"Simple" Probability and Statistics Problem

Well, that doesn't make much sense...

The answer to that equation is somewhere around 4.67%

Which, if the probability of going from 1 to 2 is 4.67%, that would mean 95% of our tries would be stopping on 1, correct?

EVERY try starts at 1, and progresses.  Goes from 1, to 2, to 3, to 4, etc.  It's not a flip of a coin that will land somewhere on a line from 1 to 20...which is what I think that last post was assuming.  Does that make sense?

actually I think the probablity will be a little greater then that what was given by ozo.
The question did not say that the try stopped at 2. but that it would go to 2. Those which stopped at 3 also went to 2

20 - 3
19 - 3
18 - 8
17 - 8
16 - 19
15 - 21
14 - 32
13 - 50
12 - 62
11 - 99
10 - 119
9 - 134
8 - 197
7 - 232
6 - 328
5 - 503
4 - 663
3 - 843
2 - 1235
1 - 1305

only accounts for 5864 tries
what happened to the other 5?

> EVERY try starts at 1, and progresses.  Goes from 1, to 2, to 3, to 4, etc.
> Those which stopped at 3 also went to 2
In that case, the number that went to 1 is
3+3+8+8+19+21+32+50+62+99+119+134+197+232+328+503+663+843+1235+1305
and the number that went to 2 is
3+3+8+8+19+21+32+50+62+99+119+134+197+232+328+503+663+843+1235
so the probability that a try that goes to 1 will also to 2 would be
(3+3+8+8+19+21+32+50+62+99+119+134+197+232+328+503+663+843+1235)
/
(3+3+8+8+19+21+32+50+62+99+119+134+197+232+328+503+663+843+1235+1305)

Close ozo, I think you're on teh right track.

But it's important to remember that 1 is where you start..your last mathematical problem basically had us starting at 20.

The number that went to 1 is 1305
The number that went to 2 is 1305 + 1235

and so on. 1 and 2 may be bad examples, lets go somewhere in the middle.

 I basically need, for example, that we are currently sitting at 5.  Given the previous set of data, what is the statistical probability that I will continue to 6?

And you're right, I'm missing 5..feel free to minus 5 from the total in your calcs :)

Oh wait, I made a small error. These numbers are actually not passing.  They are the amount of times we STOPPED on that number.

So, out of the 5 thousand whatever tries, we STOPPED on 1 1305 times.  We also, out of the 5 thousand whatever times, STOPPED on 2 1235 times. So technically, I don't think we should be adding them together.  For example, both 19 and 20 are 3.  That doesn't mean you have a 100% probability of going to the next number. It just means both had 3 stop on them...make sense?

Does stopping on 2 imply first passing 1?

20 - 3
19 - 3

3 tries stop on 20, after first passing 19
3 tried stop at 19, without going on to 20
so a total of (3+3) get to 19
of those (3) go on to 20
so the probability of going on to 20 after getting to 19 is (3)/(3+3)



The number that went to 1 is 1305
The number that went to 2 is 1305 + 1235
Does this mean that every number that went to 1 also goes to 2
and that 1235 go to 2 without going to 1?

Let us make following definitions:
call the state of your "try machine"(TM) before try as
  state 0 = state before try.
and
  state n = stop at number n

If TM is in state 0 and an event T is an entire try, then we denote
  p(n) = probability of n-th stop happened in T.

The initial table listed in problem description is a table of probabilities
p(1), p(2), ...., p(20).



If we are in try T and state is 1, then probability of next stop at 2 in the same try T is denoted as:
  p( 2 | 1 )

Probability of event of two stops 1,2 at given try, we denote as
  p( 1, 2 )


Now, the answers are:

   1) The question is about p( 2|1 ).

       As Ozo said, if e(1) and e(2) are independent, then
            p(2|1) = p(2).
       If e(1) and e(2) are dependent, than the initial table does not contain information about
       p(1,2)  and answer does not exist.  Build additional tables while making first 5... tries,
       and they  will approximate the probabilities.


  2) The question is about p(1,2).
   
      p(1,2) = p(1,2) = p(1)p(2|1)

      For p( 2|1 ),  the answer is already contained  in 1).


  3) It is possible, that we are asking about transitional probabilities or about path probabilities:

     p( 5 |  ( 1,2 ) )
   
     or

     p( 1,2,5 ).

     The answers are similar to 1) or 2)

1) Correction:  

"p(1,2)  and answer does not exist.  Build additional tables while making first 5... tries,"
must read:
"p(2|1)  and answer does not exist. ...."

2) It will be nice to know more about your TM which can lead to unexpected solutions.

For example, if one sells cookies from the bag, then probability of next stop will decrease if the bag is emptied at previous stops. However, if TM is different, like if one sells pens, then capacity of the bag is greater, and probability of next stops increases.

Or, if there are two bus routes skipping different stops, then hidden dependence should be collected in tables of conditional probabilities.

Ok, well I think to give a proper answer, we need to further divulge what my TM actually is.

I am building a rather complex indicator for forex trading markets.  Basically, I have an indicator called "heikenashi", being japanese for "average price bars".  This is a series of bars on a trading chart that are noted by color.  The heikenashi bars have two different directions -- up or down.  The attached picture is the color coded heikenashi bars. Don't worry, I know this isn't a trading forum, the math problem factors into it :)

If we are trading that chart (in the attached picture), the current number we would be on is "10"...which means we are on the 10th red bar in this given series. Each time you start a new series on a new color, your "count" starts over.

What I am looking for, is since we are on number "10" right now in this series, giving historical data, what is the probability we are going to go to the "11"th bar in this series?

I can obviously grab any sort of data need be from the chart.  What information would be needed, if the current amount of information given isn't enough?

Hello ProjectZIP.

The problem still does not look completely specified:
Is it not like "Price Drop Prediction" here:

http://www.landkey.net/Drafts/index.htm

For example, it will be nice to know are these stops equal to "lows in red bar sequence" and is "red sequence" equivalent to your "try"?

Thank you.

thehangman,

  If I use that formula, then we come out with this:

20 - 62.50%
19 -  72.73%
18 -  57.89%
17 -  70.37%
16 -  58.70%
15 -  68.66%
14 -  67.68%
13 -  66.44%
12 -  70.62%
11 -  68.06%
10 -  72.26%
9 -  76.20%
8 -  74.08%
7 -  76.61%
6 -  75.15%
5 -  72.41%
4 -  73.33%
3 -  74.68%
2 -  72.94%
1 -  77.76%

So this means, on average, I have about a 75% chance of continuing on to the next number.  Does this seem correct? o_O

-Kev

Question is still not clear.

Brainstorming.