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Topic: Probability of a Negative Number (Read 707 times) |
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Albert
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What is the probability of choosing a negative number if you are randomly choosing a number from the set of all integers.
I do not have an answer to this problem. I have come up with two completely different answers, but I cannot decide what is wrong with the logic of either of them.
attempt 1
The probability is (approaches?) 1/2. There are an equal number of positive and negative numbers, so there is an equal probability of choosing either a positive or negative number.
attempt 2
There are an infinite number of negative integers. And there are an infinite number of integers. Therefore the probability = infinity/infinity
i do not have experience dealing with infinities so i do not know what to do. but if i had to, i would guess it equals one?
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towr
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Re: Probability of a Negative Number
« Reply #1 on: Jun 26th, 2003, 2:30am » |
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If you use a uniform distribution I would say 1/2 = lim n->inf n/(2n+1)
any other probability is an indication of a non-uniform (skewed) distribution.
Also every integer -n must have the same probability as integer n. On the other hand when you're dealing with an infinite set all probabilities of getting a certain result are infinitesemal (= pretty much 0). So it's hard to define a uniform distribution in this case.
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TenaliRaman
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Re: Probability of a Negative Number
« Reply #2 on: Jun 26th, 2003, 5:08am » |
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Can't we simply say,
set of postive integers Pi is countable.
set of negative integers Ni is countable.
so let n(Pi)=n(Ni)=c (the cardinality of these sets)
So the probability of choosing a negative integer is c/2c = 1/2
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Sir Col
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Re: Probability of a Negative Number
« Reply #3 on: Jun 26th, 2003, 10:29am » |
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I don't think you can say that there are twice as many integers as there are negative numbers, but I think the following approach, similar to Towr's argument, works:
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Assume that -0 belongs to the set of negative integers, {-0,-1,-2,-3,...}, and +0 belongs to the set of positive integers, {+0,+1,+2,...}. By symmetry there exists a negative integer for every positive integer, so it follows that you are equally likely to select n or -n. Therefore, P(selecting a negative integer)=1/2.
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I think a similarly interesting problem would be to find the probability of selecting an odd number from the set of natural numbers.
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| « Last Edit: Jun 26th, 2003, 10:33am by Sir Col » |
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TenaliRaman
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Re: Probability of a Negative Number
« Reply #4 on: Jun 26th, 2003, 11:18am » |
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Well,
if total number of integers = number of positive integers + number of negative integers ,then we can argue that there are twice the number of integers as there are negative numbers.
My arguement " there exists a explicit bijection between positive and negative numbers , a simple function like y=-x does the trick." And since both are countable we can have their cardinalities equal.
Isn't this a valid argument??
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wowbagger
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Re: Probability of a Negative Number
« Reply #5 on: Jun 26th, 2003, 11:43am » |
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on Jun 26th, 2003, 11:18am, TenaliRaman wrote:| if total number of integers = number of positive integers + number of negative integers ,then we can argue that there are twice the number of integers as there are negative numbers. |
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We could if we were dealing with finite numbers. As is well known, there is a (countable) infinity of positive and negative integers. The number of positive integers equals the number of all integers, Aleph0.
Quote: My arguement " there exists a explicit bijection between positive and negative numbers , a simple function like y=-x does the trick." And since both are countable we can have their cardinalities equal.
Isn't this a valid argument??
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I'd say it is (if you don't worry about zero).
Of course, there is also a bijection between the positive integers (0,1,2,...) and all integers, e.g. (square brackets denote integer part):
(-1)n[(n+1)/2]
So there are as many integers as positive integers.
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| « Last Edit: Jun 26th, 2003, 11:45am by wowbagger » |
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TenaliRaman
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Re: Probability of a Negative Number
« Reply #6 on: Jun 26th, 2003, 11:56am » |
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Hmm i see,
Thanks for clearing that up!
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Sir Col
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Re: Probability of a Negative Number
« Reply #7 on: Jun 26th, 2003, 12:14pm » |
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on Jun 26th, 2003, 11:43am, wowbagger wrote:Of course, there is also a bijection between the positive integers (0,1,2,...) and all integers, e.g. (square brackets denote integer part):
(-1)n[(n+1)/2]
So there are as many integers as positive integers. |
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This was my objection to using a counting argument to the original question and precisely why I asked the new question, "Find the probability of selecting an odd number from the set of natural numbers."
I think there are two equally compelling arguments:
1. For every natural number, n, there exists an odd number, 2n–1. As there are infinitely many odd numbers as natural numbers, P(select odd from natural)=1.
2. For each even number, 2n, there exists an odd number, 2n–1. As the set of natural numbers are exclusively made up of odd and even numbers and there are equally many odd and even numbers, it follows that P(odd)=P(even) and as P(odd)+P(even)=1, P(odd)=1/2.
Do all these arguments fail, because probability sampling cannot be applied to infinite sets, or am I missing something?
