An infinite debate...?

Question

So I got into a debate with a friend and we had a concept we couldn't agree on...

He says if you try to choose something randomely an infinite number of times you will eventully choose the 'right one'. I gave him the following counter arguement which I believe proves him wrong, but he still seems to think it does not solve the problem...

If I was given 2 pieces of paper numbered 1 and 2 on the reverse side so i could not see the numbers and attempted to randomely select the number 2, would I, given an infinite number of tries eventually pick the number 2?

My friend says I would definitely pick 2 eventually. I said it is possible to select the number 1 an infinate number of times and explained it like this...if I could see the numbers I would be able to pick 1 an infinite number of times, therefore when not visible there is still a combination to select 1 an infinite number of times also even though it is highly unlikely.

Can you infinitely get this choice correct?

A few of you seem to be saying the same as he did but with no explanation, my explanation means the experiment would continue forever(infinitely) as you would infinitely get it correct. Why at some point would you have to get the correct number instead of infinitely getting it wrong?

Answer 1

between two numbers you definately would get both numbers ,if you didn't you would just be extremely unlucky.
more numbers would bring the odds up.

Answer 2

I think you are right and wrong.

Each time you draw a number, it is independent of the last event.

Each time, out of drawing 1 or 2, the probability would be 1/2.

So drawing a 1 and then a 2: P = .5*.5 = 1/4
Drawing a 1 and then a 1: P = .5 * .5 = 1/4

The probabilities are the same. You have just as much chance to draw 1 an infinite number of times as you do drawing a 1 an infinite number of times with a 2 in there at a certain position.

If instead, you are wondering the probability of getting a 2 at any position, then it is a higher probability.

To get an infinite number of 1's would be:

limit (1/2)^x as x approaches infinity is 0. So theoretically, not possible.

If instead, you draw an infinite number of times, and your only wondering the probability of getting a 2 somewhere, then there are an infinite number of ways to get a 2, so we would have to add the probabilities together. (Or since all have the same probability, multiply by the number of times we draw a number..) This becomes:

limit x*(.5)^x as x approaches infinity.. After doing L'Hopital's rule, I found this limit to be 0 as well.

My conclusion: Since, when doing an experiment an inifinite amount of times, there are an infinite amount of possibilties, the probability of ANY of those possibilities is zero. Although, this does not jive with common sense. But----as we cannot do any experiment an infinite number of times--it kinda makes sense.


P.S. Thanks for the interesting question.. I'm getting tired of all these homework questions.

Answer 3

Hmm. I guess it isn't possible to do it infinity times because you wouldn't finish it before you are deceased (dead). I tried this expiriment on my TI-83 Plus calculator with a program I called RANDOM. I modified it to give only the numbers 1 and 2. On a number of tries I counted the amount of 1's I get. I got 2/2 ones, then 3/3, 3/4, 3/5, 3/6, 4/7, 5/8, 6/9, 6/10, 7/11, 7/12, 7/13, 8/14, 8/15, 9/16, 9/17, 10/18, 10/19, 10/20, 10/21, 10/22, 10/23, 10/24, 11/25. I stopped there and, unfortunately, your friend was right, and not you.

Answer 4

Mathematically, the probability of picking the same number n
times is:

P[n] = 1/(2^(n-1))

So for n = infinity, P[n] = 0

However, there is never any reason why the next number should be differant from the last one so there must be a possiblity of picking the same number indefinitly.

I think this is a case of the arithmetic of probability just not working when infinity is involved so I agree with you, but the only way to be sure is to try it! Let me know when you are finished.

If the probability of always picking the same mumber is 0, that means the probability of picking a card different from the last one at some point must be 1. At what point could this happen??

Answer 5

In every textbook about probability, once you do an experiment for an infinite number of times, you will realise any theoretical probability that you have came up with will become more and more obvious in your results.

e.g. your 1 or 2 thing.
The first time you picked it it may be 1. The second time you picked it it may be 1 again. The third time may be 2. Fourth 1. Fifth 2, etc...
Eventually you will notice that you will pick 1 half the time and 2 for the other half of the time

I think your mate has understood the idea in the wrong way.

