# mmiguel1

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## mmiguel1's Activity

1. mmiguel1 added a post in a topic

The first man saw -4 coconuts.
He gave one to the monkey so there were -5.
Then he divided this by 5 to get his share of -1, putting -4 back.
This happened a bunch of times...
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2. mmiguel1 added a post in a topic

War of the egos

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3. mmiguel1 added a post in a topic

If the last man had 31 coconuts to split up, he would, give 1 to the monkey, divide the 30 into 5 groups of 6, and take 6 for himself, putting the other 24 back.
Thus, they would see 24 coconuts, not 6.

I do agree that there are multiple answers though.
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4. mmiguel1 added a post in a topic

5. mmiguel1 added a post in a topic

Is that Pirates and Gold coins?
I remembered that too.
In that puzzle however, there isn't the extra cycle in the morning after everyone has already done their misdeeds.
Because of that, I expect the answers to be different.
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6. mmiguel1 added a post in a topic

7. mmiguel1 added a post in a topic Whatchya Gonna Do (2 goats and a car)

Ahh yes, I love the math. I just usually find that people find conceptual sentences more convincing.

P[initially right] = 1/3
P[initially wrong] = 2/3

Total Probability Theorem:
P[Win|stay] = P[Win|stay and initially wrong] P[initially wrong] + P[Win|stay and initially right] P[initially right]

P[Win|stay and initially wrong] = 0 if you are wrong and you stay on the wrong door, you cannot win
P[Win|stay and initially right] = 1 you had the right door and stayed with it, you win!
P[Win|stay] = 0*P[initially wrong] + 1*P[initially right]
P[Win|stay] = P[initially right] = 1/3
P[Win|stay] = 1/3

Similarly,
P[Win|switch] = P[Win|switch and initially wrong] P[initially wrong] + P[Win|switch and initially right] P[initially right]
P[Win|switch and initially wrong] = 1, the only door you can switch to thanks to Monty is the car door
P[Win|switch and initially right] = 0, you had the right door, but you left it for a goat door
P[Win|switch] = 1*P[initially wrong] + 0*P[initially right] = P[initially wrong]
P[Win|switch] = P[initially wrong]
P[Win|switch] = 2/3

Result:
P[Win|switch] = 2/3
P[Win|stay] = 1/3

The probability of winning given that you switch is 2/3.
The probability of winning given that you stay is 1/3.

You can't really apply Baye's rule, because you cannot really calculate P[switch] or P[stay], as those are factors that you can set, not random things you only observe.
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8. mmiguel1 added a post in a topic

This reminds me of Zeno's paradox sort of.
Think about though, how many points exist on the number line between 0 and 1?
If you are an object moving on the number line from 0 to 1 in 1 second, how many points do you cross?
An infinite number of points!
But if you think about it, how many time points are there between elapsed time = 0 seconds to elapsed time = 1 second?
Also an infinite number!
As an object moving from x=0 to x=1 in the time t=0 to t=1, you are traversing an infinite number of points in space and an infinite number of points in time.
Although they are both infinite, there are just as many time points traversed as space points.
The ratio in this case is 1.

This is aspect of calculus. The derivative is one infinitesimal value divided by another infinitesimal value.
Even though these values are infathomably small, they still have a finite ratio. The space vs time case above dealt with infinities rather than infinitesimals, but the reciprocal of an infinitesimal is an infinity.

df/dx, the derivative of f with respect to f has infinitesimals df and dx.

but

df/dx = (1/dx) / (1/df)

(1/dx) and (1/df) are infinities.
And these infinities have a finite ratio, namely, the derivative.

So to sum up some stuff,

(1) an infinite sum of infinitesimals can yield finite values

and

(2) ratios of infinities or infinitesimals can yield finite values

(1) Is the principle behind integral calculus
(2) is the principle behind differential calculus

Together, these simple ideas are responsible for the modern age (in my opinion, yeah yeah, i suppose there was some other important stuff too)
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9. mmiguel1 added a post in a topic

Well... in my last post, I wasn't talking about distance at all, just time. And actually I was saying it would travel the whole way in just 2 seconds, right before our eyes. This is of course assuming that the time it takes to do a step in the +x or the +y direction is 1/N seconds, which was not stated in the initial problem, but proposed only for the sake of argument in my last post.
Basically I'm saying that you cannot assume anything about how long it will take to get to (1,1) because time was never mentioned in the original post. To demonstrate this, I say, that ok, we all know that the steps are getting smaller as N gets bigger. What if also, the time it takes to do a step also gets smaller by the same amount as the distance? In that case, the object of interest would be moving at a stable average speed.

