A bit confused

Steven G

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Many years ago while walking on campus I heard a part-time math instructor having her class recite a negative and a negative is a positive. I felt ill hearing the class recite this nonsense as I realized that a neg + a neg is never positive and that just under half the time (-) - (-) = (-). The reason I said less than 1/2 is sometimes (-) - (-) is neither pos nor neg-- it is 0.

Now this reminds me of a roulette wheel that has 18 black spaces, 18 red spaces and 2 green (at least the website I looked at say so). Let's just talk about betting on black or red. We see that black comes up 9 out of every 19 times on average (same for red). This is about half.

Now if we let the number of red and black spaces increase (but always equal to one another) and always have 2 green spaces, then the p(black) = p(red) is closer to .5 than 9/19. If we continue to increase the number of red and black spaces even more, the p(black) = p(red) gets even closer to .5. I do see that in the limit we should have p(red) = p(black) = .5

But wait a minute, this means that p(green) = 0 if we have an infinite number of red and black spaces. This bothers me since there is a chance of getting green. Actually you are as likely to get a particular green as a particular red or black.

Back to the subtraction of two negative numbers. If we randomly picked (let's stick with integers, although it does not matter) two integers there is a chance that they are both the same, isn't there?. So the p(getting a negative number when subtracting two integers) = p(getting a positive number when subtracting two integers) is less than .5. But how much less? Of course both those probabilities are exactly .5 So the probability of randomly picking two numbers and they are both the same is 0.

Wait a minute! Then the probability of picking two numbers that differ by 3 would also be 0. In fact, the probability of the two numbers differing by k, an integer, is 0. But how can all the possible differences have no chance of happening, yet the p(getting a pos difference) = .5???

I can't wait for your responses!

EDIT: I meant a (-) and (-) is a positive
 
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Many years ago while walking on campus I heard a part-time math instructor having her class recite a negative times a negative is a negative. I felt ill hearing the class recite this nonsense as I realized that a neg + a neg is never negative and that just under half the time (-) - (-) = (-). The reason I said less than 1/2 is sometimes (-) - (-) is neither pos nor neg-- it is 0.
I can't make heads or tails out of this. Are you sure it wasn't "a negative times a negative is a positive"?
(Oh, wait. Later in this post you say it was "minus", not "times".)

Now this reminds me of a roulette wheel that has 18 black spaces, 18 red spaces and 2 green (at least the website I looked at say so). Let's just talk about betting on black or red. We see that black comes up 9 out of every 19 times on average (same for red). This is about half.
Roulette wheels in the United States have 2 green spaces. European roulette wheels have only one. That decreases the house edge a little.

Now if we let the number of red and black spaces increase (but always equal to one another) and always have 2 green spaces, then the p(black) = p(red) is closer to .5 than 9/19. If we continue to increase the number of red and black spaces even more, the p(black) = p(red) gets even closer to .5. I do see that in the limit we should have p(red) = p(black) = .5
Yes, that is correct in the limit.

But wait a minute, this means that p(green) = 0 if we have an infinite number of red and black spaces. This bothers me since there is a chance of getting green. Actually you are as likely to get a particular green as a particular red or black.
Yes, but in continuous probability distributions (which you have here, taking the number of spaces going to infinity), probability 0 does NOT mean something is impossible. If you were to choose a real number, between 0 and 1, at random- every number being equally likely, then the probability of any particular number is 0. But you must have picked some number!

Back to the subtraction of two negative numbers.
Subtraction? You started this talking about "a negative times a negative is a negative". "Times" not "minus"!

If we randomly picked (let's stick with integers, although it does not matter) two integers there is a chance that they are both the same, isn't there?
What probability distribution are you assuming? Since there are an infinite number of integers, you cannot assume a uniform distribution as you did for the spaces on a roulette wheel.

So the p(getting a negative number when subtracting two integers) = p(getting a positive number when subtracting two integers) is less than .5. But how much less? Of course both those probabilities are exactly .5 So the probability of randomly picking two numbers and they are both the same is 0.

Wait a minute! Then the probability of picking two numbers that differ by 3 would also be 0. In fact, the probability of the two numbers differing by k, an integer, is 0. But how can all the possible differences have no chance of happening, yet the p(getting a pos difference) = .5???

I can't wait for your responses!
The problem is that you are trying to assume that any two integers are equally likely to be chosen- a uniform distribution. And you can't do that with an infinite set.
 
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