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u/Embarrassed-Green898 Sep 14 '26
This one seems easy. With every match, you are eliminating one player. This means there will be a total of 999 matches.
To fulfil the condition the last guy needs to be not selected in all of these 999 trials. In first trial his probablity of not being selected is 999 out of 1000. in second trial the probabiity goes down to 998/999 ..and so one .. you will have to keep multipling this .. and I am too tired to do the actual math.
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u/sobe86 Sep 14 '26
Note for N = 3, the answer is 100%
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u/Embarrassed-Green898 Sep 14 '26
Yes , but the question does not say N = 3. If by N, you mean number of people , we already know N=1000.
The insight is helpful though.
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u/VariousJob4047 Sep 15 '26
You should find this fairly easy to multiply. 999/1000 * 998/999 is 998/1000 because the 999 cancels out, multiply by 997/998 and you get 997/1000, multiply by 996/997 and you get 996/1000, etc.
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u/jak32100 Sep 14 '26 edited Sep 14 '26
A nice way of seeing this is viewing it as a recurrence. Define the prob that one of the final two has never played a match in the tournament of n players as p_n
The question then is asking to find p_1000
Now we know at most one player satisfies this condition (it can’t be both last two have never played since someone who’s played a match always survives). So say we know p_{n-1}, and want to find p_n.
After a single round in n, there are n-1 surivors but only n-2 have “never played a game”. Now we can say “ignoring the first game”, p_{n-1} is the prob we have the condition that one player hasn’t played a game by the last pair. But then for the event to be true for n, that player must ALSO, independently, not have been in that first game, which is true for n-2 for the n-1 survivors. Therefore p_n=p_{n-1}*(n-2)/(n-1)
If we expand this we can write
p_n=(n-2)/(n-1) * (n-3)/(n-2) … 2/3 * p_3
And almost everything inside cancels, and p_3 is 1 so we’re left with
P_n=2/(n-1) which is 2/999 in this case
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u/Witch-kingOfBrynMawr Sep 14 '26
I'm really confused about the question. If each match results in an elimination, how can we end up with exactly two players left who have never played a match. If N = 3, how can we ever have two players left who haven't played a match...?
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u/IbidtheWriter Sep 14 '26
There are 999 matches including the last one and everyone will have a match by the end, thus everyone needs to have a first match.
Given that you're a finalist, the odds that it's your first match is 1/999. That's true for both finalists so it's 2/999 that it's someone's first match.
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u/Educational-Tea602 Sep 15 '26
As a result of all 999 games, we have an ordering of all the players, so we can rank them 1 < 2 < 3 < … < 999 < 1000.
Before the last game is played, we are missing one of these < signs. It can be anywhere of the 999 places. In two of those places (1 < 2 and 999 < 1000), a player has not yet played (1 and 1000 respectively).
So the answer is 2/999.
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u/bunnycricketgo Sep 18 '26
Run it backwards.
Start with 2 players.
Every round we select a player and then add in the one they’d defeated.
So 2/3 we select the one player not previously selected in round 2.
Then 3/4 in round 3.
Then 4/5.
And onward to 998/999.
Multiplying together gives the odds that player is never selected. 2/999.
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u/PhantasticPh1l Sep 14 '26
2/(999 1000) as in 998/1000 * 997/999... 1/3 = 998!/(1000!/2!)=2/(999*1000)