Is there a higher chance of winning one contest if you enter many? How can we determine when it's most mathematically favorable then?

Question

I know this question sounds like others asked, but none of those answered the exact, specific question I was wondering here. I know this may sound confusing and subjective, but hear me out:

We know someone can theoretically enter one contest and win -- and it's possible to enter very, very many contests (like sweepstakes for example) and still lose in spite of all the entries/etc.

My main question is, does one become more likely to win overall -- all various factors considered -- from any sweepstakes/contest at all just solely by entering 'X' or a high number of them initially?

In other words, would someone be more likely to win if they enter more contests, versus someone entering less? Some people enter endless contests just because they feel they'll eventually win.

Is this true? Do you have a higher chance of winning any contest by entering as many contests as you possibly can, when comparing the same overall likelihood of someone entering less than you?

Because all of this can be so random that one guy can win his first entry, whereas another loses ten million of them in a row. It would seem the ten million contest guy wasted his time, correct?

So should it be considered mathematically/time valuable to enter more contests in hopes of having more chances of winning? At what point can this mathematically favor one's time investment, when the return we expect is entering as many of any contests/etc. to win in 'Y'fold favorably?

By "'Y'fold favorably" I mean any certain people when a surplus of contests entered makes said person entering have some of the best chances of winning something from ANY single one overall, so as to make their time investment in entering 'X' number of contests justifiable with math odds.

Answer 1

Let consider a simple example and then you can extrapolate.

Consider two independent lotteries $L_1$ and $L_2$ that both pays $1$ with probability $50\%$ and $-1$ otherwise. We could write it $P(L_1=1)=P(L_1=0)=0.5$

A fair ticket to enter this lottery should cost 0.5.

Let's assume you have $1\$$. You can buy two different strategies:

1. buy lottery 1 and lottery 2 and get:

   • $2$ with probability $0.5\times0.5$ = $0.25$
   • $1$ with probability $0.5$
   • $0$ with probability $0.25$

2. buy 2 ticket of lottery 1 get:

   • $2$ with probability $0.5$
   • $0$ with probability $0.5$

In both cases you will get $1\$$ on average (the price you paid to enter the deal).

You wont get any richer. The more independent you diversify your investment in the more it will just spread out/smooth your possible outcomes and their probabilities but keep the average gain the same.

In reality, some people would prefer(be willing to pay more) the first strategy and some people would prefer the second strategy. To be honest most people like their loss to be limited and their gain uncapped, potentially very high because of emotional bias. On the other hand company who sells both strategies in large amount (for many lotteries) should be indifferent between the two but adjust the price to make her profit by a trade off between attracting lottery buyers and making profit on the lottery payoff - price.

Answer 2

Suppose you enter $20$ random contests, $C_1$ to $C_{20}$, with a $1\%$ chance to win each individual contest. For all $i$ from $1$ to $20$, let $c_i$ be the number of times you win contest $C_i$ (this is a random value that is either $1$, with a $1\%$ chance, or $0$, with a $99\%$ chance).

There's the concept of the expected value $\mathbb E(x)$ of a random number $x$, which is basically the average value of $x$ weighted by probability.

For example $\mathbb E(c_i)$, the expected value of $C_i$, is equal to $99\% \cdot 0 + 1\% \cdot 1 = 0.01$.

The nice thing about the expected value is that it's additive, so we have $$\mathbb E(c_1 + \ldots + c_{20})= \mathbb E(c_1) + \ldots + \mathbb E(c_{20}) = 20\cdot 0.01 = 0.2.$$

So the total number of contests you win, $c_1 + \ldots + c_{20}$, has expected value $0.2$. On average, you'll win $0.2$ contests.

Suppose now you enter $2000$ contests instead of $20$, each with a $1\%$ chance to win.

Let's calculate the expected number of contests you win:

$$\mathbb E(c_1 + \ldots + c_{2000})= \mathbb E(c_1) + \ldots + \mathbb E(c_{2000}) = 2000\cdot 0.01 = 20.$$

So the expected number of contests you win is $20$.

But does that mean you'll always win exactly $20$ contests? No, because that $20$ is only the expected value of a random number, which is sort of an average value. In fact, there's only a $8.92828\%$ chance you'll win exactly $20$ contests, there's a decent chance you'll win $13$ or $28$ contests, there's a very small chance you win none of the contests, and there's an even smaller chance that you'll win all $2000$ contests.

But note that all these contests are independent. The chance that you win $C_1$ is still only $1\%$, no matter how many other contests you enter. Entering other contests, or winning them, or losing them, does not at all affect the outcome of $C_1$. That's a direct consequence of the fact that these contests are all independent.

Whether to enter a contest should therefore be always judged on its own merit, not based on what other contests you've also entered.

(Well, there are some finer points of risk management and nonlinear value with interesting consequences, for example that it's mathematically a good idea to buy insurance and a good idea to sell insurance, but I believe that's beyond the scope of this question.)