19BII Program Question



#5

Hello,

I would like to ask for some help. I have some general knowledge of the solver and formulas but I am stumped on how (or if) to create a simple dice rolling equation.

I am trying to create a solver function that will re-create the following.

NUM DICE? = D
HIT MIN? = H
Number of hits = X

So if I needed to recreate the rolling of 48 dice and show how many results are 5 and 6 cumulatively. or 32 dice and how many results are 4, 5 and 6 cumulatively.

I saw a post from a guy that has done the on a TI-82. He used this formula.

Quote:

ClrHome
--clear the screen

0->X
--the variable x gets reset to 0, x will hold the # of successful hits

Input ("Num Dice? ", D)
--Prompts for # of dice to roll, stores the value in variable D

Input ("Hit Min? ", H)
--Prompts for Minimum number needed for a succesful hit, usually 4,5, or 6, and stores in Variable H)

For (I,1,D,1)
--This is a loop, the Variable I is our incrememted by 1 each time through until it is equal to D, which is the # of dice rolls

If ((int (rand*6+1)>=H)
--if the randomly generated number of 1-6 is greater than variable H, your minimum hit #, then...

X+1->X
--X, your hit count variable, gets incremented by 1

END
--end the loop

Disp X
--print out the number of hits

Quote:
REPLACE THIS TEXT WITH THE QUOTE


#6

This is a bit tricky but should work:

HITS=SUM(I:1:DICE:1:MAX(0:SGN(INT(6xRAN#)+2-HMIN)))
Replace SUM by the summation sign (Sigma)

INT(6xRAN#)  returns an integer from 0 to 5
+2-HMIN converts it to a posive number if the margin is hit
MIN(0:SGN()) returns a 1 if the argument is positive, 0 else
The summation sums up the ones that indicate a hit.

#7

I just found out about the IF function:

HITS=SUM(I:1:DICE:1:IF(INT(6xRAN#)+1<HMIN:0:1))
I didn't enter this version into my calc but it should work the same way and is easier to read.

#8

Hi,

Yes that works great. I appreciate your help.

Hans


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