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If you don't remember the tail formula, you could instead calculate the CDF of min(d1, d2, d3) (as was done in the video), take the derivative to get the pdf, and then use LOTUS to compute the expected value. Nice video.
i tried to approximate it instead by using a flat continuous distribution from 0 to 1 and got 1/4 for the expected value of the lowest out of three. i multiplied by 100 to get 25 at the end. still wrong but cool that i got close
Wow that's an interesting approximation. Actually, in many interviews, expected value problems are asked where you're required to quickly derive an approximate solution rather than an exact one.
🎓 Ready to level up and gain a significant edge? Check out the Quant Interview Masterclass for an in-depth guide to mastering quant interviews (with *LIFETIME ACCESS* and 30 days *FULL REFUND* policy):
payhip.com/b/heLaQ
If you don't remember the tail formula, you could instead calculate the CDF of min(d1, d2, d3) (as was done in the video), take the derivative to get the pdf, and then use LOTUS to compute the expected value. Nice video.
1:43 Isn't it 25.5025?
yeah, i just wrote the approximate answer and forgot to put the "approximately equal to" sign
I derived this geometrically for any no. of 'dice rolls'. Really cool proof if you ask me
I think I know what you did
@@Quant_Prof relevant 3b1b video: ruclips.net/user/shortsPny70rNPJLk
87 + 5/14
i tried to approximate it instead by using a flat continuous distribution from 0 to 1 and got 1/4 for the expected value of the lowest out of three. i multiplied by 100 to get 25 at the end. still wrong but cool that i got close
Wow that's an interesting approximation.
Actually, in many interviews, expected value problems are asked where you're required to quickly derive an approximate solution rather than an exact one.