Calculating probabilities

[Raul Sousa]

I am trying to calculate probabilities in excel but I don’t really know which
functions to use.
Lets say I have only two possible outcomes 1 and 0. So far the outcome was
as follow:
1
0
1
0
1
1
1

Which are the odds to get 1 and to get 0?
Which formula can calculate this?

 

[barry houdini]

In general if there are n possible outcomes then each has a change of
1/n, so if n is 2 (in this case) then the probability of 1 or 0 is 0.5
or 50%.

Previous results make no difference to independent events

 

[Max]

Probability that its a 1:
=COUNTIF(A2:A8,1)/COUNTA(A2:A8)

Similarly, the probability that its a zero:
=COUNTIF(A2:A8,0)/COUNTA(A2:A8)

Alternatively, since Pzero + Pone = 1
as the data comprises only zeros or 1s (in this instance)
Pzero = 1 - COUNTIF(A2:A8,1)/COUNTA(A2:A8)

any good? hit the YES below
--
Max
Singapore
http://savefile.com/projects/236895
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xdemechanik

 

[Raul Sousa]

Thanks for your replay Max.
it could be a good idea. Unfortunately I think it is not correct.
The changes to have 1 or 0 are 50/50. In this case there is more 1 than 0
(5Vs2). So the chances that the next number is a zero are bigger than the
chances that the next number is 1.
I just don’t know a formula to calculate these odds.

 

[barry houdini]

"The changes to have 1 or 0 are 50/50. In this case there is more 1 tha
0
(5Vs2). So the chances that the next number is a zero are bigger tha
the
chances that the next number is 1."

No, this is a common misconception...especially amongst lotter
players. The chance of the next number being 1 is 50%, as per m
previous reply

 

[JoeU2004]

In general if there are n possible outcomes then each
has a change of 1/n, so if n is 2 (in this case) then
the probability of 1 or 0 is 0.5 or 50%.

That is not correct "in general". For example, consider:

=--(rand()<=0.75)

Clearly the probability of 1 is about 75% and 0 is about 25%.

The point is: Raul told us nothing about the distribution. You are
ass-u-me-ing a uniform distribution. Probably a good guess; but certainly
not true "in general".

Previous results make no difference to independent events

That is true __if__ the events are independent. Probably a good guess that
they are; but again, Raul told us nothing the distribution.

For example, put 0 or 1 into A1, ad enter the following into A2 and copy
down:

=if(A1=1, --(rand()<=0.75), 1-(rand()<=0.75))

This results in a random sequence of 1s and 0s, but they are not
independent. The probability of repeating the previous number is 75%.


----- original message -----

In general if there are n possible outcomes then each has a change of
1/n, so if n is 2 (in this case) then the probability of 1 or 0 is 0.5
or 50%.

Previous results make no difference to independent events


--
barry houdini
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----- message responded to -----

 

[barry houdini]

Hello Joe

Raul says

"The changes to have 1 or 0 are 50/50"

I take that to mean a uniform distribution.

Yes, he doesn't say the events are independent. I implied that the
were, I hope he will tell me if that isn't the case.

regards, barr

 

[joel]

t do reverse probablity you need to calculate the standard deviation an
get the sigma value. You could use normal distribution. The center o
the normalize curve will be the average or the numbers. The standar
deviation will determine the width of the Normal curve (Bell curve).
The sigma value is the square root of the standard deviation

 

[JoeU2004]

Raul says
"The changes to have 1 or 0 are 50/50"

Yes, he added that information in a posting dated 10/17 9:44.

I was responding to your comments dated 10/17 5:22, in response to Raul's
posting dated 10/17 5:16.

Unless you claim to be clairvoyant, my comments about your first response
were valid in that context.


----- original message -----

 

[joel]

I don't think he meant the results to be 50/50. re-read the posting

Thanks for your replay Max.
it could be a good idea. Unfortunately I think it is not correct.
The changes to have 1 or 0 are 50/50. In this case there is more 1 tha
0
(5Vs2). So the chances that the next number is a zero are bigger tha
the
chances that the next number is 1.
I just don’t know a formula to calculate these odds.




the probability could be 50/50 after a large number of trials but th
actual results with a small number of trials could be uneven. You wan
to espress the results as an average with a sttandard deviation

 

[JoeU2004]

Thanks for your replay Max.
it could be a good idea. Unfortunately I think it is not correct.

