A Cry for Help!!

Reply #8

webmark — Mon, 29 Jan 2007 13:15:06 GMT

Thanks to everyone for their response. Let me be more specific. The software I am using is Ion's MDIEditor and would like to experiment with I1 through I5 filters which can be played under or over the median. I want to play all 5 filters under or over the median or both ways at the same time. As basis of my strategy I want to use the total possible combinations.Hope that is a clearer defination of what i am trying ascertain. Good luck to everyone today.

Norm.

Reply #7

time*treat — Sun, 28 Jan 2007 19:02:19 GMT

KY is absolutely correct. The number of ways is 32.

The next step may be to find out which of those 32 ways gives the best results, and how to do those calculations. I think some of us were anticipating (jumping to) that step. As johnph77 noted, these things can have a finer granularity than on/off .

Reply #6

RJOh — Sun, 28 Jan 2007 08:25:25 GMT

I have a program that I designed myself which allows me to setup parameters that are similar to what others call filters and if they are too restrictive no combinations will be selected but I guess you'll never know until you try.

Reply #5

KY Floyd — Sun, 28 Jan 2007 07:00:19 GMT

The rest of his question wasn't well-worded, but he clearly said each of 5 filters can be played 2 ways. If he was asking how many possible ways he can use the 5 filters with two options for each filter, then it's a simple math question (maybe that's why he chose to put it in the math forum?). If that is what he was asking we have exactly the information we need, and the answer is 32. It doesn't matter what the filters do, or what games they're used for.

If the filters can be used in more... [ More ]

Reply #4

time*treat — Sat, 27 Jan 2007 21:47:37 GMT

If the filters are percentage based, I have a couple of blog entries (partial credit filtering) on doing calculations like that.

Reply #3

johnph77 — Sat, 27 Jan 2007 09:28:15 GMT

The filter webmark may be trying to use may not simply be on-off. For instance, there are in existence 90% filters on sums, numeric positions, and more.

Before anyone tries to help they would have to know specifics - what lottery, what filters?

gl

j

Reply #2

KY Floyd — Sat, 27 Jan 2007 07:09:18 GMT

For anything where you have two possible outcomes for each event, the possible outcomes for multiple events will be 2 raised to (the number of events). In your case it would be 2 to the 5th, or 2*2* 2*2*2, which is 32.

The general formula is (possible outcomes for each event) ^ (number of events). If each filter had 4 possibilities and you had 3 filters there would be a total of 4^3rd combinations or 4*4*4, which is 64.

Reply #1

CARBOB — Fri, 26 Jan 2007 18:02:38 GMT

I get confused myself.

I think it woud be 2x2x2x2x2=32 Don't hold me to this

A Cry for Help!!

webmark — Fri, 26 Jan 2007 17:56:29 GMT

I have 5 filters in my favorite software that I would like to concentrate on. Each filter can be played 2 ways. How many different would I get using 5 filters at a time? I just can't seem to get it right. Would someone please help me