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This was a 20 second job stretched out using the most tedious arithmetic. Example for the 3 bullet case: without spinning chances of survival = 4C3/5C3 = 2/5 = 40% with spinning the chances of survival = 3/6 = 50%
Yeah the only significant information is that re-spinning makes it so that there is a chance of re-selecting the same chamber. Obviously that increases your odds of survival
Can't agree more. His solution for the second question was the most redundant solution possible. It's as simple as: P(death | re-spin) = 1/3, since the bullets are completely randomized among any chamber, 4 chamber safe, 2 not safe P(death | no re-spin) = 2/5, since before 4 were safe, 2 were not safe, since you were shot at and used a safe chamber, you have 3 safe, 2 not safe. Choose re-spin since you have a better chance at survival.
For the second question, i found 4/10 without detailed calculations. So i was wondering if your way of doing wasn't overkill. I did this: there are 6 chambers, 4 emtpy and 2 with a bullet. After firing the first one, which had an empty chamber, there remains 5 chambers (3 empty chambers, and 2 loaded chambers). Therefore, isn't the probability that the next chamber is loaded 2/5, which is equivalent to 4/10 (or 40%). For the third example, based on the same reasoning, after we discover that the first chamber is empty, we have 5 chambers remaining (2 empty and 3 loaded). Therefore the probability of the next chamber being loaded is 3/5 = 60%. Therefore they should spin again as 60% > 50%. Is there a mistake in my reasoning, or did you make something simple into something more complex that it needed to be?
The quickest way to solve this problem is to use the hyper geometric distribution if I have n chambers and r bullets With spin considered the highest entropy solution it is r/n. This is intuitive but can be proven with a hypergeometric distribution. After the click the number of combinations of bullets being chambered is (n-1)Cr which is the denominator of the hypergeometric distribution knowing that chamber 1 is empty. The number of ways that the next chamber has a bullet is (n-2)C(r-1) x 1C1 being the numerator 1C1 is 1 so we can ignore it for simplicity where our probability of getting shot without spin is (n-2)C(r-1)/(n-1)Cr Which simplifies to r/(n-1) Since r/n is always smaller then r/(n-1) you should always spin for the non adjacent case
Nice vid as always. A friend of mine asked me this question which I thought seemed simple but actually took some work. You have a deck of 52 cards and you draw them one by one. A red card pays you a dollar and a black one fines you a dollar and you can stop anytime you want. Cards are not returned after they are drawn. What’s optimal stopping rule to maximize expected payoff and what is that payoff? Card has 26 red and 26 black cards. Thought I would share since it seems so simple
Blabber. Jargon. What does maximise mean? What does payoff mean? What does stopping rule mean? This is all just madeup nonsense to make yourself feel clever, not an actual explanation of a problem. Instead of appearing smart you appear clumsy and lacking clarity.
For the adjacent case where all r bullets are adjacent in n chambers the position of the first bullet can be anywhere from 2 to n-r+1 meaning there is n-r choices for the first bullet. This means I would compare a spin being r/n versus 1/(n-r). For that case the spin would be preferable if n
Hi Mr.Quant. I still don't get why in the 1st simplest case the probability that the next chamber has the bullet is 2/4? Since the previous chamber was empty, there are now only 5 chambers remaining with 2 still having the bullets. Shoudn't the probability of the current chamber having the bullet be 2/5=0.4. Where 2=no. of chambers with bullets, and 5=no. of chambers remaining?
Not in this case, because we have the information that the bullets are placed adjacently. Therefore, there is only one empty chamber out of four that are followed by a bullet.
It's explained very badly by someone who doesn't really understand the maths or the problem. There are lots of hidden extra rules they keep mentioning in passing.
This is a classic example of why people find maths difficult. It could be explained clearly in twenty seconds but instead it's minute after minute of blabber, jargon, mixed metaphors and hidden extra rules which are never explained at the start and are just thrown in at random. It's a very very clumsy person making something simple look incredibly difficult so they feel smart.
Copy my EXACT resume and cover letter, Kick start your quant applications with a 25%-DISCOUNT Summer Prep Sale using code DE29J7CY
buymeacoffee.com/myquantitative/extras
Need interview advice or want to chat? Drop me an email at myquantitative@gmail.com
Practise makes Perfect, more interview questions: ua-cam.com/play/PLaOlxMtlLosX8V6f3U_wZeMgNftNW_RYz.html
Unconventional advice aspiring Quants are not using enough! : ua-cam.com/video/BKs_1lVTNrs/v-deo.html
This was a 20 second job stretched out using the most tedious arithmetic.