It is correct for what Max was trying to do -- which might not be what you
need.

You told us nothing about the distribution of 1s and 0s.

So Max's approach attempts to derive the probabilities based on observed
results. Since you have 5 1s and 2 0s, we might reasonably conclude that
the probablilities are 5/7 and 2/7 respectively -- that is, until you
provide additional information.

The changes to have 1 or 0 are 50/50.

New information breeds new solutions.

But you must also tell us whether the events (1 and 0) are independent or
not.

It would help if you explained how you are generating the 1s and 0s. For
example, flipping a coin, or --(RAND()<=0.5)?

the chances that the next number is a zero are bigger
than the chances that the next number is 1.
I just don’t know a formula to calculate these odds.

Assuming independent events....

If you had not generated any numbers, the chances of getting a __specific__
sequence of 7 1s and 0s followed by either a 1 or 0 would be 1/(2^8).

The chances of getting __any__ sequence of 6 1s and 2 0s is COMBIN(8,6)/2^8
(about 11%), and the chances of getting any sequence of 5 1s and 3 0s is
COMBIN(8,5)/2^8 (about 22%).

So if you had not generated any numbers, you would be correct that the
probablility of 5 1s and 3 0s is higher than 6 1s and 2 0s.

But having gotten any sequence of 5 1s and 2 0s, the probability of getting
a 1 or 0 next is still 50% each, for the very reason (assumed above) that
the choice of 1 or 0 is independent of the past.

These are difficult concepts to grasp. You should take an intro course in
probability.


----- original message -----

 

[barry houdini]

Unless you claim to be clairvoyant

I knew you were going to say that.......

Joe,

Of course I made some assumptions based on the limited informatio
available. It seemed to me that Raul was trying to predict the outcom
of an event based on previous independent events, I may well be wrong
it wouldn't be the first time....

regards, barr

 

[Mike H]

Hi,

The changes to have 1 or 0 are 50/50. In this case there is more 1 than 0
(5Vs2). So the chances that the next number is a zero are bigger than the
chances that the next number is 1.

That's incorrect because it implies that (say) a coin has memory and if we
flip 10 consecutive heads the next is more likely to be tails which it isn't,
the chance of the next being tails is exactly the same as for the previous 10
flips 50/50.

Mike

 

[Raul Sousa]

If you flip 9 times a coin and get 9 faces then at the 10Th time you flip it
the odds of getting a tail are much higher than the odds of getting a face.
At any flip the chances are 50/50. But, the chances to get 10 faces straight
are lower than the chances to get 9 faces and 1 tail in 10 coin flips.
I am sure there must be a way to calculate this probabilities in excel. Just
don’t know how to and appreciate any help.

 

[David Biddulph]

When you say:
"If you flip 9 times a coin and get 9 faces then at the 10Th time you flip
it the odds of getting a tail are much higher than the odds of getting a
face."
you are entirely wrong.

When you say:
"the chance to get 10 faces straight are lower than the chances to get 9
faces and 1 tail in 10 coin flips."
you are right, but only because there are 10 sequences in which you can get
9 faces and 1 tail.
If you have had 9 faces in the first 9 flips, the probability of getting 10
faces is exactly the same as getting 9 faces followed by 1 tail.

 

[Mike H]

Hi,

If you flip 9 times a coin and get 9 faces then at the 10Th time you flip it
the odds of getting a tail are much higher than the odds of getting a face.

Incorrect!!

The odds of 9 straight head flips is easy to calculate

=0.5^9

or around 0.19% chance

But having done that the odds of the tenth are 50/50 because the coin has no
memory and the result of the next event is not influenced by the previous
events. You're actually referring to the 'gamblers fallacy' which is believed
to suggest that ultimately the number of events will even out. Ten
consecutive black on the roulette wheel so lets bet large on red is a very
quick way indeed to end up bankrupt.

What will move near to 'normal' is the percentage for each event (50%) but
there could be a very large difference in the frequency of each that equate
to 50% especially when dealing with large numbers.

The odds of doing 10 straight head flips is
=0.5^10

Mike

 

[Raul Sousa]

Based on your anwser I realized that I must learn some more about ...
probabilities.

My idea was to know more about statistical functions in Excel and after that
calcuate what I need.

I think, know, that I will first take a look at a book about probabilities.

Thanks again for your anwser.