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| « Last Edit: Jun 26th, 2003, 12:16pm by Sir Col » |
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TenaliRaman
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Re: Probability of a Negative Number
« Reply #8 on: Jun 26th, 2003, 12:42pm » |
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Sir Col,
you have made some very good points..
i once had a chat with a PhD student(doing his PhD in algebraic topology and measures).He said to me once the normal probability rules that one applies assumes that the all events are equally likely but this may not be the case everywhere.(i think i now understand the gist of towr's post).
Now he also gave some technique of mapping the entire sample set we are working in the region (0,1) such that the distribution is uniform and then we may proceed with the application of our probability principles.....
(if i am wrong its prolly my mistake,i am just putting whatever i could understand from the student and he is not to blame for my mistakes)
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Icarus
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Re: Probability of a Negative Number
« Reply #9 on: Jun 26th, 2003, 6:54pm » |
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The first and most accurate answer to the question is "there is no such probability, because you CAN'T 'randomly choose' an integer from [bbz], the set of all integers."
[bbz] does not admit a uniform distribution (one where the probability of picking a number from a set is proportional to the size of the set) - the common interpretation of the phrase "randomly choose". There are several other puzzles in this forum that are based on this fact.
So before you can calculate a probability, you have to specify more exactly how the choosing is to be done.
IF the method used is independent of sign of the number, (so that the probability of picking a number < 0 is the same as picking a number > 0), and IF the probability of getting 0 is zero, then a simple calculation shows that P(x<0) = 1/2:
P(x<0) + P(x=0) + P(x>0) = 1
P(x<0) = P(x>0) and P(x=0) = 0, so
2P(x<0) = 1, and P(x<0) = 1/2
Another problem with "attempt 2" is that you cannot evaluate [infty]/[infty]. This expression cannot be given a consistent meaning. For example:
[infty]/[infty] = limitx[to][infty] x/x = 1
[infty]/[infty] = limitx[to][infty] (2x)/x = 2
[infty]/[infty] = limitx[to][infty] (x2)/x = [infty]
[infty]/[infty] = limitx[to][infty] x/(x2) = 0
So your calculation in attempt 2 can tell you nothing.
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| « Last Edit: Aug 17th, 2003, 9:15pm by Icarus » |
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towr
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Re: Probability of a Negative Number
« Reply #10 on: Jun 27th, 2003, 1:40am » |
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on Jun 26th, 2003, 12:14pm, Sir Col wrote:
2. For each even number, 2n, there exists an odd number, 2n–1. As the set of natural numbers are exclusively made up of odd and even numbers and there are equally many odd and even numbers, it follows that P(odd)=P(even) and as P(odd)+P(even)=1, P(odd)=1/2. |
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But also for each even number 2n there exist TWO odd numbers 4n-3 and 4n-1, so the amount of odd numbers is now twice that of even numbers and the chance would now be P(odd) = 2/3
There are also 4 odd numbers for each even number 2n, namely 8n-7, 8n-5 ,8n-3 and 8n-1. So now P(odd) = 4/5
Since there are an infinite amount of these kinds of mappings this approach won't work.
On the other hand you could count between -n and n, and take the limit for n to infinity. In which case you find odd numbers make up half of the set.
You could also look at what properties a uniform distributuion should have. You may not get all properties on any infinite set, but you may get some. For instance the chance of odd/even quite reasonably should be equal for a uniform distribution, the same goes for positive and negative.
It's also easy to make sure you get this result, by flipping a coin for odd/even and one for positive/negative, then choose a number n from the natural numbers. And combine the results to get your random number
sign*(2*n+odd).
The only problem is you have a +0 and -0, but on an infinite set chances of getting 0 is 0.
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| « Last Edit: Jun 27th, 2003, 1:41am by towr » |
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Sir Col
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Re: Probability of a Negative Number
« Reply #11 on: Jun 27th, 2003, 9:05am » |
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on Jun 27th, 2003, 1:40am, towr wrote:
But also for each even number 2n there exist TWO odd numbers 4n-3 and 4n-1, so the amount of odd numbers is now twice that of even numbers and the chance would now be P(odd) = 2/3
There are also 4 odd numbers for each even number 2n, namely 8n-7, 8n-5 ,8n-3 and 8n-1. So now P(odd) = 4/5
Since there are an infinite amount of these kinds of mappings this approach won't work. |
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Not quite, the method I used ensures a complete and unique one-to-one mapping. For each (and every) even number, 2n, there exists a unique odd number, 2n–1. What you're really saying is that for every even number, 4n, there exists two odd numbers, 4n–3 and 4n–1. I would argue that to form a complete set, you would need to map 4n–2 to 4n–3 and 4n to 4n–1, otherwise you're missing every other even number. For n=k, I have formed the complete set of natural numbers from 1 to 2k, whereas you've formed the complete set of odd numbers and only half the even numbers.
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