My colleague had a similar assignment on the final interview to Jane Street (Quantiative Researcher Intern position). The only difference was that there were three points instead of two. Logic is almost the same though and calculating the simpler case with two points is essential. I love your content ❤❤❤
@@atypicalquant yeahhh and the problem was formulated in a slightly different way. There are three points, X, Y, Z ~ Unif(0, 1). Croupier shows u a random point and u have to tell if it's the smallest one, highest or the one in the middle. Then croupier shows you the second random point and u have to guess and at the end he shows you the last point (you have only one option left). If you guess correctly (all three points) you win 10$. How much would you pay to play this game?
@@g1rlss1mp This question was very interesting. I figured that similar to the two points, your first point would be the lowest in bottom 1/3, middle in middle 1/3 and highest in top 1/3. However, that means that if the first number is 2/3 then you should be able to choose either top or middle with same results. However the top would be (3/4)*(x)^2=1/3 while the middle is 2*x*(1-x)=4/9. It turns out that making (3/4)*x^2=2*x*(1-x) you get 8/11 and symmetrical 3/11. This means you should take any number between 0.272727 and 0.727272 as the middle. By doing this you can integrate from 0 to 3/11 (3/4)(1-x)^2 + integrate from 3/11 to 8/11 2x(1-x) + integrate from 8/11 to 1 (3/4)x^2 and get the result as 377/726 or about 51.92%.
@@peterboylan8560 @UCmYCRqOomNbUC9T-V13zY2QI might be being dense, but if either of you could elaborate on how you obtained the relation (3/4)*(x)^2 = 2*x*(1-x). what do the left and right hand sides correspond to in terms of conditional probabilities. It is not immediately apparent to me what is being computed here and why? And how it relates to the video.
Great video, as always! Many thanks! There's actually a good brain teaser that u might think about: how many times in a day we couldn't tell the exact time if arrows of a clock would be identical.
Thank you for these videos! I've been watching them recently, and they have been really great. I was interested in finding the maximum of the quadratic expression for P(win|X=x) to see if you could optimize your choice of x. The maximum is at x=1/2, so does that mean S1 actually gives the best odds you could possibly get? I am curious if there could be a strategy that is better, but it seems unlikely.
Glad you like them! It's not unlikely that there exists a better strategy than the one in the video, for any value of x. This is part of the reason why I asked for one such strategy, not all of them, since the proof that you maximise the odds of winning for every possible value of the first number looks quite difficult. Also, it's possible that such a strategy does not even exist (one for which you are achieving the global maximum probability of winning for each value of the first number chosen).
I don't know about anyone else, but the fact that the advanced stratergy underperforms the "random" strategy feels wrong. Is there an inutitive way to understand why that would be the case? I can imagine that if I was rolling a 2 die, and the first roll was a 1, it is more likely that I will win the bet if I guess that the next die roll will be greater than 1 than if I were to randomly guess? Is my thought process wrong? If so, what is wrong?
Hi, Julien! In this video we look at three strategies. The "random" one (R, 1:31), the "easy" one (S1, 4:07), and the "complex" one (S2, 6:44). Their respective win probabilites are 0.5, 0.75 and 0.666... As you can see, both S1 and S2 outperform random, as they leverage the knowledge of the outcome of the first die. Now, for some intuition on why S2 is worse than S1, let's look at the game you proposed. We throw a die twice. Let's say that our first roll is actually 2. With S1, we always bet that the 2nd roll is higher, while with S2, with a probability, we expect that it could be lower. While, indeed, at times, the 2nd roll will be lower, getting those right is like trying to time the stock market; you're better off with just sticking with the most likely outcome; that's what you're doing and that's what S1 is doing.
Hey Ankit! I'm not sure what you mean. You can check that the result for the integral at 6:47 is correct: www.wolframalpha.com/input?i=integral+from+0+to1+%281%2B2x-2x%5E2%29%2F2 .
Hi, just a quick query, with the first strategy, always picking a2>a1, isn’t it a lower than 50% win rate, since there is also a non negative probability of a1=a2?
Hi! While, intuitively, you would think that the probability is greater than 0, but, it actually is exactly equal to 0. One way to think about it is to take two random numbers between 0 and 1. The probability that their first n decimal digits are equal is 1/10*1/10*...*1/10 = 1/10^n. But we want all of them to be equal, so we do a limit as n goes to infinity. This gives you exactly 0. A different way of thinking is you can consider X,Y two RVs, U([0,1]). Now look at Z=X-Y which follows an Irvin Hall distribution. Z is a continuous random variable, so P(Z=0) can be compute as an integral from 0 to 0 of the pdf of Z and again, you get exactly 0.
@@atypicalquant Oh I see, thank you so much, I guess it would be a non 0 probability if we were taking a discrete distribution from 0-1, such as 0.01 intervals, but the 0 probability comes from the fact it is a continuous distribution?
Wait, you assumed that a_1 and a_2 are random. No distribution was mentioned neither if they are independent. Then you picked the uniform distribution and assumed independence. Then, for the better strategy you allowed sampling the distribution again which was not mentioned that it is allowed in the description of the problem. What's going on here atypicalquant? 🙂
While not the intention of this video, it happens in practice that the question you receive in an interview is ambiguous, and you are left to make assumptions about the details of it. These assumptions can then be validated or invalidated by the interviewer. I chose the most common assumptions for this video, but I can try to clarify them, probably in the description, for future videos
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This one felt like common sense? No real calculation needed, only takes a second to convince yourself that the immediate intuition was correct
My colleague had a similar assignment on the final interview to Jane Street (Quantiative Researcher Intern position). The only difference was that there were three points instead of two. Logic is almost the same though and calculating the simpler case with two points is essential. I love your content ❤❤❤
Happy to hear! That's an interesting twist to the problem :).
