

Stevie
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posted March 10, 2016 02:33 PM 

Edited by Stevie at 14:34, 10 Mar 2016.

Think it's just a trick. Give it a fixed time period, like once a month, and you have a 100% probability to receive the money on the last day. Yet because there's days before that with a high enough probability, the chance to get the money on the very last day would be pretty low. So I was thinking along the lines of finding a peak within that timeframe, but it was not exactly the case in question and it would've took math too...
Just my train of thought.
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Corribus
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The Abyss Staring Back at You

posted March 10, 2016 03:02 PM 

Edited by Corribus at 15:06, 10 Mar 2016.

Pawek did a nice mathematical explanation. Here's a less mathematical one.
The problem has a lot of misleading information included (the actual probability per day of receiving a gift, the amount of the gift, and even the fact that I received a gift today). Regardless of when I got the last gift, or what the probability is of receiving a gift on any day, the next day always has the highest probability of getting a gift. If I had asked what day I was most likely to receive a gift, the answer is of course that every day has equal probability.
But this isn't what was asked.
What was asked was, what day is most likely to be the one where I receive my next gift?
The answer to this question will always be the next opportunity I have to receive a gift, which is tomorrow  even if I hadn't received a gift today.
To see why this is the case, you can look at it this way. What is the probability that the next time I receive a gift is 10000 days from now? It can't be 3%, because there's a 3% chance I'd get a gift on any particular day. For the next gift to be received 10000 days from now, there has to be 9999 consecutive days of not receiving gifts, which is very unlikely indeed, because there is a 3% chance on any day of receiving a gift. I.e., the odds are I'd get some gift between now and 10000 years from now, so the odds are very low that it would be 9999 days between now and the next gift I receive. The statistics work out that the highest probability of receiving the next gift is tomorrow, because any other possibility includes at least 1 day where I don't receive a gift, a probability that is less than 3%.
For the mathematically inclined, this is an example of Poisson distribution, which results in the somewhat paradoxical observation that random independent events tend to be spatially and temporally clustered together rather than evenly spaced apart. There is an old saying that "bad things come in threes". This is actually a manifestation of Poisson statistics. While there is not anything special about the number three, random events do tend to cluster together, and we notice bad events more often than innocuous ones.
And Stevie, statistics does not work like that at all. There is not 100% probability that you will receive a gift every month  not getting a gift for 29 days does not ensure you will a gift on the 30th. The 1 gift per month is an average value.
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fred79
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posted March 10, 2016 05:27 PM 


Corribus said: The statistics work out that the highest probability of receiving the next gift is tomorrow, because any other possibility includes at least 1 day where I don't receive a gift, a probability that is less than 3%.
i don't understand. why would the probability be less on a day that you don't get a gift? does that ungifted day equate to 0%, thus subtracting from the 3%? how could it even subtract from the 3% possibility at all, when every day is a guaranteed 3% probability?
i should warn you, i suck at math. trying to work puzzles like this makes me feel utterly retarded.


JollyJoker
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posted March 10, 2016 05:47 PM 


He asks for a specific day of the NEXT event. NEXT is decisive here, because not only it has the actual event probability included, but also the probability for the event NOT happening for days after the first.
Take a coin toss. You toss a coin ONCE each day. Head pays, tail doesn't. So, question on which day do you have the highest probability for your NEXT win (NEXT is the important thing).
Tomorrow is 50/50. However, for the day after tomorrow to be your NEXT win, it's necessary that you get
a) Tail tomorrow (50/50) (so that you NOT win) AND
b) Head the day after tomorrow (again 50/50)
all in all only 25%, not 50%.
NEXT event is the key word here. Not ANY event.
EDIT: maybe it works better with Russian Roulette as an example. One bullet in a 6shoot drum, new turn after every shot. Which shot has the highest probability to kill you? Answer: THE NEXT ONE!
Right?


Humanoid
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Rest in Peace Juvia (48499)

posted March 10, 2016 05:50 PM 


Finaly! Here im on higher grunds!


fred79
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posted March 10, 2016 05:59 PM 


JollyJoker said: Tomorrow is 50/50. However, for the day after tomorrow to be your NEXT win, it's necessary that you get
a) Tail tomorrow (50/50) (so that you NOT win) AND
b) Head the day after tomorrow (again 50/50)
all in all only 25%, not 50%.
this part confuses me. why would you have to lose once, to win the next day? couldn't you just win or lose 3 days in a row; or win twice in a row and lose once, etc.?
JollyJoker said: EDIT: maybe it works better with Russian Roulette as an example. One bullet in a 6shoot drum, new turn after every shot. Which shot has the highest probability to kill you? Answer: THE NEXT ONE!
Right?
lol.


