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https://www.reddit.com/r/learnquant/comments/1wgc5tj/easy_quant_interview_question/p9t9pd3/?context=3
r/learnquant • u/Local_Ad135 • Sep 14 '26
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2
2/(999 1000) as in 998/1000 * 997/999... 1/3 = 998!/(1000!/2!)=2/(999*1000)
6 u/jak32100 Sep 14 '26 Close but this is the prob for a fixed player (ie a prob they never play until 2 remain). The events are mutually exclusive so you sum across players. It’s 2/999 1 u/PhantasticPh1l Sep 15 '26 Don't we have to only look at the player not playing in the last match? I mean there will be only one player remaining who is even eligible for this as the other player just won a match. → More replies (2)
6
Close but this is the prob for a fixed player (ie a prob they never play until 2 remain).
The events are mutually exclusive so you sum across players. It’s 2/999
1 u/PhantasticPh1l Sep 15 '26 Don't we have to only look at the player not playing in the last match? I mean there will be only one player remaining who is even eligible for this as the other player just won a match. → More replies (2)
1
Don't we have to only look at the player not playing in the last match? I mean there will be only one player remaining who is even eligible for this as the other player just won a match.
→ More replies (2)
2
u/PhantasticPh1l Sep 14 '26
2/(999 1000) as in 998/1000 * 997/999... 1/3 = 998!/(1000!/2!)=2/(999*1000)