@@atypicalquant yeahhh and the problem was formulated in a slightly different way. There are three points, X, Y, Z ~ Unif(0, 1). Croupier shows u a random point and u have to tell if it's the smallest one, highest or the one in the middle. Then croupier shows you the second random point and u have to guess and at the end he shows you the last point (you have only one option left). If you guess correctly (all three points) you win 10$. How much would you pay to play this game?
@@g1rlss1mp This question was very interesting. I figured that similar to the two points, your first point would be the lowest in bottom 1/3, middle in middle 1/3 and highest in top 1/3. However, that means that if the first number is 2/3 then you should be able to choose either top or middle with same results. However the top would be (3/4)*(x)^2=1/3 while the middle is 2*x*(1-x)=4/9. It turns out that making (3/4)*x^2=2*x*(1-x) you get 8/11 and symmetrical 3/11. This means you should take any number between 0.272727 and 0.727272 as the middle. By doing this you can integrate from 0 to 3/11 (3/4)(1-x)^2 + integrate from 3/11 to 8/11 2x(1-x) + integrate from 8/11 to 1 (3/4)x^2 and get the result as 377/726 or about 51.92%.
@@peterboylan8560 exactly, this intuition and result are correct 👏
@@peterboylan8560 @UCmYCRqOomNbUC9T-V13zY2QI might be being dense, but if either of you could elaborate on how you obtained the relation (3/4)*(x)^2 = 2*x*(1-x). what do the left and right hand sides correspond to in terms of conditional probabilities. It is not immediately apparent to me what is being computed here and why? And how it relates to the video.
Great video, as always! Many thanks! There's actually a good brain teaser that u might think about: how many times in a day we couldn't tell the exact time if arrows of a clock would be identical.
Thank you, and thank you for the suggestion of problem! I will add it the list of future problems I'm considering for videos :D
Just discovered your channel and you're amazing! Thank you so much for posting these videos :D
Thank you for watching! It's great to hear you are enjoying them!
Thank you for these videos! I've been watching them recently, and they have been really great.
I was interested in finding the maximum of the quadratic expression for P(win|X=x) to see if you could optimize your choice of x. The maximum is at x=1/2, so does that mean S1 actually gives the best odds you could possibly get? I am curious if there could be a strategy that is better, but it seems unlikely.
Glad you like them! It's not unlikely that there exists a better strategy than the one in the video, for any value of x. This is part of the reason why I asked for one such strategy, not all of them, since the proof that you maximise the odds of winning for every possible value of the first number looks quite difficult. Also, it's possible that such a strategy does not even exist (one for which you are achieving the global maximum probability of winning for each value of the first number chosen).
I don't know about anyone else, but the fact that the advanced stratergy underperforms the "random" strategy feels wrong. Is there an inutitive way to understand why that would be the case? I can imagine that if I was rolling a 2 die, and the first roll was a 1, it is more likely that I will win the bet if I guess that the next die roll will be greater than 1 than if I were to randomly guess? Is my thought process wrong? If so, what is wrong?
Hi, Julien! In this video we look at three strategies. The "random" one (R, 1:31), the "easy" one (S1, 4:07), and the "complex" one (S2, 6:44). Their respective win probabilites are 0.5, 0.75 and 0.666... As you can see, both S1 and S2 outperform random, as they leverage the knowledge of the outcome of the first die.
Now, for some intuition on why S2 is worse than S1, let's look at the game you proposed. We throw a die twice. Let's say that our first roll is actually 2. With S1, we always bet that the 2nd roll is higher, while with S2, with a probability, we expect that it could be lower. While, indeed, at times, the 2nd roll will be lower, getting those right is like trying to time the stock market; you're better off with just sticking with the most likely outcome; that's what you're doing and that's what S1 is doing.
@@atypicalquant Thank you so much for clarifying this point, you're totally right.
Your integration calculation (6:47) may be wrong in my opinion, when you differentiate the integral value to get the final x, it will come as 1/2.
Hey Ankit!
I'm not sure what you mean.
You can check that the result for the integral at 6:47 is correct: www.wolframalpha.com/input?i=integral+from+0+to1+%281%2B2x-2x%5E2%29%2F2 .
Hi, just a quick query, with the first strategy, always picking a2>a1, isn’t it a lower than 50% win rate, since there is also a non negative probability of a1=a2?
Hi!
While, intuitively, you would think that the probability is greater than 0, but, it actually is exactly equal to 0.
One way to think about it is to take two random numbers between 0 and 1. The probability that their first n decimal digits are equal is 1/10*1/10*...*1/10 = 1/10^n. But we want all of them to be equal, so we do a limit as n goes to infinity. This gives you exactly 0.
A different way of thinking is you can consider X,Y two RVs, U([0,1]). Now look at Z=X-Y which follows an Irvin Hall distribution. Z is a continuous random variable, so P(Z=0) can be compute as an integral from 0 to 0 of the pdf of Z and again, you get exactly 0.
@@atypicalquant Oh I see, thank you so much, I guess it would be a non 0 probability if we were taking a discrete distribution from 0-1, such as 0.01 intervals, but the 0 probability comes from the fact it is a continuous distribution?
Exactly.
Wait, you assumed that a_1 and a_2 are random. No distribution was mentioned neither if they are independent. Then you picked the uniform distribution and assumed independence. Then, for the better strategy you allowed sampling the distribution again which was not mentioned that it is allowed in the description of the problem. What's going on here atypicalquant? 🙂
While not the intention of this video, it happens in practice that the question you receive in an interview is ambiguous, and you are left to make assumptions about the details of it. These assumptions can then be validated or invalidated by the interviewer.
I chose the most common assumptions for this video, but I can try to clarify them, probably in the description, for future videos
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