikan
  • 374,822
  • What I mean by "better" is that you have a greater chance of winning the jackpot (matching all numbers) if you play all your money in one draw. – user159407 Jun 21 '14 at 21:20
  • I'm not sure the 12/36 value above is correct. When throwing a die, playing one number on two separate rolls gives a 1/36 chance of winning both times, 10/36 chance of winning exactly once, and 25/36 chance that you lose both times. The total chances of winning at least once are 11/36, not 12/36. Playing two numbers on one roll gives a 2/6 (1/3) chance of winning (exactly once), which is "better" than 11/36. (Unless I've completely misunderstood the comment) – user159407 Jun 21 '14 at 21:33
  • Ah - and the "other side of the 'discussion'" is focused exclusively on winning the jackpot at least one time. It's presumed that the value of the jackpot is a few orders of magnitude greater than K, and that winning twice has no greater value than winning once (in fact, one presumption is that once you've won once, you'll stop playing). – user159407 Jun 21 '14 at 21:38
  • @user159407: On the die roll, I said the chance of winning at least once betting on two rolls was $\frac {11}{36}$ The mathematics is not hard and you seem to understand it clearly. Those arguing one way of playing is "better" than another should define "better" carefully. You have done so-by saying the chance of winning at least once. Having done so there is a clear answer and it is to play all your funds on on draw. If the chance of more than one win is very small, the difference between playing one draw or more will be very small. – Ross Millikan Jun 21 '14 at 23:50