The reason 5 or 6 does not matter is because there are two cases where rolling a 6th dice makes a difference to the outcome of our game : 1. the first 5 dice did not contain any 6 2. the first 5 dice contained exactly one 6 For case 1, which happens with probability (5/6)^5, if we roll a six, with probability 1/6, we turn a "loss" into a "win", and therefore increase the probability of winning by (5/6)^5 * 1/6 For case 2, which happens with probability 5/6 * (5/6)^4 = (5/6)^5, if we roll a six, with probability 1/6, we turn a "win" into a "loss", and therefore decrease the probability of winning by (5/6)^5 * 1/6 Notice these are the same amounts! Therefore the "marginal benefit" (difference in probability of success between rolling 5 and 6 dice) is exactly 0.
It's interesting because the probabilistic binomial distribution solutions are the same as the expectation of negative binomial distributions, which is where I assume people get their "intuition" from. (Intuition seems to come much more quickly from expectations). Like on a nuanced level probability maximisation in one distribution shouldn't necessarily be linked to expectation in a different distribution, but intuitively in your mind it feels much more appropriate initially (at least to me) to calculate the expectation.
when considering n=5, we are (approximately) choosing a subset of 5 of the set {1, 2, 3, 4, 5, 6}. There are 6C5 subsets possible; only one of these {1, 2, 3, 4, 5} doesn't have the outcome we desire. Therefore 5/6 subsets have the outcome of having at least one six; this makes up for the 5/6 probability term contributed by the extra dice.
1. As we show, it is a binomial distribution, and the peak is actually coming at 5.5 (if n is assumed continuous). Then, we can find 6 and 5 are equal distances from the peak. Considering the symmetry in the distribution of the random variable around the peak( binomial distribution), we can say 5 and 6 have the same probability (maximum if n is discrete, taking only integer value). 2. in the second question, at n=mr & n = mr-1 the probability will be maximum!😉😉 btw I am a big fan of optiver, here in my college ( IITD) you guys hired intern , unfortunately I couldn't appear due to low cg😅😅😅, it's pretty cool to solve problem in this series, keep bringing more. Hope I will get into optiver someday!
Pr (1 six) = n * 1/6 * (5/6)**(n-1)
Maximize n (5/6)**(n-1)
Max at biggest n s.t. (n+1)/n * (5/6)
The reason 5 or 6 does not matter is because there are two cases where rolling a 6th dice makes a difference to the outcome of our game :
1. the first 5 dice did not contain any 6
2. the first 5 dice contained exactly one 6
For case 1, which happens with probability (5/6)^5, if we roll a six, with probability 1/6, we turn a "loss" into a "win", and therefore increase the probability of winning by (5/6)^5 * 1/6
For case 2, which happens with probability 5/6 * (5/6)^4 = (5/6)^5, if we roll a six, with probability 1/6, we turn a "win" into a "loss", and therefore decrease the probability of winning by (5/6)^5 * 1/6
Notice these are the same amounts! Therefore the "marginal benefit" (difference in probability of success between rolling 5 and 6 dice) is exactly 0.
It's interesting because the probabilistic binomial distribution solutions are the same as the expectation of negative binomial distributions, which is where I assume people get their "intuition" from. (Intuition seems to come much more quickly from expectations). Like on a nuanced level probability maximisation in one distribution shouldn't necessarily be linked to expectation in a different distribution, but intuitively in your mind it feels much more appropriate initially (at least to me) to calculate the expectation.
This is one is quite straightforward…
Love it :) Good luck on my Optiver OA today :)
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@@AshmitIITH Just go to optiver website > careers, and submit your resume and transcript
when considering n=5, we are (approximately) choosing a subset of 5 of the set {1, 2, 3, 4, 5, 6}. There are 6C5 subsets possible; only one of these {1, 2, 3, 4, 5} doesn't have the outcome we desire. Therefore 5/6 subsets have the outcome of having at least one six; this makes up for the 5/6 probability term contributed by the extra dice.
Divide both sides by P_n, then it can be seen that P_6/P_5 = 1
Interesting. I simulated this process and got the same maximum probability when n is 5 and n is 6.
You should roll r*m or r*m - 1 dice (both maximise the probability).
What type of problems this is called?
1. As we show, it is a binomial distribution, and the peak is actually coming at 5.5 (if n is assumed continuous). Then, we can find 6 and 5 are equal distances from the peak. Considering the symmetry in the distribution of the random variable around the peak( binomial distribution), we can say 5 and 6 have the same probability (maximum if n is discrete, taking only integer value).
2. in the second question, at n=mr & n = mr-1 the probability will be maximum!😉😉
btw I am a big fan of optiver, here in my college ( IITD) you guys hired intern , unfortunately I couldn't appear due to low cg😅😅😅, it's pretty cool to solve problem in this series, keep bringing more. Hope I will get into optiver someday!
For those interested, this is a hypergeometric distribution problem.
en.wikipedia.org/wiki/Hypergeometric_distribution