1/7000000000000 or 50/50

[Ruwrak]

Okay I've been trying to wrap my head around this and cannot make exact logical sense in it. If I were to push a button and the result of doing so shall randomly kill a person in the world what are the chances of it being my best friend ?

My argument is that it would 50/50 of it being my friend since the process is random and no one is favoured in anyway so it is either it is him or it is not him.

2 friends of mine instead say that the chances are 1 to 7 billion (assuming there are still 7 billion people on earth) that someone I know shall get axed from the random process.

I ask you escapist which one is it and why ?

Depends is on how you ask the question

What is the chance I kill my friend? ( 1 / 7 billion, as he is 1 person on a scale of 7 billion)
Does this button kill my friend? (50/50 as it is a 'it does' or a 'it does not')

Depends on how you ask the question. In a way you are all three correct.

 

[WolfThomas]

OP 1/7Billion

Monty Hall Problem time:
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1 [but the door is not opened], and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

I'd say no. It's 50/50 whether or not you're going to get a car or a goat now. Unless I'm missing something...

Spoiler: You are indeed missing something

When you first picked a door it was more likely that you picked a goat door, there being two goats, one car. By eliminating a goat, the remaining unchosen door is more likely to contain the car. See the example below:

 

[Caravelle]

I know that chance is actually really low but normally chance could be proven only when tested. If you would have done it multiple times and the chance would be proven, you would not have a problem pressing the button again however mental state of person who is going to press this button for a first time would be that anything can happen and subjectively for him/her it would be 50/50 . I know we are talking about two different things and and that 50% isnt chance but the thing is if someone told you they will give you 1000? if you press a button that has 0.0001% chance to kill you or do nothing, would you press it?

How people deal with risk is a completely different issue, which is only remotely related to the actual probabilities involved.
I wouldn't press the button in the situation you describe, but it's not because the probability is 50/50 in my mind. It's because dying is too bad an outcome and wtf would I do with 1000 dollars ? Lower that probability a few zeros and make it a billion dollars and you might see me turn around.

And then there is the fact that people are very bad at distinguishing between very low probabilities, tend to have biased risk-averse decision-making and do not evaluate the quality of outcomes on a simple money-based scale...

 

[drosalion]

You see there being 2 different variables:
1. Your friend dies
2. Your friend lives

And in your example thats obviously correct as your only concerned with your friend.

You, however, have taken it 1 step further and implied that both variables have an equal chance of occuring (hence 50/50) which is simply incorrect.

 

[Flare Phoenix]

OP 1/7Billion

Monty Hall Problem time:
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1 [but the door is not opened], and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

I'd say no. It's 50/50 whether or not you're going to get a car or a goat now. Unless I'm missing something...

Spoiler: You are indeed missing something

When you first picked a door it was more likely that you picked a goat door, there being two goats, one car. By eliminating a goat, the remaining unchosen door is more likely to contain the car. See the example below:

Ah I see. I didn't think of it like that.

As for the original topic, I really don't see how you came to conclusion that it would be 50/50 if everyone has an equal chance of getting killed. Everyone else has already made all the arguments I can think of, so I won't bother repeating them though.

 

[Major_Tom]

You are using long scale, in that case the number of people isn't 7 billion (7x10[sup]12[/sup]) but 7 milliard or thousand millions (7x10[sup]9[/sup]). If you are using short scale, that number you wrote is 7 trillion (7x10[sup]12[/sup]) not 7 billion (7x10[sup]9[/sup]).

OT: It's 1/7x10[sup]9[/sup].

 

[Mr Pantomime]

If the button algorithm has 7 billion different possible outcomes, and your friend is merely 1 of those, then its 1 in 7 billion.

If it only has two outcomes, one that kills your friend, and one that kills a random person, its 50/50

 

[PH3NOmenon]

OP 1/7Billion

Monty Hall Problem time:
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1 [but the door is not opened], and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

I'd say no. It's 50/50 whether or not you're going to get a car or a goat now. Unless I'm missing something...

Actually, this is a problem that gets explained in most courses on probability. It actually is in your best interest to switch your choice.

Offhand the best explanation I could give this is that when you made your initial pick, you had a 2/3 chance of picking a goat. Given that you most likely chose a goat on your first pick and that there's only two choices left, it's in your best interest to switch your pick, as it'll increase the likelyhood of getting the car.