JollyJoker
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posted March 10, 2016 06:12 PM 


You have to look at the QUESTION!
Take the coin tosses. The question is in which toss you have the highest probability for getting THE NEXT Heads (not just Heads AT ALL). But the NEXT Heads.
1st toss, well, 50/50.
But 2nd toss NEXT, means, you cannot have Heads in the 1st toss, because in that case your NEXT Heads would come in the first toss.
And of course you still must get Heads in the 2nd toss and so on.
If you move further  how likely is it, you get your next Heads only in 5 or 6 or 7 tosses?
You can make it a rule of thumb: if you have a series of eventy with equal chances, your best chance is ALWAYS the next one (which is true for plane crashes and other things as well).
Actually, that makes a lot of sense, if you think about it.


fred79
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posted March 10, 2016 06:21 PM 


i'm trying, and i'm not getting anywhere. i've been up since midnight(with hardly any sleep before then, considering) and i'm tired though, so i'll come back and ponder after i've had some sleep(sometime). i don't want you blowing a circuit by attempting to help me when i'm not at my optimal functioning. thanks thus far, though.


Maurice
Hero of Order
Part of the furniture

posted March 10, 2016 06:41 PM 


It's quite simple, Fred, but let me try and cut it down to the essence a bit more.
For the NEXT payment to occur, you already don't consider today. Whether it happened or not, doesn't matter; it's about the next one.
So, assuming 50% chance each day to get paid, you also have a 50% chance not to get paid. So, for tomorrow, the chance to get paid is 50%. In short:
Chance to get next payment tomorrow: 50%
And hence, chance to not get paid tomorrow: 50%
But it might not happen tomorrow, there's a 50% chance it doesn't. Instead, it might be further into the future. So, let's say it's not tomorrow, but the day after tomorrow. For that to happen, we have pretty much the same requirements as above, with one caveat: if we're considering the NEXT payment from the here and now and we assume it's going to occur on the day after tomorrow, we immediately say that we won't get paid tomorrow. The chance to not get paid tomorrow, is 50%.
So we get combined odds. For the NEXT payment to be the day after tomorrow, we first need to not get paid tomorrow (50% chance on that) and then get paid on the next day (50%). You multiply those odds: 50% chance on a 50% chance to occur is 25%.
Now assume it's not the day after tomorrow, but the day after that. The odds of that being the NEXT payment is 50% (for not getting paid tomorrow) x 50% (for not getting paid the day after tomorrow) x 50% (for getting paid the day after that) = 12.5%
If you then make the list, you get the following odds of getting the NEXT payment from this day to be ...
... tomorrow = 50%
... the day after tomorrow = 25%
... the day after that = 12.5%
etc.
Of course each individual day has a 50% chance on payment, but we're not considering individual days; we're considering the next payment in a chain of days.


fred79
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posted March 10, 2016 07:09 PM 


*smacks forehead, after a handful of sentences in*
it just clicked. what the snow is wrong with me. i didn't pay attention to one word, and i couldn't assimilate because of it.
i'd get some sleep, but i have to stay up until tonight. i have things to do during the day, damnit. too tired to run the errands i need to, today.


Stevie
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posted March 11, 2016 12:24 AM 

Edited by Stevie at 00:25, 11 Mar 2016.

Corribus said: And Stevie, statistics does not work like that at all. There is not 100% probability that you will receive a gift every month  not getting a gift for 29 days does not ensure you will a gift on the 30th. The 1 gift per month is an average value.
Maybe you misunderstood. I was thinking on my own scenario drawing inspiration from yours. I was saying if the problem read like this: "A day once a month, the kid gets 10$. When does he have the highest chance of getting his buck?" Which I would say is the last day because the chance would be 100%, since you have the certainty that he would get the money at least once a month. A day earlier it would be 50%, right? But now think that the starting chance for him to get it in day 1 would be 3%, with each unlucky draw the chance of getting a lucky one would only increase in probability, right? And from then on I was thinking in terms of what is the likeliest day within that month to get the 10$, and it seemed to me that even if you were bound to get the money on the last day with 100% certainty, there would still be many days before when you could get lucky, so the probability would actually be low. I don't know if I'm making any sense, it might appear like my mind's flying all over the place. Never liked math anyway.
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RedSoxFan3
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posted March 17, 2016 06:02 AM 


