WHAT IS THE PROBALITY OF A MOTHER HAVING A GIRL AFTER HAVING THREE BOYS IN A ROW.?
OR WHAT IS THE PROBABILTIY OF HAVING FOUR BOYS AFTER HAVING THREE BOYS IN A ROW.

Hi!!

The answer is quite simply for both cases: 50%

We can assume that the chances to have a boy or a girl are the same,
50%. Then GIVEN the fact that the family has three boys in a row, the
event "sex of new baby" is completely independent from the event "sex
of previous babies".
The problem is similar to toss a coin and, after getting three tails
you wonder which is the probability to get a head in the next toss.
Because each toss is independent from the other (the coin still have
two sides), the probability is 50%.

For further reading and references see the following pages:
"Chapter 6 - PROBABILITY ":
"However, one part of the traditional material is useful and easily
mastered.  It concerns successful repetitions of an event.  Suppose
there is a 50% chance of having a girl baby and a 50% chance of having
a boy.  The question is:  What happens to the 50% as we specify longer
and longer strings of girls born to the same mother?  Some students
have had enough instruction in probability to remember something about
chances remaining the same.  Partial memories will lead them astray if
they believe that the chances of two girls (and no boys) in a family
of 2 children are the same as the chances of one girl in a family of
one child.  There is something that remains the same and it is just
that initial 50%.  We get the probability of an event happening ?in a
row? by using the base probability, the 50%.  But whatever we do, we
clearly must arrive at smaller chances for more difficult or more
unlikely events.  It is harder to get two girls in a row, with no
intervention from boys, than to get a single girl at the beginning. 
It is harder to get five heads in a row with a coin than to get one
head with one flip.
If half the babies born are girls, and all those with daughters have a
second child and the chances of a girl are always 50%, then half of
those with a girl the first time will have a girl the second time. 
Half had a girl the first time and half of that half will have a girl
the second time. ... "
http://www.uwsp.edu/education/wkirby/t&m/6Probability.htm


"Chapter 4 - Probability":        
See the 'Another two-outcome probability model' paragraph at page 5 of
this document.
http://www.stat.auckland.ac.nz/~wild/ChanceEnc/Ch04.pdf

You need Acrobat Reader to see the above page, you can download it from:
http://www.adobe.com/products/acrobat/readstep2.html
or
you can see the cached version by Google:
http://64.233.161.104/search?q=cache:U1YxM8B4MSQJ:www.stat.auckland.ac.nz/~wild/ChanceEnc/Ch04.pdf+PROBABILITY++MOTHER+HAVING+A+GIRL+%22IN+A+ROW%22&hl=en


I hope that this helps you. if you need a clarification do not
hesitate to request it before rate this answer.

Regards.
livioflores-ga

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Clarification of Answer by livioflores-ga on 30 Jan 2005 06:46 PST

Thank you for the comment dragon_2!!

In that case we must solve the following system to find the
probability to having a girl:

If M is the probability to having a boy and F is the probability to having a girl:
M + F = 100%
M / F = 1.05

From the second equation we have that:
M = 1.05 * F 

Then:
100% = M + F = 1.05*F + F = 2.05*F ==> F = 100% / 2.05 = 48.78%

and M = 100% - 48.78% = 51.22%

The answer for your question (if the sex ratio at birth male/female is
1.05) is 48.78% . The reasoning of the main problem is the same.

Regards.
livioflores-ga

I HAVE BEEN OUT OF CURCULATION WITH A TUMOR OPERATION AND A AN I.T.A..
 YOU ANSWERE WAS QUITE COMPREHENSIVE AND MORE THAN APPRECIATED.  (IN
GEST--I WOULD A HAVE PREFERRED IT SAID A 60-70% CHANCE OF HAVING A
GIRL AFTER A BOY.  I WOULD GIVE A FIVE STAR RATING

The only problem with your analysis is that gender is NOT 50/50.

http://education.yahoo.com/reference/factbook/invert/sex_ratio.html

In the United States, there are 1.05 males to each female at birth.

It is interesting to compare this to China (selective abortion)where
the ratio is 1.15 males to each female.

It seems that most countries have a ratio somewhere between 1.04 and
1.05 males to each female.

To be exact, you may want to refigure your analysis, although,
functionally, the answer remains the same.

Ed.

I don't think it's as simple as this.  One could imagine that there
are biological factors predisposing a couple to producing either boys
and girls.  In this case, one would expect to find that single gender
offspring families occur more often than would be calculated by the
simple probabilities.

Given the wealth of census data available on the web, I'd hoped to be
able to locate these statistics, but so far I've had no luck.

I'm no good at math, but it seems to me, say in a coin-flip situation,
there would be statistics/odds of flipping 4 tails in a row.  I agree
that at every flip there is a 50/50 chance of tails, but the
succession of 4 in a row has odds, doesn't it?

Just curious.

Yes cynthia, but note that the situation here is the following:
- the woman already had three children in a row, so having this 3
boys, the probability for the fourth baby to be a girl is 50%.

I've always been confused by this question.
You take your coin and flip it, having a 1/2 chance of it landing on heads.
You flip it again, having a 1/2 chance of it landing on heads.
So on and so forth until your 100th flip.
Now assume you have managed to get 99 heads up to this point. What is
the probability to get another head in the 100th toss?
Theoretically, because each toss is independent, it should be again
50%. But probability theory also tells us that in the long run, the
tendency should be half heads and half tails. So, we got 99 heads
already, but it doesn't increase the chance of getting a tail next
toss. Then, how could the probability structure of half-half be
demonstrated anyway?

Some things to remember and consider:
for each coin toss, there is a 50% chance that you will get heads,
assuming we are using a "fair" coin (not a double headed coin).  The
probability that you will get two heads in a row is 25% (.5 x .5 = .25
= 25%).  The probability of three in a row is 12.5% (.5 x .5 x .5 =
.125 = 12.5%).  Four in a row is 6.25%.

Lets say you flipped 3 heads in a row.  The odds that your next flip
will be heads are still 50%.  Remember, there was only a 12.5% chance
that you would flip those first 3 heads.  Same logic applies to the
100 head flips in a row scenario.  Just because you may have been
luckily enough to flip those first 99 heads in a row doesn't change
the odds for that next flip.  BTW, the odds of flipping 99 heads in a
row are 1 / (633825300114114700748351602688), a ridiculously small
number.  You question how can the odds be 50% for that next flip and I
question how are you going to get the first 99 in a row to begin with.
  
This is all cookbook discrete math.  Don't let your intuition mislead
you.  Using an example such as a coin toss is sometimes too trivial
(50/50) of an example to demonstrate probability calculations.

In answer to readerlife's suggestion that once you have 99 heads, the
law of averages should weigh the tendency of the next batch of flips
to favor tails:

You are correct, over time, the tendency of coin tossing will approach
50%.  However, each toss remains independent.  The tendency you speak
of will (most likely) be demonstrated over a large number of tosses. 
E.g., if the next two million tosses come out to their "expected"
outcome of one million heads and one million tails, the score will now
be: 1,000,099 heads to 1,000,000 tails, a ratio of 1.000099 (or a
percentage of tails of .49997525), which is extremely close to 50%. 
Even if the first million tosses were somehow to come up heads (and
assuming you were somehow still satisfied you were tossing a fair
coin),  the next trillion tosses would likely bring you just as close
to 50-50.  So you see, over time the "law of averages" will play out
and the outcomes will approach the predicted probability.