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Q: Calcularing odds of a single event occuring from multiple attempts ( Answered 5 out of 5 stars,   0 Comments )
Subject: Calcularing odds of a single event occuring from multiple attempts
Category: Science > Math
Asked by: thog-ga
List Price: $15.00
Posted: 06 Apr 2004 19:45 PDT
Expires: 06 May 2004 19:45 PDT
Question ID: 326367
Odds calculation. $15.00
I'm trying to calculate the eventual (or overall)
odds of an single occuring based on multiple events occuring. For
example; the is a 20% chance that an event will turn up positive each
time the dice are rolled (so think of having to roll 1 or 2 on a d10).
This 20% chance comes up 5 times (you get 5 rolls). What are the odds
that one of those 20% chances will turn up positive (it's not 100%)?
What are the odds that two of them will turn up positive?

I need the specific, workable formula for both the single event
occuring (5x 20% = ?). Again, preferably a formula that will also
predict the odds of mulitple 'hits' (so 80% odds of one hit based on a
base 20% x 5 rolls, and 30% odds of a second hit, 10% odds of 3 hits,

Bonus points if it will work when the odds may vary between each roll
(so 5 rolls, at 25%, 15%, 8%, 30%, and 22%).

I'm relatively math illiterate (three cheers for public education), so
please write out the formula in full and as simply as possible.
Subject: Re: Calcularing odds of a single event occuring from multiple attempts
Answered By: mathtalk-ga on 06 Apr 2004 21:55 PDT
Rated:5 out of 5 stars
Hi, thog-ga:

When your chances of "success" vs. "failure" are the same in each
repeated attempt, the formula which is probably easiest to apply and
remember is the "binomial theorem".

Bear with me and I think I can help stir up some of those dusty
memories of high school math for you!

With each attempt there is either success or failure, so in absolute
terms the probability of success in a single attempt must be 1 (or
100%) minus the probability of failure in a single attempt.  These
"complementary" events must have probabilities that add up to 1 (or to
100%, if that's more familiar).

Okay, now let's put this into a bit of algebra.  Let X be the
probability of success and Y the probability of failure.  What we've
just observed is:

( X + Y ) = 1

The left hand side of this simple equation is a "binomial", fancy talk
for "two terms".  The "binomial theorem" tells us what happens when we
raise such a thing to a power.  In the present context, the power
represents the number of trials.  For example, to plug in the numbers
you threw out in your Question, let's suppose five rolls of a
(ten-sided?) die, where the chance of success is 20% or 1/5.  Then:

( 0.2 + 0.8 ) = 1

and with five repetitions, all the possible outcomes are compactly given by:

( 0.2 + 0.8 )^5 = 1

The trick here is to expand the expression ( X + Y )^N using the
binomial theorem, which tells us the coefficients for particular "like
terms" based on powers of X and Y.  These coefficients are
"combinations of N things taken k at a time" and we'll review them in
just a moment.

But first let's see how this would help to pin down the chance of one
success in five tries, versus two successes in five tries.

The entire expansion for N = 5 is this:

( X + Y )^5 

  = X^5 + 5(X^4)Y + 10(X^3)(Y^2) + 10(X^2)(Y^3) + 5X(Y^4) + Y^5

Recall that the term X by itself is 0.2 or 20%, meaning the chance of
success in a single attempt.  The X^5, the first term in the
expansion, corresponds to getting a success in each of the five
attempts!  That is:

X^5 = (0.2)^5 = 0.00032

At the other end of the expansion, the term Y^5 represents failure in
each of the five attempts, or:

Y^5 = (0.8)^5 = 0.32768

The in between terms all represent a mixture of successes and
failures, with the particular case of one success (and four failures)
given by:

5X(Y^4) = 5(0.2)(0.8)^4 = 0.4096

and the particular case of two successes (and three failures) by:

10(X^2)(Y^3) = 10(0.2)^2(0.8)^3 = 0.2048

As an exercise, work out for yourself the other two terms in the expansion:

10(X^3)(Y^2) = ?

5(X^4)Y = ?

and then confirm that all six terms in the expansion add up to 1! 
Thus, in five repeated trials, we must have exactly one of these six
outcomes, and the sum of their probabilities is 100%:

no successes
one success
two successes
three successes
four successes
five successes

To vary the number of "rolls" or repetitions, one varies the exponent
N while the chance of success (on a single trial) is X (and Y the
corresponding chance of failure).  The only ingredient missing for
this "constant odds" setup is the explanation of the binomial
coefficients or "combinations of N things taken k at a time".

