Citadel quant interview problem | Direction Change II

[@Quant_Prof]

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[@praaan0308]

Not able to pay, PayPal not working and credit card not supported for India

[@praaan0308]

"The merchant isn’t able to accept PayPal payments at this time. Please try again later or return to the merchant and choose another way to pay."

[@alina2200]

Can you include the option of paying through UPI?

[@lightningranger8161]

the payments using credit card is not being processed for india, please do it

[@Quant_Prof]

@@lightningranger8161 sorry for this, please give us some time. There is a problem with the platform we are now, we are moving to a new platform and that will take about 10 days.

[@el-swago8427]

a much simpler way could be: minimum no. of turns required=1 (when traveling along the edge only) max no. of turns = 15 ( when traveling down stepwise) by symmetry and intuition, we say that she can choose her path containing any no. of turns between 1-15. there are 8 odd nos. between 1-15 (both included). ans= 8/15

[@Quant_Prof]

After arriving at the answer, I had the same thought. But I believe this reasoning isn’t correct. You are right that the number of turns can vary between 1 and 15 (inclusive), and there are indeed 8 odd numbers within this range. However, you cannot directly apply the probability formula and conclude that the probability is 8/15. ​For the formula to work, all outcomes in the denominator must be equally likely. For example, consider the scenario of tomorrow either having an alien attack or not. While there are two possible outcomes, it doesn’t mean the probability of an alien attack is 1/2, because the likelihood of these outcomes isn’t equal.

[@postselector]

Expected value ought to be zero, right? Since given any configuration, it’s equally as probable as the same configuration but with the first two rows swapped, and they have negated determinants.

[@Quant_Prof]

Correct

[@sagnikbiswas3268]

Similar to 2024 AIME I #6

[@kanshank]

I guess so, you can find the det by summing the det of all 2x2 matrice in a specific dev, and you can check the event are independent Then E(det(2x2)) = E(X1*X2 -X3*X4) = E²(X)-E²(X) =0, So Sun(E(det2x2)) = 0, E =0 It mean the value of det have a average value of 0 but we have no info how often it will have exactly 0 as a value. Also notice if X = [0; inf [ we will have some issue to calculate the E(X = inf) bc of an infinite sum Maybe if use 1+2+3+4+5+6... =-1/12 we can have some clue that proof is stable in infinite dim.

[@AdoptedPoo]

this is a back tracking problem