Current Events > What is this maths paradox called?

(+) Yeah! (+)

Cant remember the exact premise so bear with me.

Basically you have a rocket taking people into space. There is only a 1% chance it will explode on take off.

First take off takes 1 person. Second 2 people. Third 4 people and so on.

Despite there only being a 1% chance of it blowing up your actual chances of dying due to it blowing up are something 99.999%.

Hows it work?

Bc after you land safely, you die of old age 60 years later

lilORANG posted...

Bc after you land safely, you die of old age 60 years later

Dying during the rocket exploding I should say.

That's two seperate things, one of them being conditional on the other. Not a paradox.

1. There's a 1% chance the rocket explodes.

2. In the event that the rocket explodes, there is a 99.999% that anyone inside dies.

Aressar posted...

That's two seperate things, one of them being conditional on the other. Not a paradox.

1. There's a 1% chance the rocket explodes.

2. In the event that the rocket explodes, there is a 99.999% that anyone inside dies.

Nah thats not what Im on about. Im on about even though there is only a 1% chance of someone dying die to the rocket exploding in reality the chances of any one person dying die to the rocket exploding is a lot higher.

SHRlKE posted...

Cant remember the exact premise so bear with me.

Basically you have a rocket taking people into space. There is only a 1% chance it will explode on take off.

First take off takes 1 person. Second 2 people. Third 4 people and so on.

Despite there only being a 1% chance of it blowing up your actual chances of dying due to it blowing up are something 99.999%.

Hows it work?

I may not be understanding this problem correctly but the chance is near 100 because even if it's .000001 that something MIGHT happen, if you repeat the same thing indefinitely it will eventually happen. Meaning, if you send the same person up 1 million times, and the odds are only 1/100 the person dies you will have died a bunch of times over 1 million launches.

Uh, does the explosion kill people outside the rocket?

That's not a paradox, just how probabilities work. 1% chance for each instance independently. Higher percent for it happening at least once in N number of trials.

WrkHrdPlayHrdr posted...

I may not be understanding this problem correctly but the chance is near 100 because even if it's .000001 that something MIGHT happen, if you repeat the same thing indefinitely it will eventually happen. Meaning, if you send the same person up 1 million times, and the odds are only 1/100 the person dies you will have died a bunch of times over 1 million launches.

nah its different people all the time

COVxy posted...

That's not a paradox, just how probabilities work. 1% chance for each instance independently. Higher percent for it happening at least once in N number of trials.

I dont get it could you explain it like Im five.

Dunno how it would ever get to 99.999%. Whenever the rocket explodes, a bit more than half the people who have rode it die

uwnim posted...

Dunno how it would ever get to 99.999%. Whenever the rocket explodes, a bit more than half the people who have rode it die

It might not have been that high maybe like 50% I dunno

If I'm understanding that correctly, the amount of people double every trip. That means eventually it will get to millions. I think the idea is that when that 1% happens, the rocket will be carrying millions or billions of people, and your more likely to be on that trip than the 2 person ship.

markconig posted...

If I'm understanding that correctly, the amount of people double every trip. That means eventually it will get to millions. I think the idea is that when that 1% happens, the rocket will be carrying millions or billions of people, and your more likely to be on that trip than the 2 person ship.

yeah it was something like that I think its actually a well known problem but I cant recall what. It may have been another thing not a rocket maybe a bomb or something.

This result is arrived at by combining the total number of survivors with the total number of fatalities. Let's say the rocket explodes on the 10th trial for example.

1+2+4+8+16+32+64+128+256 = 511 survivors vs 512 dead. Rocket is also exploded so it can't take off anymore.

Your individual odds of dying are basically 50% because you have roughly the same chance as being in the survivor group as you do the fatalities even though the rocket itself only has a 1% chance of exploding.

The numbers stay about the same regardless of how many trials you do before failure.

