10 replies to this topic

[w0uter]

Made a diceroll func in java.
Wonderd if it was harder in autoit.

and it was SOOO easy.
but i still decided to post it

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Func _DiceRoll($i_Dices = 1, $i_Sides = 6)
    
;check for valid numbers
    If Not IsInt($i_Dices) Or Not IsInt($i_Sides) Or Round($i_Dices) <> $i_Dices Or Round($i_Sides) <> $i_Sides Then
        SetError(1)
        Return 0
    EndIf
    
;filter out technical impossibilitys
    If $i_Dices <= 0 Or $i_Sides <= 3 Then
        SetError(2)
        Return 0
    EndIf
    
;add the buffer
    Local $i_rnd = 0
    
;roll the dices and add em to the buffer
    For $i_count = 1 To $i_Dices
        $i_rnd += Random(1, $i_Sides, 1)
    Next
    
;return the buffer
    Return $i_rnd
    
EndFunc;==>_DiceRoll

Edited by w0uter, 30 May 2005 - 08:57 PM.

[MSLx Fanboy]

Is there a reason why you multiply $i_Dices by 1 in the first argument in the Random() func

[GaryFrost]

seemed useless to me

$a_dice = _DiceRoll(2)
$s_text = ""
For $i = 1 To $a_dice[0]
    $s_text = $s_text & "Die #" & $i & ": " & $a_dice[$i] & @LF
Next
$s_text = $s_text & "Total: " & $a_dice[UBound($a_dice) - 1]
MsgBox(0,"Rolled",$s_text)

Func _DiceRoll($i_Dices = 1, $i_Sides = 6)
    Dim $a_dices[$i_Dices  + 2], $i
    $a_dices[0] = $i_Dices
    For $i = 1 to $a_dices[0]
        $a_dices[$i] = Random(1, $i_Sides, 1)
        $a_dices[$i_Dices + 1] = $a_dices[$i_Dices + 1] + $a_dices[$i]
    Next
    Return $a_dices
EndFunc

Edited by gafrost, 23 May 2005 - 03:15 PM.

[w0uter]

Is there a reason why you multiply $i_Dices by 1 in the first argument in the Random() func

<{POST_SNAPBACK}>

no not really. first it had 1.
but then i added multiple dices.
and forgot to remove the 1.

ty for pointing that out

Edited by w0uter, 23 May 2005 - 03:39 PM.

[Smed]

gafrost corrected for it, but he didn't point out that yours will return each possible value with a linear probability rather than the nice bell curve you expect from rolling multiple dice. I.e. if you roll 2, 6-sided dice, you are more likely to get a 7 than a 2 or a 12.

[GaryFrost]

took about 30 seconds to put the below together, probably need more items on the wheel, lower the chances, but see lots of potential for the function for making games like below, even the old game of yatzee

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dim $slots[7]
$slots[1] = 'cherry'
$slots[2] = 'bananna'
$slots[3] = 'coin'
$slots[4] = 'watermelon'
$slots[5] = 'lemon'
$slots[6] = 'Lucky 7'
$a_dice = _DiceRoll(3)
$s_text = ""
For $i = 1 To $a_dice[0]
    $s_text = $s_text & "Slot #" & $i & ": " & $slots[$a_dice[$i]] & @LF
Next
MsgBox(0,"1 armed bandit",$s_text)

Func _DiceRoll($i_Dices = 1, $i_Sides = 6)
    Dim $a_dices[$i_Dices  + 2], $i
    $a_dices[0] = $i_Dices
    For $i = 1 to $a_dices[0]
        $a_dices[$i] = Random(1, $i_Sides, 1)
        $a_dices[$i_Dices + 1] = $a_dices[$i_Dices + 1] + $a_dices[$i]
    Next
    Return $a_dices
EndFunc

Edited by gafrost, 24 May 2005 - 12:53 AM.

[w0uter]

gafrost corrected for it, but he didn't point out that yours will return each possible value with a linear probability rather than the nice bell curve you expect from rolling multiple dice.  I.e. if you roll 2, 6-sided dice, you are more likely to get a 7 than a 2 or a 12.

<{POST_SNAPBACK}>

i think this is incorrect. since dices have the same chances for evry side.

Edited by w0uter, 24 May 2005 - 10:32 PM.

[/dev/null]

i think this is incorrect. since dices have the same chances for evry side.

<{POST_SNAPBACK}>

Nope, it is correct. Dices have the same chances if you throw them independently but not if you add the values of each dice, because then you have dependent probabilities.

See the following numbers for a "visual" proof ;-))

1+1 = 2
1+2 = 3
1+3 = 4
1+4 = 5
1+5 = 6
1+6 = 7
2+1 = 3
2+2 = 4
2+3 = 5
2+4 = 6
2+5 = 7
2+6 = 8
3+1 = 4
3+2 = 5
3+3 = 6
3+4 = 7
3+5 = 8
3+6 = 9
4+1 = 5
4+2 = 6
4+3 = 7
4+4 = 8
4+5 = 9
4+6 = 10
5+1 = 6
5+2 = 7
5+3 = 8
5+4 = 9
5+5 = 10
5+6 = 11
6+1 = 7
6+2 = 8
6+3 = 9
6+4 = 10
6+5 = 11
6+6 = 12

value=2 : count= 1
value=3 : count= 2
value=4 : count= 3
value=5 : count= 4
value=6 : count= 5
value=7 : count= 6
value=8 : count= 5
value=9 : count= 4
value=10 : count= 3
value=11 : count= 2
value=12 : count= 1

There are simply more combinations to get a 7 (6/36) than a 2 or 12 (1/36), so the chances to throw a seven are much higher.

Cheers
Kurt

Edited by /dev/null, 24 May 2005 - 11:57 PM.

[MSLx Fanboy]

God bless Mutually-Exclusive probabilities. Hehe, just covered that a few weeks ago in PreCalc, I miss them now that we did limits of functions and such

[w0uter]

ahh i see /dev/null. ty.

so this would be better ?

(also added some checks)

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Func _DiceRoll($i_Dices = 1, $i_Sides = 6)
    
;check for valid numbers
    If Not IsInt($i_Dices) Or Not IsInt($i_Sides) Or Round($i_Dices) <> $i_Dices Or Round($i_Sides) <> $i_Sides Then
        SetError(1)
        Return 0
    EndIf
    
;filter out technical impossibilitys
    If $i_Dices <= 0 Or $i_Sides <= 3 Then
        SetError(2)
        Return 0
    EndIf
    
;add the buffer
    Local $i_rnd = 0
    
;roll the dices and add em to the buffer
    For $i_count = 1 To $i_Dices
        $i_rnd += Random(1, $i_Sides, 1)
    Next
    
;return the buffer
    Return $i_rnd
    
EndFunc ;==>_DiceRoll

Edited by w0uter, 26 May 2005 - 10:36 AM.

[/dev/null]

ahh i see /dev/null. ty.

so this would be better ?

Yes, running your func 10.000 times gives:

value=2 count=259
value=3 count=525
value=4 count=834
value=5 count=1138
value=6 count=1443
value=7 count=1671
value=8 count=1323
value=9 count=1134
value=10 count=819
value=11 count=530
value=12 count=324


See also the post of gafrost in this thread.

Cheers
Kurt