<p>well, i was recently going through the list of common misconception page in wikipedia, and realized this was listed. (misconception being 0.999...not =1 when in fact it is)</p>
<p>i knew this beforehand, but when i looked at its wiki discussion page, i was surprised what kind of intense argument was going on there w/ ppl who just could't accept the fact that .999... = 1 Talk:0.999.../Arguments</a> - Wikipedia, the free encyclopedia</p>
<p>i think everyone has heard of monty hall problem already, but here it goes:</p>
<p>there are 3 doors labeled A,B,C.
[A] ** [C]</p>
<p>one of the doors holds a luxury sports car.
two of the doors hold useless goats.
but you don’t know which door holds which.</p>
<p>after you pick one of the doors to open, (e.g. door A)
before you actually open the door,
out of the two doors left, I know which door holds which, and i will open a door that holds a goat. (e.g. let’s say i opened door B and show you a goat inside)</p>
<p>now, at this point, you have a choice:
either keep your original choice, (e.g. door A in our example above)
or change your decision. (.e.g. door C in our example above)</p>
<p>which is better?</p>
<p>==== </p>
<p>this is a popular problem and the answer is, changing the decision is better.</p>
<p>just wondering if anyone has problem w/ this either.</p>
<p>There are a variety of reasons why 0.9… does equal 1. Questions such as this are only mildly more interesting than questions such as “1 - 1 = 0?” and “2 / 3 = 4 / 6?”. There is a lot of room for disagreement in most things in life, but the degree to which these questions are debatable is almost negligibly small.</p>
<p>"Probability proves that .9999999->n = 1 "
I would be curious to see this proof, member. Are you talking about taking a limit?</p>
<p>As for the Monty Hall problem… while it may be less than intuitive, the proof is iron-clad and the result can be, and has been, experimentally verified (not that true statements need to be, but - for whatever reason - your garden variety proles prefer evidence to proof.)</p>
<p>I have a problem with the monty hall problem. My ap stats teacher in high school was telling the class this and I had trouble believing it. My thoughts on it are that when you choose a box, theres a 33% chance it has the car. Once you find out B has a goat in it, theres a 50% chance of A having the car and a 50% chance of C having the car. That simple. If that’s not true, whats the percentage of the car being in A and C. 60% and 40% or what. My stats teacher wasn’t able to explain why and just told us the problem cause he saw it on the internet.</p>
<p>and I don’t think .999=1. I’m pretty sure .999 = .999. So I guess I disagree with both of these problems. Maybe I’m just stupid</p>
<p>I too had a problem with the Monty Hall, but it was recently explained to me very well. The key to the problem is that the announcer KNOWS which one the car is in, so he isn’t randomly picking a door to open.</p>
<p>It is much easier to see if you imagine 1,000,000 doors with one car. You pick one, and the announcer then gets rid of 999,998 doors he knows do not have the car. Then he asks if you want to switch to the other door. If you switch in this scenario, you have a 999,999/1,000,000 chance of getting the car. The 1/1,000,000 chance of not getting it on the switch stems from the probability that you picked the car in the first place–very unlikely. The key is that it does not become 50/50 because the announcer knows which door holds the car. You picked a door, having a pretty much sure chance it was not going to be the car door…but then the announcer who knows which door has it gets rid of all the rest except one other one. Basically, he is telling you which door the car is in (with a slight chance that he is tricking you and you actually picked the right door to begin with).</p>
<p>norris, your insticts were correct. However, the key element of the Monty Hall problem is that the host opens a door with a goat behind it, NOT a random door. This detail is why the probability is not 50%.</p>
<p>Regarding your second paragraph, .999 obviously does not equal 1. But .99999… (repeating forever) does equal 1.</p>
<p>Thanks audiorage18 for that explanation. It’s easier to see with 1,000,000 doors. So would the probability that the car is in A be 33% and C 66% for the 3 door example</p>
<p>my thinking is that .99999…n approaches 1 but never quite gets there.</p>
<p>(1/3 is not .3333333333 it’s .3333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333…)</p>
<p>Same problem if you take a distance and move 1/2 way closer n times. You’re always just a bit away. </p>
<p>At some point it becomes meaningless - certainly so in the real world. I wonder if nano-technologists or quantum physicists would agree???</p>
<p>well for .999999999999 to not equal 1, .333333333333 would have to not equal 1/3. So I guess .333333333333 doesn’t = 1/3, it’s just the best way to describe what 1/3 is equal to. Haha this is confusing. I can see it either way.</p>
<p>It all depends on your understanding of infinity:</p>
<p>.99999… = 9/10 + 9/100 + 9/1000 + … + 9/10^n
= Σ[k=1 to ∞] of 9/10^n
= lim Σ[k=1 to n] of 9/10^n
n -> ∞
= lim 1 - 1/10^n
n -> ∞
= 1 - 1/∞
= 1 -0
= 1</p>
<p>This is valid if you accept that 1/∞ = 0, which is the real crux of this argument. If you use an extended number line, we can actually calculate limits at infinity and the above is correct.</p>
<p>^ …unless you use an extended number line, in which case the limit can be found. I made a typo above when I wrote “1 - 1/∞” instead of just going straight to “1 - 0” - that first step would occur as part of taking the limit, not afterwards.</p>
<p>Good catch.</p>
<p>EDIT: Let me put this another way. 0.999… can only exist as the sum of an infinite number of increasingly minuscule values; in other words, as a limit. You can’t accurately express 0.999… without the concept of a limit because it continues on to infinity.</p>
<p>I think that 1/3 doesn’t equal .3333333. When you go from fraction to decimal you lose a little bit of accuracy. So we assume it equals .33333333 since it’s so close, but really its not actually equal.</p>
<p>On an extended number line, 1/3 DOES equal 0.333…, better expressed as Σ[k=1 to ∞] of 3/10^n. 0.333… is not a value unless you allow limits at infinity to be evaluated (which is the whole point of an extended number line).</p>
<p>I have to get a copy of my prob text, il see if anybody has one. I remember my professor doing the poof, and it was quite long.</p>
<p>As for those who don’t understand the monty hall problem, im pretty sure you need to use multiple conditional probabilites like P(A|B) = P(AB)/P(B) or Bayes formula which I forget.</p>