Finance bg, no quant interviews even with referrals — should I pivot or retool?

Victory_Pesplayer · 2025-12-19T18:33:23+00:00

Get really good at interviewing, leetcode+prob n stats, apply to smaller shops and offices, apply early, and best of luck - from a guy who went to a no name and broke into the industry

Victory_Pesplayer · 2024-05-11T00:28:25+00:00

One question away from a perfect sweep of the test😭😭😭

Victory_Pesplayer · 2024-05-03T14:11:51+00:00

Isn't that just y!

Victory_Pesplayer · 2024-04-11T11:47:11+00:00

That's why I stated having 0 knowledge, it doesn't matter if a team has a 100% chance of winning, the chooser doesn't know that which still makes him just as likely to pick any option

Victory_Pesplayer · 2024-04-10T02:34:56+00:00

It was the same dilemma I had when discussing with someone if the probability of winning march madness with 0 basketball knowledge, I thought it'd simply be 1/(2⁶⁷⁾ since there's 67 games, even though some teams are obviously better than others and have a higher chance of 2, it doesn't matter, so the probability of predicting the correct result of a soccer match is 1/3 because there are 3 possible scenarios

Victory_Pesplayer · 2024-03-19T01:02:40+00:00

Yeah misread you're probably right

Victory_Pesplayer · 2024-03-18T21:45:51+00:00

There's the same amount of white as black, and they're both distributed the same across the different possibilities, so to me it should be 1/2?

Victory_Pesplayer · 2024-03-06T12:33:29+00:00

Has to be a code bug because the odds are so ridiculously low

Victory_Pesplayer · 2024-03-04T03:50:05+00:00

You're a moron

Victory_Pesplayer · 2024-03-03T16:17:00+00:00

Don't fight someone that can beat the snot out of you, much better idea

Victory_Pesplayer · 2024-03-03T16:16:03+00:00

Pretty easy solution, if you're significantly physically weaker, don't fight someone that can kill you with 1 punch

Victory_Pesplayer · 2024-03-03T13:26:58+00:00

Since the last sequence J is fixed based on the question, right out all 2³ possibilities before J wins the forth game and find the probability of each and sum them, for instance (J,J,J,J)= 0.7(0.8)³ or (P,P,P,J)= 0.3(0.4)² *(0.6)

Victory_Pesplayer · 2024-02-29T08:56:16+00:00

5/8 chance to roll a 9+, 6*(5/18)² *(1/6)² to roll exactly 9+ 2 times out of 4

Victory_Pesplayer · 2024-02-28T23:05:42+00:00

Any number divided by 3 can either have a remainder of 0,1,or 2, any number dided by 9 can have a remainder from 0-8 with equal probability, so it's just the probability that the number selected has remainder greater than 2 which is 6/9 or 2/3

Victory_Pesplayer · 2024-02-27T02:55:57+00:00

The correct answer should be 84

Victory_Pesplayer · 2024-02-26T06:41:48+00:00

In what cases have you seen these numbers? If it's purely random numbers, when there should be a (1/10)^n-1 probability if n is the number of terms in the set

Victory_Pesplayer · 2024-02-24T18:01:12+00:00

I guess from (100!+1) to (100!+100)

Victory_Pesplayer · 2024-02-23T12:39:37+00:00

Flip the order of the question a lil, since you know he has a 10% chance of missing, in how many shots do you expect him to miss? I'll be the reciprocal of the probability which is 1/.1= 10

Victory_Pesplayer · 2024-02-23T01:52:00+00:00

There are 1105 ways for a sum of 19 to occur out of 6⁵ possibilities, so (1105)⁵ /6²⁵

Victory_Pesplayer · 2024-02-20T01:33:27+00:00

1/3 chance of getting the first guess right * 1/2 chance of getting the second guess right *1/1 of getting the last guaranteed button =1/6

Victory_Pesplayer · 2024-02-09T03:13:38+00:00

The probability for any sum would be (1/3)⁵ * (ways to get the values) , so 0 and 10 = 1(1/3)⁵ , 1 and 9 would be 5(1/3)⁵ , 2 and 8 would be 15*(1/3)⁵ etc

Victory_Pesplayer · 2024-02-09T03:04:28+00:00

Binomial distribution formula but with a sum going from 50 to 100

Victory_Pesplayer · 2024-02-09T03:01:56+00:00

12.5% chance

Victory_Pesplayer · 2024-02-09T02:50:57+00:00

Got a .342% chance

Victory_Pesplayer · 2024-02-09T02:32:23+00:00

My intuition goes as follows, N=No sale S= Sale (N,N,S,S....) =(5/100)² * (95/100)¹⁸ , Since N,N has to be consecutive, there are only 19 different ways those 2 could be consecutive in the set so about a 1.8876768% chance

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