# texas holdem combination checker

• ### Question

• Hi all,

I am trying to make a texas holdem game using VB. I have come accross an issue of finding the highest combination between the players hand and the 5 cards available. Combinations can be found here.

I have tried using this type of method:

```Dim year As Integer = dob.Year 'sets year as constant to disregard the year
Dim zodiacs = {
New With {.first = New Date(year, 1, 1), .[second] = New Date(year, 1, 19), .Zodiac = "Capricorn"},
New With {.first = New Date(year, 1, 20), .[second] = New Date(year, 2, 18), .Zodiac = "Aquarius"},
New With {.first = New Date(year, 2, 19), .[second] = New Date(year, 3, 20), .Zodiac = "Pisces"},
New With {.first = New Date(year, 3, 21), .[second] = New Date(year, 4, 19), .Zodiac = "Aries"},
New With {.first = New Date(year, 4, 20), .[second] = New Date(year, 5, 20), .Zodiac = "Taurus"},
New With {.first = New Date(year, 5, 21), .[second] = New Date(year, 6, 20), .Zodiac = "Gemini"},
New With {.first = New Date(year, 6, 21), .[second] = New Date(year, 7, 22), .Zodiac = "Cancer"},
New With {.first = New Date(year, 7, 23), .[second] = New Date(year, 8, 22), .Zodiac = "Leo"},
New With {.first = New Date(year, 8, 23), .[second] = New Date(year, 9, 22), .Zodiac = "Virgo"},
New With {.first = New Date(year, 9, 23), .[second] = New Date(year, 10, 22), .Zodiac = "Libra"},
New With {.first = New Date(year, 10, 23), .[second] = New Date(year, 11, 21), .Zodiac = "Scorpio"},
New With {.first = New Date(year, 11, 22), .[second] = New Date(year, 12, 21), .Zodiac = "Sagittarius"},
New With {.first = New Date(year, 12, 22), .[second] = New Date(year, 12, 31), .Zodiac = "Capricorn"}}

Return (From q In zodiacs Where (q.first <= dob And dob <= q.[second])).Single.Zodiac 'tests date through parameters```

obviously applying it to my needs as it is from a previous program I made (found it in these forums some time ago, if its yours, shout and i'll reference you ;) )

To identify the cards I have a structure including each cards value, suit and image.

Any help is appreciated, I have tried applying the previous code however, I don't think it is really the right way to do it (if possible??).

Josh

Thursday, July 19, 2012 2:51 PM

### All replies

• Hi Josh,

I didn't found what do you mean.

Have a nice day.

Ghost,
Call me ghost for short, Thanks
To get the better answer, it should be a better question.

Friday, July 20, 2012 6:25 AM