The first step in the +x direction travels 0.1 meters and takes 0.1 seconds.
The +y direction step travels 0.1 meters and takes 0.1 seconds.
The speed of the object here is 0.1 meters / 0.1 seconds = 1 meter/second

Let's make N bigger, like N = 10000
The first step in the +x direction travels 0.0001 meters and takes 0.0001 seconds.
The +y direction step travels 0.0001 meters and takes 0.0001 seconds.
The speed of the object here is 0.0001 meters/0.0001 seconds = 1 meter/second

Notice that the speed did not change, even though N got a thousand times bigger!
In this scenario, the speed is independent of N. If N were to go to infinity, the speed of the dot would still be 1 meter/second.

It was already proved that the distance traveled is 2 meters, so the time taken would be 2 meters / 1 meter/second = 2 seconds

This is just a counter example for your claim that the dot would take an infinite amount of time to move. It was never actually specified how time depends on N, so in fact we can't really say anything about how long it will take.
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10. mmiguel1 added a post in a topic Whatchya Gonna Do (2 goats and a car)

What justification do you have for changing these probabilities?
If I am pointing at a car before Monty opens a goat door, then I am still pointing at a car after Monty opens a goat door.
Monty opening some other door has no effect on what is lying behind my door.
If I stay with this door and win, it is only because the car was always behind the door and I got lucky when I picked this door out of the 3 choices. If I definitely choose to stay, then whether or not I win or lose is decided only at the time when I make the initial pick of the 3 choices. Although I may not know if I have won or lost yet, it is decided at this point. And the probability of me winning at this point is 1/3. As long as I stay with this door, Monty can do whatever he wants (within reason), and it will not affect my odds of winning. Thus, by staying, I win ONLY if my initial guess was right and I lose if my initial guess was wrong. My initial guess is right with probability 1/3. Therefore by staying, I win ONLY if a certain event happens whose probability is 1/3, and I lose otherwise. This certain event being that my initial choice of 3 doors was correct.

If I decide to switch, that is a different story. If my initial guess was right and I switch, then no matter what Monty does, I will lose. If my initial guess was wrong, then I point at one goat, Monty shows the other goat, and the only possible door left I can switch to is the one holding the car. It HAS to be the case that I win if I was initially wrong and I switch. Monty's only purpose here is to make sure that it HAS to be the case that I win if I was initially wrong and I switch.
Therefore, if I do in fact switch, I will win if and only if I was wrong in my initial guess of 3 doors. Winning becomes an equivalent outcome to being initially wrong as long as I decide to switch. There is no randomness to that idea.

Given that I switch:
I was wrong initially = I Win
I was right initially = I Lose

Therefore the probability that I win is equal to the probability that I was initially wrong BECAUSE those two ideas are equivalent, thanks to Monty. No one debates that the probability that I was initially wrong is 2/3. Because of Monty, given that I switch, Winning is the same outcome and must have the same probability of 2/3.

I hope that was clear.
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11. mmiguel1 added a post in a topic

That is not necessarily true, I never said anything about how time depends on N.
If the time it takes to perform a step is independent of N and is also positive and finite, then you are correct.

But if the time it takes to do a step is 1/N seconds, then you will reach your destination in 2 seconds.

All of integral calculus and basically all technological advancements that can be called "modern" (which I don't think would be possible without calculus) rely on the fact that the sum of an infinite number of infinitesimal values can produce a finite result. The dx is the infinitesimal in an integral case, and you are summing an infinite number of these to produce a finite value.
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12. mmiguel1 added a post in a topic Whatchya Gonna Do (2 goats and a car)

Thanks matt, (my name is matt too, last name is miguel).
Looking at the length of this thread, I would bet that it's been proven before somewhere in here at least 10 times over.
I don't know if they stated it the same way as me already, but I wouldn't be terribly surprised if it was already stated like this. (Bonanova likes to convey the logic in multiple perspectives)
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13. mmiguel1 added a post in a topic

I do not understand the question.
If you want her to put it in 4, then ask her to put it in 3.
According to your table, if you ask her to put it in 3, she will always put it in 4, where you actually want it.

If you want her to put it in 3, ask her to put it in 1... etc,

I don't understand what you are asking if it is not this.
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14. mmiguel1 added a post in a topic

how will they explain their prison outfits?
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15. mmiguel1 added a post in a topic