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A person is hosting a raffle event. There are 1000 participants in the raffle. The raffle draw will produce one winner.

The Special Rule

The host is also 1 of 1000 participants, but he announces he will not claim the prize, so, if the host wins the raffle, a re-draw will happen and he will be removed from the second draw which makes the total number of participant to 999.

Question

What is the probability of me winning the raffle?

Note

This might be a really silly question, but i cant seem to come up with an answer :).

My attempts:

Answer 1. Chance is $1/1000 + 1/1000 * 1/999$ = (chance of me winning first round + chance of host winning * chance of me winning second round)

Answer 2. Chance is $1/999$ because logically speaking, there are 999 people who can win, chance is just 1/999.

Edit: just did a calculation to actually calculate above 2 answers, they have the same result :)

maxi C
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  • Does your calculation in your edit answer your question? $\frac1{999}$ looks correct to me – Henry Oct 25 '19 at 00:18
  • @Henry - Yes, i am curious on what is the most common or efficient to solving problems like this. Because obviously the 2nd way is very quick without even thinking mathematically. – maxi C Oct 25 '19 at 00:20
  • Common and efficient methods are sometimes the same and sometimes different. I would have chosen your second method here, since the chances for the $999$ non-hosts are equal to each other and add up to $1$ – Henry Oct 25 '19 at 00:23
  • I would like to believe that answer 1 is more appropriate way to describe the solution. 999 is NOT really the total number of choices . Would love to hear comments on this. – kgkmeekg Oct 25 '19 at 01:11
  • @smkj33 - answer 2 is a completely different style of argument, i.e. a symmetry argument. Henry already mentioned the gist of it, and I expanded on it in my answer below. Hope you find it helpful. – antkam Oct 25 '19 at 01:45

1 Answers1

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Well, as you found out:

$${1 \over 1000} + {1\over 1000}{1\over 999} = {1\over 1000}(1 + {1\over 999}) = {1\over 1000}{1000\over 999} = {1 \over 999}$$

so the two answers agree, and both answers are correct.

In a sense, answer 1 is more mechanical and is completely non-controversial. Answer 2, however, is actually a symmetry argument - it is in fact arguing that (1) someone among the $999$ must win, and (2) each of them has the same chance.

In particular, answer 2 is not concerned about certain details of the drawing, as long as symmetry is preserved. E.g. imagine a modified process: if the host draws his own number, he puts it back into the hat and redraws, and if he redraws his own number, he puts it back into the hat and redraws, ... and he only throws away his own number (in exasperation) after $17$ consecutive draws of it. Then you can still write a stupidly long expression for answer 1, and it will evaluate to $1/999$, or you can argue, again by symmetry, that it is $1/999$.

In fact, if the host never throws away his own number, you can still write an infinite expression for argument 1, and it will (in the limit) evaluate to $1/999$.

IMHO the symmetry argument, i.e. answer 2, is obvious and preferred. However, the problem with "proof by obviousness" is that not everyone agrees it is obvious. :) So if you don't like answer 2, then please consider answer 1 as a way to justify answer 2.

BTW, not "seeing" an "obvious" symmetry happens somewhat often when dealing with probability questions. Here and here (esp. Ned's answer) are more examples where there is a very simple symmetry argument, but if you don't see it or disagree with it, then you need more complicated algebraic manipulations.

antkam
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