@OP When asking these kinds of things, always make sure of what exactly it is what you're asking.

The fact that your friend would either die or not die and you have two possible outcomes doesn't automatically imply that each of these outcomes is equally likely: i.e. it's not a 50/50 deal. In this case, you're friend will die in 1/7 billion scenarios. Hence why the chance of your friend keeling over is 1 in 7 biollion.

 

[Jimbo1212]

I agree. If you did it 1000 times and your best friend would be still alive then the truth would be noticable howover when you do it just once you have no idea what might happen because there always is chance that it would be him so it would be 50/50.

So so wrong.

It is not him and someone else, but him and 7 billion other people, each just as likely as each other to die making it 1:7000000000

All the maths fail in this thread makes me want to cry =(

 

[Caravelle]

The "either you do or you don't = 50/50" argument is actually stupid enough to be rightfully called retarded, and you need to be embarrassed that you would even consider it being even remotely correct.
I'm serious, this is shit a 6 year old can figure out.

... I've taught math to 6 year olds. They really, really can't.
Cut him some slack, probabilities are something human brains aren't naturally good at. They're better with frequencies apparently, which is why it's sometimes helpful to recast the question "what are the odds of X in situation Y ?" as "out of a million cases of Y, how often does X happen ?".

 

[Teh Jammah]

Okay I've been trying to wrap my head around this and cannot make exact logical sense in it. If I were to push a button and the result of doing so shall randomly kill a person in the world what are the chances of it being my best friend ?

My argument is that it would 50/50 of it being my friend since the process is random and no one is favoured in anyway so it is either it is him or it is not him.

2 friends of mine instead say that the chances are 1 to 7 billion (assuming there are still 7 billion people on earth) that someone I know shall get axed from the random process.

I ask you escapist which one is it and why ?

You assume only 2 choices - him or not him. But according to your wording there would actually be 3 potential outcomes

1) Kills Best Friend
2) Kills You (It says would randomly kill a person - doesn't say excluding the button presser)
3) Kills someone else.

The odds however are the 1 in x billion, which are the same odds as you killing yourself or any other specific individual.

 

[Feylynn]

I think 1/7b personally.
If you invented a button that had a chance to randomly kill 'anyone' it would be 50/50.
But if this button is constrained to killing only one then the options are not binary: He dies/He does not die.

The options are: [Insert list of every name of every living being in specified Death Button effect radius]

So proper theory but you are examining the wrong result pool if the button can only kill one.

 

[WolfThomas]

Ah I see. I didn't think of it like that.

As for the original topic, I really don't see how you came to conclusion that it would be 50/50 if everyone has an equal chance of getting killed. Everyone else has already made all the arguments I can think of, so I won't bother repeating them though.

It's a good example of how even the most obvious probability question can be confusing.

 

[willsham45]

There is a 1 in 7 billion chance if it will kill anyone in the world.
Just like there is a 1 in 60 million I will be the next person to die in the UK

 

[Firetaffer]

I don't see how this thread is still going.

If you roll a die, the odds of getting a 4 is 1/6, not 50/50.

When I say the odds are 1/6, I mean that if you roll it 6 times you should get a 4 once.
By saying your friend has a 50/50 chance means that if you pushed the button twice the most likely scenario is that your friend would die one of those two times; which is NOT true.

 

[NickCaligo42]

1/7 billion, unless someone rigged the machine so that half of the time it picks your best friend and half the time it picks somebody else.

 

[Abengoshis]

There is a 1 in 7 billion chance he will be killed unless your friend represents 3.5 billion people, if so he is clinically obese.

 

[Sniperyeti]

If it chooses randomly from a pool of 'the population of earth', the chance will be
1/Earth pop. = 1/7,000,000,000

If it choose specifically from a set of either 'your friend', or 'another', the common denominator is 2 because it is choosing randomly between two sets. But that is not the situation, you said "anyone in the world". Just because the situaion is 'your friend' or 'not your friend' doesn't mean its a 50/50 chance.
You 'may' be assassinated by the Swedish bowling team's collective malice personified into a grapefruit wielding hitman before you finish reading this, or you may not. I would hope that doesn't mean you only have a 50% chance of making it to the end of this post.

 

[honestdiscussioner]

Really? Okay . . . it isn't 50\50. It's 1\7 billion. Is this serious here?

"It's him or it's not him" is inaccurate. It's him, or someone else, or someone else, or someone else.