Boy this thread got derailed pretty fast.
Let's get this thing back on track.
Here's an old math joke that I decided to turn into a riddle.
A group of Psychologists team up to do some a study on how people of various professions react to certain situations.
They put 3 people into this enormous gymnasium sized room.
At one end of the room they put 2 naked young women sitting on a bed who are ready for some adult fun. They are told that if they leave the bed the test will end immediately.
At the other end of the room they put an engineer and a mathematician. They are told that every minute they can move halfway towards the beautiful naked woman, but if they move a step further, the entire test is over and everyone will immediately be removed from the test.
The mathematician crosses his arms and shakes his head, but the engineer thinks for a minute and smiles. After a minute the person leading the test announces that they may move halfway towards the bed. The engineer moves forward, but the mathematician shakes his head and walks out of the room.
Why did the mathematician leave the room?
Why did the engineer decide it was worth staying?
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Elvin
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posted March 17, 2016 06:18 AM 

Edited by Elvin at 06:19, 17 Mar 2016.

The mathematician perceives halfway as a variable distance and he'd always have another half to cross. Unsure about the engineer.
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artu
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My BS sensor is tingling again

posted March 17, 2016 06:25 AM 


Yes, the mathematician is definitely Zenon. The engineer thinks he'll get close enough to make things work?
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RedSoxFan3
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posted March 17, 2016 06:54 AM 


x = 100a + 10b + c
x = a + b^2 + c^3
x = 100a + 10b + c
x = a + b^2 + c^3
100a + 10b + c = a + b^2 + c^3
99a = b^210b + c^3c
99a possibilities 99,198,297,396,495,694
c^3c possibilities 720,504,336,210,120
b^210b possibilities 25,24,21,16,9
Observations:
1.) The bgroup is by far the smallest number. Maximizing the Cgroup is what we are trying to do.
2.) We know that the Cgroup will be larger than the A group, because Bgroup is always negative.
3.) Best strategy is to find a Cgroup that is within 25 of an Agroup. Rest is trial and error.
720694 = 26 Nope
504495 = 9 This one works!
c = 8, a = 5, b = 9 or 1
This leads to 598 and 518.
5 + 1^2 + 8^3 = 5 + 1 + 512 = 518
5 + 9^2 + 8^3 = 5 + 81 + 512 = 598
This means the answer is 598.


RedSoxFan3
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posted March 17, 2016 06:58 AM 


artu said: Yes, the mathematician is definitely Zenon. The engineer thinks he'll get close enough to make things work?
Correct. The mathematician thinks that halfway will "never get there".
But the engineer knows that eventually you will be close enough for all practical purposes!
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Go Red Sox!


artu
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My BS sensor is tingling again

posted April 06, 2016 12:48 AM 


Okay, I just read this online, since I suck at math, I don't know if it's actually hard or easy but it sounds interesting, so here I go:
A man comes into a chicken store and tells the owner:
 Give me half the chicken in here and half a chicken.
The owner gives him the chickens and he leaves.
A woman comes in to the same store and says again:
 Give me half the chicken in here and half a chicken.
She also takes her food and leaves.
A third guy then comes in and asks for the same:
 Give me half the chicken in here and half a chicken.
And when the owner does that, there is no more chicken left in the store. So, how many chickens were there in the store, before the first guy went in?
____________
I admit it, I like it when they are bombastic  Neraus


fred79
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posted April 06, 2016 01:03 AM 

Edited by fred79 at 01:04, 06 Apr 2016.

the store started out with half a chicken in his store already; that much i know. or maybe not. because i suck at math too.
who the hell sells chicken halves, anyway?


Stevie
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posted April 06, 2016 01:07 AM 



Corribus
Hero of Order
The Abyss Staring Back at You

posted April 06, 2016 05:45 AM 


Concur.
____________
I'm sick of following my dreams. I'm just going to ask them where they're goin', and hook up with them later. Mitch Hedberg



