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I've entered sweepstakes and contests, and recently they all have the option to refer a friend. If your friend enters the contest, you get an extra entry for yourself, "increasing your odds of winning", as the contest likes to proclaim.

Is this always true? Sure, I get an extra entry, but that's somewhat negated by the fact that someone else has an entry for themselves.

With some quick math, I found this may be a true statement, assuming there are more than 2 people already entered.

  • It certainly does not hold true if you're the only entrant. You have a 1/1 chance of winning, and your friend enters, reducing your odds to 2/3.

  • Similarly, with 2 entries, you have a 1/2 chance. With your friend entering, you still have a 50% chance, 2/4.

Aside from those two edge cases, which are extremely unlikely in any contest, are you always going to have a higher chance of winning by getting a referral entry? For the general case, assume you get one entry for every one entry of your referral. However, in other cases, you get 2 or 3 entries for a single entry of someone else.

Stephen S
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  • You seem to have answered your own question. But of course it would depend on how the contest rules are set up, which you haven't rigorously defined. If you're asking about actual contests in the society and the rules they use, it's hardly a mathematical question anymore, is it? – Wildcard Oct 25 '16 at 19:11

2 Answers2

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Yes - if there are $3$ or more people and you have $1$ ticket, you will increase your odds of winning. Suppose you had $n$ tickets out of $m$ tickets. Then if you refer a new person, you will have $n+1$ tickets out of $m+2$ tickets. We see that you increase your chance of winning if and only if $\frac{n+1}{m+2} > \frac{n}{m}$. With some algebra, we see that:

$$\frac{n+1}{m+2} > \frac{n}{m}$$

$$\frac{m(n+1)}{m(m+2)} > \frac{n(m+2)}{m(m+2)}$$

$$mn + m > mn + 2n$$

$$m > 2n$$

$$\frac{n}{m} < \frac{1}{2}$$

Then we see that $\frac{n+1}{m+2} > \frac{n}{m}$ if and only if the probability of you winning without the referral is less than $\frac{1}{2}$.

user2825632
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  • Interesting. So if my odds were ever higher than 50%, say if I had other non-referral ways of entering, then a 1-for-1 referral entry would reduce my odds. – Stephen S Oct 25 '16 at 19:23
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    Yep! This is because the overall probability will be a weighted average of $\frac{n}{m}$ and $\frac{1}{2}$. If $\frac{n}{m} < \frac{1}{2}$, the $\frac{1}{2}$ will bring the overall probability up. Otherwise if $\frac{n}{m} > \frac{1}{2}$, it will bring the probability down. – user2825632 Oct 25 '16 at 19:27
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Say there are $n$ participants, including you. The chance of winning is: $$\dfrac{1}{n}$$ Now suppose your friend joins and you get an extra entry. your chance is now: $$\dfrac{2}{n+2}$$ Say $n>2$. Now: $$n>2$$ $$2n>n+2$$ $$\dfrac{2n}{n^2+2n}>\dfrac{n+2}{n^2+2n}$$ $$\dfrac{2}{n+2}>\dfrac{1}{n}$$ So with more than 2 participants (that's including you, but excluding your friend), your chances will be higher)

Mastrem
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