Answer 6

well, u r wrong.
it is not possible to keep getting either 1 or 2 infinitely.
the probability of getting atleast one 1 and rest 2 or vice versa is not zero although it is very very less.
the greatest probability is of getting equal no. of 1's and 2's.
it means it is impssible to always get a 1 or a 2 becauz if u select a no. n times, the probability of getting a 1 or 2, throughout is 1/(2^n) which approaches 0 as n goes till infinity.
so u will not select just 1 no. all the time, u have to select another one as well.

the probability of getting a single 2 and rest 1 will be,

nC1 * 1/(2^n) n→∞
similarily as u go on till an equal no. of 1's and 2's
the probability will increase upto

nC(n/2) * 1/(2^n)
and as the nCr term →∞ as n→∞ and also 1/(2^n)→0
the product will be definite one.
but in case of 1/(2^n), that is keeping selecting just a single no.,
the probability is 0.

if you don't know about nCr and nPr(combinations and permutations), then u cannot understand this one.

do you know that keeping selecting a single no. in 100 tries out of 1 and 2 has a probability of
1 in 1.2676*10^30

and that of selecting 50 1's and 50 2's
is 1 in 12.56

sorry, but your friend was right.

Answer 7

Well, I think you are right. You can choose "1" forever. Although it wouldn't be that random.

A computer selecting a number randomly would certainly choose 2 at least once. But that's about as certain as the sun rising tomorrow.

Right now, someone could invent a device that would keep us warm and give us light. And it could replace the sun, so we wouldn't be sucked into a black hole. Then they would get rid of the sun.

I suppose the computer could choose it randomly and get one.

"proceeding, made, or occurring without definite aim, reason, or pattern" is dictionary.com's definition of random.

It's about as likely as plucking the number 1 eternally from a hat with both 1 and 2, but it could happen.

Extremely, extremely, EXTREMELY rare, almost to the point of impossibility, but not quite.

There is half a chance, and each time you get that.

If you infinitely tossed a penny, you COULD get heads everytime.

Answer 8

i think it all comes down to how you define infinity
i'm no math genius, but i would argue

1 minus (.5) ^ infinity power = x, which stands for the chance of you picking the second piece paper

since x can never ever be negative (domain x>0, for whole numbers), there automatically exists an absolute possibility that 2 would ultimately be chosen, given an infinite number of tries

Answer 9

Your pal is right. Sorry. Because if you picked the number 2 an infinite number of times, the number 1 could still come up next! That's why it's called infinity!

Answer 10

is it possible, to pick 1 over and over? sure.
is it likely? probably not.

you have a (1/2) chance of picking 1 (or not picking a two) the first time
(1/4) chance of picking it two times in a row (11,12,22,21)
1/8 of picking it three times in a row
1/2^n for picking it n times in a row... you're letting n approach infinity

lim n->inf (1/2^n) = 0

as you approach infinity you're approaching a 0% chance of not picking a 2, however you're only approaching.

So the chance of not picking a 2 is possible, but extremely unlikely. This means the chance of picking a 2 is extremely likely, but not definite, therefore you're correct, if only for the fact that your friend used the term definitely, when talking about random things

Answer 11

Hmmm.

Think about one of those fraction tree diagrams to show all of the outcomes of something:

_______________________ start
_____________________/_______\__
_______________ pick 1________pick 2
______________ /_______\______/_____\
___________ pick 1____pick 2__pick 1___pick 2

and so on, until infinity.

So, if you follow the left hand branch, with an outcome of one every time, the probability for that is a half multiplied by a half multiplied by a half.......

Now, by my reckoning, if you multiply by a fraction over and over again the answer gets smaller and smaller and smaller, infinitely small you could say. Over time therefore, the chance of repeatedly picking a 1 tends to zero. And the chance of picking a 2 tends to certainty.

And that's what would happen at infinity. The chance of picking a one becomes zero.

Sorry!

Mind you, probability was never really my thing and part of me wants to argue that there's an infinite number of combinations and one of them must be repeatedly choosing one. All my maths lessons/lectures taught me was that infinity is a troublesome so-and-so, defying my best logic at times.