When we have k successes in N trials, those k "good rolls" can be any
subset of the collective N trials of the right size.  So the number of
ways to pick k rolls of a "pool" of N things is the number of
different subsets of size k.

Here it will help to memorize a formula, together with one or two
properties that come in handy.  The formula for combinations of N
things taken k at a time is:

C(N,k) = N!/(k!(N-k)!)

Okay, what's up with the exclamation marks, you may be asking.  Did we
get to the exciting part of the math?  No, sorry, the exclamation
marks are simply a shorthand notation for factorials.  N "factorial"
for a positive integer N is just the product of N time (N-1) times
(N-2) and so on down to "times 1":

N! = N * (N-1) * (N-2) * ... * 1

In particular here are the first few factorials:

1! = 1
2! = 2
3! = 6
4! = 24
5! = 120

Indeed we have skipped here the first and most surprising factorial, which is:

0! = 1

This may be a little hard to accept, but it makes the formulas come
out correctly.  For example, how many combinations of 5 things taken 5
at a time are there?  We expect the answer to be 1, because given 5
things to choose from the only way to take 5 is to take all of them.

In a math formula perspective, we should have:

C(5,5) = 5!/(5!(5-5)!) = 1/(0!)

From this we see that in order to get 1, we need 0! to be 1.  And so
by convention that is how 0! is defined.  It's the definition that
makes the formulas work.

One other property to remind you of is that:

C(N,k) = C(N,N-k)

whenever k is between 0 and N.  You may recall "Pascal's triangle" and
how one can obtain these binomial coefficients by adding two entries
in the row above to get the one between and below them in the next row
(I hope these are not terribly unpleasant memories for you; I'm quite
fond of Pascal's triangle, although it does turn out to have been
discovered earlier by a Chinese mathematician).

At any rate with the binomial theorem and the formula for binomial
coefficients (aka the combinations of N things taken k at a time), you
have the tools to find the answers to all sorts of probability
questions involving repeated trials with constants "odds".  Such
experiments are sometimes called Bernoulli trials:

(X + Y)^N 

   = C(N,0)Y^N + C(N,1)X Y^(N-1) + C(N,2)X^2 Y^(N-2) + ...

      ... + C(N,N-1)X^(N-1) Y + C(N,N)X^N

Note that the binomial coefficient in each term is C(N,k) where k is
the exponent on X, and that the powers k of X and N-k of Y always add
up to N (the overall degree of the polynomial) in each "monomial"
(single term) of the expansion.

Now you've also asked how to deal with cases in which the chances vary
from one trial to the next.  Because these chances can be arbitrary
numbers (subject to their being between 0 and 1 as all good
probabilities must), the computations are necessarily "messier" as we
must allow for different numbers with each succeeding trial.  However
I will show you a simple (if tedious) way to organize these
computations, requiring nothing more than the high school algebra
we've already touched upon.

What we need to do is provide a different binomial expression for each
different trial, rather than repeating the same factor (X+Y) to the
power N.

As a formality I'll suggest writing the probability of success and the
complementary probability of failure for each trial, e.g. in two
columns.  Then form a binomial for each trial using these
complementary chances as coefficients, say (to make similar but
distinct notation) the coefficient of little x for success and the
coefficient of little y for failure.

To illustrate using the "varying odds" of your Question for five rolls, we'd have:

(0.25x + 0.75y)(0.15x + 0.85y)(0.08x + 0.92y)(0.3x + 0.7y)(0.22x + 0.78y)

Multiplying these five factors out, while not very rewarding in
itself, will give us a fifth degree polynomial in x and y.  The
coefficient of x^5 will be the chance of five successes, and the
coefficient of y^5 the chance of five failures.  Intermediate terms
will represent the collective mixtures of success and failure,
according the represented powers of x and y, very much like they did
before with the binomial expansion of (X+Y).

Subject to rounding in the arithmetic, our expansion of those factors
shown above produces this sort of result:

          5               4             2  3             3  2
0.320229 y  + 0.418662 x y  + 0.207542 x  y  + 0.048212 x  y

            4               5
+ 0.005157 x  y + 0.000198 x

As described above, the coefficient 0.320229 corresponds to the chance
of five failures (no successes), the coefficient 0.418662 to one
success (four failures), and so forth.

Please feel free to ask for further clarification of any points
presented here.  I'd love to help you master the concepts of this

regards, mathtalk-ga

Request for Answer Clarification by thog-ga on 19 Apr 2004 11:46 PDT
Unfortunately, I can't follow this. I have no idea what the carrot ^
symbol is supposed to stand for, don't know the use of the word
'factorial', and can not follow the algebriac forms once you start to
compress them (so pretty much everything from 1/3rd of the way through
the entry is unintelligible to me).

