Just a question for you, have you learned how to use a normal approximation to a binomial distribution? Usually that is what is done when the value of n is larger than 20.

To use the normal approximation to the binomial distribution, consider the following:

n = number of trials = 1000

p = probability of a head = 0.5

q = probability of a tail = 0.5

m = mean = n*p = 1000*0.5 = 500

j = standard deviation = sqrt(n*p*q) = sqrt(1000*0.5*0.5) = 15.8114

Now, find the probability of getting between 450 and 550 heads, or P(450<x<550).

Using the normal approximation, we have to calculate the z-score of each condition. First, P(x < 550):

z = (550 - m) / j = (550 - 500) / 15.8114 = 3.16

From normal probability tables, a z-score of 3.16 shows as 0.99921

This means that the probability of x >550 is 1 - 0.99921 = 0.00079

Next, find P(x < 450):

z = (450 - m) / j = (450 - 500) / 15.8114 = -3.16

This z-score gives a probability of 0.00079

Finally, add up the probabilities of getting the outcomes:

0.00079 + 0.00079 = 0.00158

This means the chances of getting MORE than 550 heads and LESS than 450 heads adds up to 0.158%

So, the probability that the number of heads will be between 450 and 550 is simply:

P(450 < x < 550) = 1 - 0.00158 = 0.99842 or 99.842%

While the probability of getting one head is 50% per throw, this does not answer the question. I have thought about this question since I posted it and now have a better idea myself. The solution is to use the binomial theorem. This will give the probability of getting any number of heads after 1000 coin tosses. E.g. getting exactly 1000 heads is vanishingly small, (0.5)^1000. Getting exactly 500 heads is a lot more likely. The trouble it I do not know how to do a binomial expansion when the number of tosses, in this case 1000, is large. If I use a spreadsheet the numbers rapidly go into overflow so using brute force is not on.

...My calculations are 550/1000= .55 or 55%. This doesn't mean you WILL hit in this range after 1000 flips, it simply means you have a 55% chance of it. To obtain either heads or tails 50% of the time depends on how many flips you make. The more flips you make, the closer to the true 50% odds you will become. Because your range is 450 to 550 out of 1000 we can just say you need to hit heads no less than 45%-55% of the time. I say your chances of hitting your range is 55% because you have a 55% chance of getting at least 450 heads and a lesser chance of obtaining 550 heads out of 1000 flips at 45%.

...For the least number of favorable tosses (450)

Total number of tosses = 1000

Number of favorable tosses = 450

Probability = 450/1000 = 9/20

For the most number of favorable tosses (550)

Total number of tosses = 1000

Number of favorable tosses = 550

Probability = 550/1000 = 11/20

Find the average of the two probabilities: (9/20 + 11/20)/2 = (20/20)/2 = 1/2 = 50%

So the probability of getting between 450 to 550 heads from 1000 tosses is 50%

You may be right and the normal distribution is the better way to go. Your logic seems OK up to a point anyway. Nearly 99% seems too high. I will use your approach and see what I get. On the right track certainly.

...The odds of flipping a coin to obtain either a heads or tails outcome is 2 to 1, or 50% of the time. To obtain between 450 and 550 heads out of 1000 tosses the odds of you hitting in this range is 55%.

...I'm going to go out on a limb here and say that if you flip that coin 1,000 times the probability that it will land on heads approximately half the time is about a 95% to 100% chance.

...The chance is and will always be 50/50. it could be tossed 3 times or 3 hundred million times, the chance is still the same due to a coin having only two sides.

...When flipping a coin, all possible outcomes are 50/50.

2 possible outcomes. half and half chance = 50/50