Don't think too quickly

Okay you bit twiddler...

I have a random number between one and three, inclusive. What's the chance that the number is a three?

Same question reduced.

Flip two binary coins, the possible results are
00
01
10
11
.... or convert to decimal 0, 1, 2, 3

We know that there is a head, so 00 (or 0) is out.

That one gives more leeway to "creative interpretation".

If you are asking "what is the probability with the following rules" it's 1/3:

a) If 00, flip again.

AND

b) Both coins are flipped before the question is asked and both flips are repeated if the result is 00.


If you do it this way it is 1/2:

Flip one until you get heads.

Then flip the other.

edit: added quotation marks to correct grammar.

Now that that one is cleared up, here's another "don't think too quickly" type of riddle...

Compare the numbers 0.99999... (infinitely many 9s) and 1. Which of the following statements is true? Why?

0.99999... < 1
0.99999... = 1
0.99999... > 1

<, it'll never quite get there

Bill, there's a method to remove recurrence from a decimal:



2 other simple examples:

1/3 = .333333......
2/3 = .666666......
3/3 = .999999......

And,
1 - 0.9999... = 0.0000... (If you take the difference between two numbers, and the difference is 0, then they have to be the same number.)

Ah, but that's rounding - still didn't get there, but we'll just say it did! What's the largest number?

None of the explanations I gave use rounding. Remember the ... means it's continued to infinity. The first example, to me, is the most clear. The second example shows 1/3 = .333....... exactly. The .... is an indication that the 3's continue on to infinity. This is not an aproximation, it is the exact value, represented by the dots to save time from writing an infinite number of 3's. But it is the exact answer.

And 3 x .333... = .999..... which is not = 1, it approaches 1.

What is the largest number?

Monger, In bookkeeping, it's equal to 1.0.

c.i., that's because we record at 2 decimals and round at 3!

There is no largest number.
If the sum to infinity of 3/10 + 3/100 + 3/1000 + ... is not equal to 1/3, what is it equal to? If it is EXACTLY equal to 1/3, then 3/3 = 0.999...

Don't confuse "doing something many, many times" and "doing something an infinite number of times." Yes you get closer every step you do, and as you are approaching infinity you are approaching 1, that is true. However, when/if you hit infinity you hit 1 also.

Math works on paper. Sometimes you have to believe in what your calculations prove and show, otherwise we have to imagine that math works completely differently than it does.

Again, take a look at the first example I gave:
Note that no rounding or approximation is used there at all.

Here's a good link I found on this topic. http://mathforum.org/dr.math/faq/faq.0.9999.html

Here's an interesting excerpt from one of the pages at that web site:

It's =
Think this way:

1/1 = 1
2/2 = 2
3/3 = 3
.
.
.
999..../999.... = 1

1/3 = 0.33...

2/3 = 0.66...

3/3 = 0.99...

therefore, with calculus it can be proven that the limit as n approaches infinity of n/n = 1, which means 3/3 = 1 for certain, and therefore 3*1/3 = 3/3 = 1!

Welcome, TechnoGuyRob! :-D

TechNoGuy, WELCOME to A2K. Now that I understand a little math, and 3/3 = 1 or .99, because 2/3 = .66, is it still accurate to say 2/3 = .67? c.i.

N/A
1 is not equal to 0.99, nor is it equal to 0.9999999999999999. 1 is only equal to 0.999..., and 2/3 is only equal to 0.666...
Saying that 2/3 = 0.66 or 0.67, is only an approximation. But for example, for anyone who knows calculus, you can see this is a geometric serie:

0.9 + 0.09 + 0.009 + ....

A geometric series is a series that is defined as:

a + a*r + a*r^2 + a*r^3 + ....

In this case, a = 0.9, and r = 0.1, and according to definition, a geometric series approaches

a/(1-r)

when it's sum is added (assuming r < 1). Therefore, in this case

.9/(1 - 0.1) = 0.9/0.9 = 1

I really have to give it to Enthusiast !!!!!! Bravo!!!
1/3. You might initially think that the answer is 1/2, but not so. For two coins, there are four possible outcomes: HH, HT, TH, TT, since we know that at least one was a head, we can eliminate TT. Of the remaining three possibilities, only 1 allows the second head: HH. As mentioned by Enthusiast.