Background; I moved every 6 months or so of my childhood. This wrecked
my education (opportunities for which were not rich in the areas I
lived in any case) to the point where I wasn't taught any algerbra (or
even dividing fractions for that matter). As I said, I've taught
myself quite a bit, but it's very patchy. This is absolutely madding
to me, as the skills would be useful for both my work and my personal
interests, but even the local community collages don't have programs
that would be helpful in this case (assuming I know too much or too

So we can either start over, or we can go through this paragraph by
paragraph as I try to map it. It's up to you which you think will work

Clarification of Answer by mathtalk-ga on 19 Apr 2004 12:52 PDT
Hi, thog-ga:

It's good to hear back from you.  If the polynomial algebra is
confusing to you, I think it would probably make the most sense to
tackle the question from a purely arithmetic approach.  It's a
tradeoff between being able to "solve" a lot of problems quickly with
an approach that is "abstract" (algebra) versus hitting it at a
tedious but concrete level (arithmetic).

Let's start with a simply case of two tries, but as long as we are
doing it this step-by-step way, let's go ahead and include the
possibility that the chances of success/failure are not the same for
both tries.

Suppose the first try has success 10% of the time and failure 90% of the time.

Then _before_ the first try, we know:

Pr( 0 successes in 0 tries ) = 100%
Pr( 1 success   in 0 tries ) =   0%
Pr( 2 successes in 0 tries ) =   0%

and so on.  [Obviously with 0 tries we are certain to 0 successes.]

Okay, _after_ the first try, we know:

Pr( 0 successes in 1 try ) = 90%
Pr( 1 success   in 1 try ) = 10%
Pr( 2 successes in 1 try ) =  0%

and so on.  Of course this seems pretty clear, but that may be just
because this is the first step.  The second step will start to make it
clear that there's a rule for computing a new sequence of
probabilities following each additional try.

For the sake of illustration, I'm going to let you pick the
probability of success (and thus of failure) on the second try, and
then apply those figures to the results above.  This will hopefully
let you see where the numbers are used more clearly.  Remember that
the chance of success can be anything between 0 and 1 (or between 0%
and 100%, if that sounds more familiar).

regards, mathtalk-ga

Request for Answer Clarification by thog-ga on 19 Apr 2004 13:27 PDT
This sounds like a better approach for me. I work well with the
concrete examples, and once down, can often learn the abstract
terminology handling.

Ok, so in example 1 you have zero attempts, so zero chance of success.
In example 2 we have 1 attempt at 10%, so a total of 10% chance of

If for example 3 we had an additional 27% chance of success (.27),
we'd have one chance at 10%, one at 27% and two attempts total. If you
could illustrate how that looks and works, and how it would look in
comparision to two attempts at 10% each, that would help.

For future attempts, throw in either flat odds and multiple attempts,
or the following sets;
(starting with current)


Clarification of Answer by mathtalk-ga on 19 Apr 2004 20:40 PDT
Hi, thog-ga:

Okay, let's take a look at how to advance from one try to two, given
that the second try will have a 0.27 chance of success (and thus a
0.73 chance of failure).

Here's the probability distribution of outcomes after one try:

#successes   Probability

    0            0.9

    1            0.1

    2            0.0

Now the second try is considered "independent" of the first try, which
means its chance of success or failure is the same regardless of how
the first try turned out.

The only way to have _no_ successes after the second try is to have no
successes after the first try and then to fail as well on the second
try.  Thus:

Pr( 0 successes after 2 tries ) = 0.9 * 0.73 = 0.657

Now there are two ways you could have (exactly) one success after two
tries.  You might have no success on the first try and then succeed on
the second try, or you might succeed on the first try and fail on the
second.  These two possibilities are "disjoint" (there's no overlap),
so the probability of either of them is just the sum of their

Pr( 1 success   after 2 tries ) = (0.9 * 0.27) + (0.1 * 0.73)

                                = 0.316

Finally there is positive probability for 2 successes after 2 tries,
i.e. succeeding both times:

Pr( 2 successes after 2 tries ) = 0.1 * 0.27 = 0.027

Let's incorporate these results into an expanded table, with one more
row and one more column.