Maybe you are referring to balance between risk and number of people? Given that with each ride the number of individuals doubles at a certain point a 1% chance of a rocket crashing represents a mean loss of 20.48 people (in the case that 2^11 = 2048 people are on board) on each ride whereas with 1 person it represents a mean loss of 0.01 individuals every time you send a rocket into space.

Also reminds me of the martingale strategy if that was what you were referring to maybe, in which you double up on bets every time you lose because you wager that it is increasingly unlikely to say lose 8 bets in a row (of course if you do lose 8 bets you lose a crapton which is why I think mathematically it's been proven to not be viable).

Tyranthraxus posted...

Your individual odds of dying are basically 50%

I think this is the crux of the problem. I think you are wrong though and that actually my chance of dying is only 1%. This is different to the chances of me being part of the group that explodes.

I think its like the goat problem. The odds are different if you look at it before and after the explosion.

Well, there are two options
Either it explodes, or it doesn't

Therefore, it's 50/50

SHRlKE posted...

It might not have been that high maybe like 50% I dunno

This is correct, the limit the percentage approaches as the number of days goes to infinity is 50%, because the number of passengers doubles each day, and the number of survivors is the sum of all the people who rode the previous days.

So if it explodes the third day, 4 people die, 3 people survived the first two days (1+2), so your chance of having been one of the people who died is 4/7. On day five its 16/31. On day 12 its 2048/4095. On day n its (2^[n-1])/([2^n]-1).

The paradox is that since the odds of the of rocket exploding seem low, you would expect that the odds of dying in an explosion are low, but because the number of passengers doubles every day, the odds of the rocket exploding do not matter. The death rate will always be slightly over 50% as n approaches infinity.

SHRlKE posted...

Cant remember the exact premise so bear with me.

Basically you have a rocket taking people into space. There is only a 1% chance it will explode on take off.

First take off takes 1 person. Second 2 people. Third 4 people and so on.

Despite there only being a 1% chance of it blowing up your actual chances of dying due to it blowing up are something 99.999%.

Hows it work?

Because odds are you're gonna be on the rocket that contains 6 billion people.

If the first rocket starts with 2 people, there will be 30 rockets total, with the 30th rocket containing 96% of the entire population of all the rockets.

MrToothHasYou posted...

This is correct, the limit the percentage approaches as the number of days goes to infinity is 50%, because the number of passengers doubles each day, and the number of survivors is the sum of all the people who rode the previous days.

So if it explodes the third day, 4 people die, 3 people survived the first two days (1+2), so your chance of having been one of the people who died is 4/7. On day five its 16/31. On day 12 its 2048/4095. On day n its (2^[n-1])/([2^n]-1).

The paradox is that since the odds of the of rocket exploding seem low, you would expect that the odds of dying in an explosion are low, but because the number of passengers doubles every day, the odds of the rocket exploding do not matter. The death rate will always be slightly over 50% as n approaches infinity.

Thst makes sense in terms of the total number of people dying but if I were going to board that rocket its still only a 1% chance it will blow up with me on it. I think this is looking at the odds before and after its exploded. If Im sitting here now and about to join the rocket its still only a 1% chance it will exploded.

SHRlKE posted...

I think this is the crux of the problem. I think you are wrong though and that actually my chance of dying is only 1%. This is different to the chances of me being part of the group that explodes.

I think its like the goat problem. The odds are different if you look at it before and after the explosion.

It seems like the key is that the group of dead will always be greater than the group of survivors because 2^n = 2^n-1 + ... + 2^0 + 1
When the rocket explodes, the number of casualties is always 1 more than the number of survivors who came before, so the odds are roughly equal that you'll survive or die (50%).

EmbraceOfDeath posted...

It seems like the key is that the group of dead will always be greater than the group of survivors because 2^n = 2^n-1 + ... + 2^0 + 1
When the rocket explodes, the number of casualties is always 1 more than the number of survivors who came before, so the odds are roughly equal that you'll survive or die (50%).

Each shuttle launch is independent of all the others.

