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If a fair coin is flipped 100 times, what is the probability that it will come
up heads 50 times and tails 50 times, in total ?

If a fair coin is flipped 100 times, what is the probability that is will
come up heads (or tails) the
first 50 and then tails (or heads) the last 50 ?

Thanks ArnieM. I don't have the ambition or math skills for that however.
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traderwerks

I don't understand something here though.

If someone actually did flip a fair coin 100 times and the outcome was
that there were 50 heads and 50 tails, then most people would say
that this was a highly likely outcome. (Or at least a 50:50 outcome,
I really know nothing about probability.)

But if that person showed a video, with several eyewitnesses, of the
actual coin flipping series, and the outcome was 50 heads then 50 tails,
I don't think anyone would believe the video.

The odds possibly are: 2^50
then 2^50 again. (Not positive on these calculations.)

In probability theory it's said that each individual coin flip is a separate event
with 50:50 chance of heads or tails...but...if you group a 100 coin flips...well, as you
can see I don't have the knowledge of what's going on with this stuff.

Lets take a look. Each flip is a seperate event. The odds of a binary outcome in a singe coin flip being heads or tails is 50%. We can write this a 0.50^1 = 0.50 = 50%. Now for two coin flips it is 0.5^2 = 0.25 = 25%. For 100 coin flips the answer would be 0.50^100 = 7.8886090522101180541172856528279e-31

It does not matter if it is 50 heads then 50 tails it does not matter since the probability is that you will follow a path of 100 coinflips. So the percentage would be the same for any sequence of 100 coin flips, heads, tails, whatever.

traderwerks wrote:So the percentage would be the same for any sequence of 100 coin flips, heads, tails, whatever.
=======================================================================

This I don't get, but thanks. Isn't there a bell curve in effect here where the likely hood
of even 25 (or more) identical faces of the coin in a row are at the tail ends of the bell?

If you have a bell curve on say the number of heads to tails then it would be closer to what you are thinking. However, the question was 'sequence' . 50 heads followed by 50 tails. Not that there would be 50 heads or tails in any sequence.

Of course the probability of getting 50 heads and 50 tails in ANY sequence is higher than 50 heads and 50 tails.

The thing is, that the outcome that is the most likely:

A total of 50 heads, with a total of 50 tails,
out of the 100 flips,
can arise from an extremely unlikely series of events
(that is, a certain series of coin flips) all the way up to an extremely likely series
of events (that is, a certain series of coin flips).

An outcome that is likely might only appear likely until an analysis
of the individual events of it are seen. That's my take on it.
It's not based on math though.

Last edited by stephenszpak; October 31st, 2010 at 03:11 AM.
Reason: edit

Ok, not really sure what your question is anymore. But Google "Probability Density Function" and that will probably get you headed in the right direction.

If you flip your coin 100 times, one event would be any result describing the 100 coin flips such as (heads, tails, tails, tails .... , heads, tails). Each single event has a probability of (1/2)^100 as @traderwerks explained. Your second example is just a single event, which has exactly this probability.

To answer your first question, you would have to count all the permutations of a series made up of 50 heads and 50 tails. This is a classical problem of probability theory.

Consider the 100 flips as mechanical boxes. How many different ways exist to put 50 eggs (=heads) into those boxes? For the first egg there is a choice of 100 boxes, for the second egg a choice of 99 boxes, etc. In the end the number of permutations are

100 x 99 x 98 x 97 x 96 x 95 x ......... x 52 x 51

Unfortunately the eggs are not numbered or coloured but all identical (heads is heads, there is nothing such as heads up), so you have counted too many different options. Laying an egg into box 3 and another egg into box 5 is the same as if you first put an egg into box 5 and then another into box 3. So you have to divide by the number of identical results that the first process generates. The number of permutations for 50 objects is

50 x 49 x 48 x 47 x 46 x 45 x ......... x 2 x1

The number of different ways to put 50 identical eggs in 100 boxes in the end becomes therefore

(100 x 99 x 98 x 97 x 96 x 95 ........ x 52 x 51) / (50 x 49 x 48 x 47 x46 x ....... x 2 x1)

= 100! / (50! x (100-50)!)

This is not so easy to write, so mathematicians have invented a short form for this, which is the binomial coefficient. Definition from Wikipedia below.

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So if you set n = 100 (number of coin flips) and k = 50 (cannot write that here) and multiply this with the probability of a single event, which is (0.5)^100, you get the probability that you asked for in question 1. So your probability for 50 heads and 50 tails is

(0.5)^100 x 100! / (50! x 50!)

This would be the peak value of the binomial distribution, which approximates the bell curve of the Gauss distribution. The binomial distribution is quite important for option pricing modells. The Cox-Rubinstein model - which is simpler than the Black-Scholes model - relies on the binomial distribution.

Hope I did not get anything wrong here, comments appreciated.

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