Here's the probability distribution of outcomes after one try and after two tries:

               One Try     Two Tries
#successes   Probability  Probability

    0            0.9         0.657

    1            0.1         0.316

    2            0.0         0.027

    3            0.0         0.0


There's a fairly simple pattern to the calculations which can be
repeated over and over, to obtain chances of a smallish number of
successes for any number of tries (if the calculations are only
repeated enough to cover the number of tries).

To illustrate this, I'm going to back up and do the first two tries
using _equal_ chances of 0.1 (success).

            No Tries    One Try    Two Tries
#success  Probability Probability Probability

    0         1.0    -    0.9    -   0.81
                     \           \
    1         0.0    -    0.1    -   0.18
                     \           \
    2         0.0    -    0.0    -   0.01
                     \           \
    3         0.0    -    0.0    -   0.0


The pattern here is to find a new entry, you look to the column just to the left:

Multiply the entry above and to the left by
the chance of SUCCESS in the current trial,
and add to that the entry on the same row to
the left times the chance of FAILURE in the
current trial.

In other words, a success in the current trial will increase the
number of successes overall by 1, while a failure keeps the number of
successes the same as before.

Compare those results with the ones for the unequal probabilities that
we started with:

            No Tries    One Try    Two Tries
#success  Probability Probability Probability

    0         1.0    -    0.9    -   0.657
                     \           \
    1         0.0    -    0.1    -   0.316
                     \           \
    2         0.0    -    0.0    -   0.027
                     \           \
    3         0.0    -    0.0    -   0.0


We can set up a table in which the same odds of success are repeated
over and over, or we can change the odds of success from one column to
the next.  If we repeat the 0.1 chance of success four times, we'd
have this table:

            No Tries    One Try    Two Tries  Three Tries Four Tries
#success  Probability Probability Probability Probability Probability

    0         1.0    -    0.9    -   0.81    -   0.729   -  0.6561
                     \           \           \           \
    1         0.0    -    0.1    -   0.18    -   0.243   -  0.2916
                     \           \           \           \
    2         0.0    -    0.0    -   0.01    -   0.027   -  0.0486
                     \           \           \           \
    3         0.0    -    0.0    -   0.0     -   0.001   -  0.0036
                     \           \           \           \
    4         0.0    -    0.0    -   0.0     -   0.0     -  0.0001


If instead we have the unequal probabilities that you suggested (of
success: 0.1, 0.27, 0.18, 0.56), then:

            No Tries    One Try    Two Tries  Three Tries Four Tries
#success  Probability Probability Probability Probability Probability

    0         1.0    -    0.9    -   0.657   -  0.53874  - 0.2370456
                     \           \           \           \
    1         0.0    -    0.1    -   0.316   -  0.37738  - 0.4677416
                     \           \           \           \
    2         0.0    -    0.0    -   0.027   -  0.07902  - 0.2461016
                     \           \           \           \
    3         0.0    -    0.0    -   0.0     -  0.00486  - 0.0463896
                     \           \           \           \
    4         0.0    -    0.0    -   0.0     -  0.0      - 0.0027216


The tables we've worked out here have enough rows in them for all the
possibilities to be accounted for.  The sum of each column of
probabilities will add up to 1 in this case.  However if _only_ the
chances of one success were desired, it would be unnecessary to
compute anything beyond the top two rows (which give the chances of
cumlative no successes and one success, respectively).  In other
words, this approach allows us to economize on computation if only a
most two or three successes need to be accounted for, because then we
only retain rows that correspond to that many or fewer successes.

I hope these tables will assist your understanding, though of course
it will be tedious to go through a representative sampling of the
arithmetic behind them.

regards, mathtalk-ga

Request for Answer Clarification by thog-ga on 20 Apr 2004 12:00 PDT
Ok, I think I've got it (hell, I could easily make an Excel
spreadsheet to do this). And I've mostly sorted out the 'why' as well
as the 'what. My thanks. You have the distinction of being the first
person to teach me anything useful in terms of math since around the
6th grade. *grin*

I'll let you go now, and if I've further questions I'll post another
bid. Thanks again.

Duri / Thog

Clarification of Answer by mathtalk-ga on 20 Apr 2004 13:23 PDT
Uhh, I guess I should confess that I _did_ use an Excel spreadsheet to
whip out the very last table.  I was terrified that I'd get an
arithmetic or clerical mistake in the mix, just adding to the innate

best wishes, mathtalk-ga
thog-ga rated this answer:5 out of 5 stars and gave an additional tip of: $10.00
Excellent. I have an effectively non-existant math education, which
required the researcher to explain in step-by-step, easy to follow
terms a problem that many have failed to clarify in the past. The
researcher was quite patient and thorough. Highly recommended.

Great service Google!

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