Take the nightingale thing as an example and swap out the money for people. When I finally lose then the chance of any 1 money being lost is with the last bet. But each bet still has its own odds. Lets assume you are just betting on red or black its still 50/50.

Say they were going to keep putting people into the rocket and launching it until it explodes. If you were to wake up and find yourself in the rocket, it would be slightly more likely that you are in the rocket launch that explodes than the one that doesnt. Over half the people who ever ride the rocket will be on the final one, and with no way to place yourself at any particular moment youd assign slightly more than 50% chance of dying.

uwnim posted...

Say they were going to keep putting people into the rocket and launching it until it explodes. If you were to wake up and find yourself in the rocket, it would be slightly more likely that you are in the rocket launch that explodes than the one that doesnt. Over half the people who ever ride the rocket will be on the final one, and with no way to place yourself at any particular moment youd assign slightly more than 50% chance of dying.

you would have no way of knowing if you were on the last rocket or not though

SHRlKE posted...

you would have no way of knowing if you were on the last rocket or not though

Yes. Thats the point. If you knew you were on the last rocket it would be 100%, not about 50%. In that situation, the relevant probability isnt the chance any particular launch explodes, but the probability that you are on the one that does explode.

Why would it be 50% if you didnt know it was the last one? Its no different picking one ball out of one hundred out of a bag or flipping a coin. Its still a 50/50 chance per individual flip or 1/100 in the case of a launch.

If I were to step foot on a rocket it would be a 1 in 100 chances providing I didnt know any other information.

SHRlKE posted...

Why would it be 50% if you didnt know it was the last one? Its no different picking one ball out of one hundred out of a bag or flipping a coin. Its still a 50/50 chance per individual flip or 1/100 in the case of a launch.

If I were to step foot on a rocket it would be a 1 in 100 chances providing I didnt know any other information.

The rocket will eventually blow up. It's not a matter of if, just when.

And when it blows up, how likely are you to be on the rocket?

SHRlKE posted...

Thst makes sense in terms of the total number of people dying but if I were going to board that rocket its still only a 1% chance it will blow up with me on it. I think this is looking at the odds before and after its exploded. If Im sitting here now and about to join the rocket its still only a 1% chance it will exploded.

No it's not. The way probabilities work is that the closer you come to infinity, the more likely that the rocket will explode. You have a one out of one hundred chance of dying statistically, however probability wise you will eventually die when the rocket explodes, assuming you keep repeating launches.

You have to look at more than just the statistics when trying to find the probability of the even actually occurring.

itachi15243 posted...

No it's not. The way probabilities work is that the closer you come to infinity, the more likely that the rocket will explode. You have a one out of one hundred chance of dying statistically, however probability wise you will eventually die when the rocket explodes, assuming you keep repeating launches.

You have to look at more than just the statistics when trying to find the probability of the even actually occurring.

as I said below each person only takes one trip

SHRlKE posted...

Why would it be 50% if you didnt know it was the last one? Its no different picking one ball out of one hundred out of a bag or flipping a coin. Its still a 50/50 chance per individual flip or 1/100 in the case of a launch.

If I were to step foot on a rocket it would be a 1 in 100 chances providing I didnt know any other information.

It is more like rolling a die. Like say you had a fair 7 sided die, if you roll a 1, 2 or 3 you live. But if you roll a 4, 5, 6 or 7 you are dead. All of the numbers are equally likely, but your chance of dying is higher than your chance of living. In this situation you can consider it to be rolling a die with an unknown number of sides with a rule that half(rounded up) of the results result in death.

uwnim posted...

It is more like rolling a die. Like say you had a fair 7 sided die, if you roll a 1, 2 or 3 you live. But if you roll a 4, 5, 6 or 7 you are dead. All of the numbers are equally likely, but your chance of dying is higher than your chance of living. In this situation you can consider it to be rolling a die with an unknown number of sides with a rule that half(rounded up) of the results result in death.