Example for the 3 bullet case:
without spinning chances of survival = 4C3/5C3 = 2/5 = 40%
with spinning the chances of survival = 3/6 = 50%
Yeah the only significant information is that re-spinning makes it so that there is a chance of re-selecting the same chamber. Obviously that increases your odds of survival
Can't agree more. His solution for the second question was the most redundant solution possible.
It's as simple as:
P(death | re-spin) = 1/3, since the bullets are completely randomized among any chamber, 4 chamber safe, 2 not safe
P(death | no re-spin) = 2/5, since before 4 were safe, 2 were not safe, since you were shot at and used a safe chamber, you have 3 safe, 2 not safe.
Choose re-spin since you have a better chance at survival.
For the second question, i found 4/10 without detailed calculations. So i was wondering if your way of doing wasn't overkill. I did this: there are 6 chambers, 4 emtpy and 2 with a bullet. After firing the first one, which had an empty chamber, there remains 5 chambers (3 empty chambers, and 2 loaded chambers). Therefore, isn't the probability that the next chamber is loaded 2/5, which is equivalent to 4/10 (or 40%).
For the third example, based on the same reasoning, after we discover that the first chamber is empty, we have 5 chambers remaining (2 empty and 3 loaded). Therefore the probability of the next chamber being loaded is 3/5 = 60%. Therefore they should spin again as 60% > 50%.
Is there a mistake in my reasoning, or did you make something simple into something more complex that it needed to be?
The quickest way to solve this problem is to use the hyper geometric distribution
if I have n chambers and r bullets
With spin considered the highest entropy solution it is r/n. This is intuitive but can be proven with a hypergeometric distribution.
After the click the number of combinations of bullets being chambered is (n-1)Cr which is the denominator of the hypergeometric distribution knowing that chamber 1 is empty.
The number of ways that the next chamber has a bullet is (n-2)C(r-1) x 1C1 being the numerator
1C1 is 1 so we can ignore it for simplicity where our probability of getting shot without spin is (n-2)C(r-1)/(n-1)Cr
Which simplifies to r/(n-1)
Since r/n is always smaller then r/(n-1) you should always spin for the non adjacent case
Nice vid as always.
A friend of mine asked me this question which I thought seemed simple but actually took some work. You have a deck of 52 cards and you draw them one by one. A red card pays you a dollar and a black one fines you a dollar and you can stop anytime you want. Cards are not returned after they are drawn. What’s optimal stopping rule to maximize expected payoff and what is that payoff? Card has 26 red and 26 black cards. Thought I would share since it seems so simple
Blabber. Jargon. What does maximise mean? What does payoff mean? What does stopping rule mean? This is all just madeup nonsense to make yourself feel clever, not an actual explanation of a problem. Instead of appearing smart you appear clumsy and lacking clarity.
Actually not sure about this one, the EV at the initial point is simply 0, so do you limit the optimal stopping rule when EV < 0 ?
For the adjacent case where all r bullets are adjacent in n chambers the position of the first bullet can be anywhere from 2 to n-r+1 meaning there is n-r choices for the first bullet.
This means I would compare a spin being r/n versus 1/(n-r). For that case the spin would be preferable if n
Hi Mr.Quant. I still don't get why in the 1st simplest case the probability that the next chamber has the bullet is 2/4? Since the previous chamber was empty, there are now only 5 chambers remaining with 2 still having the bullets. Shoudn't the probability of the current chamber having the bullet be 2/5=0.4. Where 2=no. of chambers with bullets, and 5=no. of chambers remaining?
Not in this case, because we have the information that the bullets are placed adjacently. Therefore, there is only one empty chamber out of four that are followed by a bullet.
It's explained very badly by someone who doesn't really understand the maths or the problem. There are lots of hidden extra rules they keep mentioning in passing.
This is a classic example of why people find maths difficult. It could be explained clearly in twenty seconds but instead it's minute after minute of blabber, jargon, mixed metaphors and hidden extra rules which are never explained at the start and are just thrown in at random. It's a very very clumsy person making something simple look incredibly difficult so they feel smart.
What hidden extra rules? What’s unclear
Yes I'm sure you can't see it. I've shown in your own comment how unclear your mind is.
@@daveblack2602 you’re comment got blocked under my other one I can’t see it. But u are likely a troll 🤣🤣
I can’t agree more
i dont get why he has to go through all that case by case analysis just to get to 2/5 which is the number of bullets over number of chambers left/