I dont get it. From my logic its no different to this

If I flip a coin the probability of it landing on a heads is the same regardless of how many times I roll it or how much money I have bet on it or regardless of how many times Ive previously guessed correctly previous.

Ok, so lets say after the rocket finally explodes, you put the names of everyone who has ever rode it in a box and pull out a name. What is the chance you picked the name of someone who died?

This is a variation on the Monte Hall problem. And a very poorly demonstrated one.

uwnim posted...

Ok, so lets say after the rocket finally explodes, you put the names of everyone who has ever rode it in a box and pull out a name. What is the chance you picked the name of someone who died?

Pretty damn high. 50% +1.

But this isnt what Im on about. Im on about if I were to get in a rocket right now without knowing anything of all the rockets before or any of the rockets to come after to me it would be 1%.

From my perspective. I agree that if you were to review it afterwards it would be 50%+1.

The odds of me getting heads 8 times in a row is 2^8 but the odds of any one flip is still 50/50.

warlock7735 posted...

This is a variation on the Monte Hall problem. And a very poorly demonstrated one.

Yeah I did mention that above. I have heard this before but I couldnt remember the specifics. If you read the first post I made I was trying to find out if it had a proper name.

Ok. So you get that.

So being on the ticket with no info about possible past or future flights is basically the same thing. Replace the names with your seat/slot whatever you want to call it.

If the launches all has the same number of people, then your odds would be 1% to die because the odds of being in the last flight are the same as the odds of being on say the first. But because the passengers double each time, it is more likely you are on the last flight than it is that you are on any previous flights. You can consider yourself to be a random member of everyone who rides the rocket and as you agreed, a random person is going to be on the last rocket over half the time.

SHRlKE posted...

Pretty damn high. 50% +1.

But this isnt what Im on about. Im on about if I were to get in a rocket right now without knowing anything of all the rockets before or any of the rockets to come after to me it would be 1%.

From my perspective. I agree that if you were to review it afterwards it would be 50%+1.

The odds of me getting heads 8 times in a row is 2^8 but the odds of any one flip is still 50/50.

Yes but you're not flipping 8 times. You're flipping infinite times and stopping as soon as you get your first tails. Your chances of getting 100% heads are 0 where as if you're only flipping 8 times, your chances of getting 100% heads is small, but not 0.

Right, I'm going to answer you OP, because no one else has.

It's not a Paradox, it's a "Fallacy", and it's called the "Gamblers Fallacy":
https://en.wikipedia.org/wiki/Gambler%27s_fallacy
"The gambler's fallacy, also known as the Monte Carlo fallacy or the fallacy of the maturity of chances, is the incorrect belief that, if a particular event occurs more frequently than normal during the past, it is less likely to happen in the future (or vice versa), when it has otherwise been established that the probability of such events does not depend on what has happened in the past. Such events, having the quality of historical independence, are referred to as statistically independent. The fallacy is commonly associated with gambling, where it may be believed, for example, that the next dice roll is more than usually likely to be six because there have recently been fewer than the usual number of sixes."

coolguyjimmy posted...

Right, I'm going to answer you OP, because no one else has.

It's not a Paradox, it's a "Fallacy", and it's called the "Gamblers Fallacy":
https://en.wikipedia.org/wiki/Gambler%27s_fallacy
"The gambler's fallacy, also known as the Monte Carlo fallacy or the fallacy of the maturity of chances, is the incorrect belief that, if a particular event occurs more frequently than normal during the past, it is less likely to happen in the future (or vice versa), when it has otherwise been established that the probability of such events does not depend on what has happened in the past. Such events, having the quality of historical independence, are referred to as statistically independent. The fallacy is commonly associated with gambling, where it may be believed, for example, that the next dice roll is more than usually likely to be six because there have recently been fewer than the usual number of sixes."

That isnt really what it is. That would be thinking that if the rocket has launched successfully like 150 times, then it is more likely to fail next time. This topic though relies on the passengers doubling each launch which causes the death rate to be much higher than the rockets failure rate.