slide1 l.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
Is it Yahtzee? PowerPoint Presentation
Download Presentation
Is it Yahtzee?

Loading in 2 Seconds...

play fullscreen
1 / 15

Is it Yahtzee? - PowerPoint PPT Presentation


  • 135 Views
  • Uploaded on

Is it Yahtzee?. Exploring Probability with Dice. Yahtzee Probability Times 3. James Jacobs Rebecca Raulie Ron Saucier Elvira Crockett. AIS Summer Teacher Institute June 2002. What is the Probability of Rolling 3 Dice the Same?. In 1 Roll ? In 2 Rolls ? In 3 Rolls ?. Probability .

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'Is it Yahtzee?' - neena


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
slide1

Is it Yahtzee?

Exploring Probability with Dice

yahtzee probability times 3
YahtzeeProbability Times 3

James Jacobs

Rebecca Raulie

Ron Saucier

Elvira Crockett

AIS

Summer Teacher Institute June 2002

what is the probability of rolling 3 dice the same
What is the Probability of Rolling 3 Dice the Same?
  • In 1 Roll ?
  • In 2 Rolls ?
  • In 3 Rolls ?
slide4

Probability

1 Roll = 2.77 %

2 Rolls = 11.26 %

3 Rolls = 21.68%

slide6

import UserInput;

public class Yahtzee

{ public static void main (String args[])

{double a,b,c;

double success=0;

double prob;

for(int i=1; i<=10000; i=i+1){

a=1+(int)(Math.random()*6);

b=1+(int)(Math.random()*6);

c=1+(int)(Math.random()*6);

if(a==b && b==c)

success=success+1;

System.out.print(a);

System.out.print(b);

System.out.print(c);

System.out.println();

}

System.out.println("success"+success);

prob=(success*100)/10000;

System.out.println("prob");

System.out.println(prob);

System.exit (0);

}

}

Java Program

3 Dice

One Roll

slide12

mport UserInput;

public class Yahtzee2

{ public static void main (String args[])

{double a,b,c,d,e,f,g;

double success=0;

double prob;

for(int i=1; i<=2000i00; i++){

a=1+(int)(Math.random()*6);

b=1+(int)(Math.random()*6);

c=1+(int)(Math.random()*6);

if(a==b && b==c)

success=success+1;

else

if(a==b || a==c || b==c)

{d=1+(int)(Math.random()*6);

if(a==d || b==d)

success=success+1;}

else

{e=1+(int)(Math.random()*6);

f=1+(int)(Math.random()*6);

g=1+(int)(Math.random()*6);

if(e==f && f==g)

success+=1;}

System.out.print(a);

System.out.print(b);

System.out.print(c);

System.out.println();}

System.out.println("success"+success);

prob=(success*100)/200000;

System.out.println("prob");

System.out.println(prob);

System.exit (0);

}

}

2 Rolls

slide13

Now for the second roll

Case 1: given that 2 were same from 1st roll

we want Prob (1st roll 2 same) * Prob (new roll matches)

90/216 * 1/6

Prob(Case 1 roll2)=.069

Case 2: given that all three different from 1st roll

we want Prob (1st roll different) * Prob (new roll all 3 same)

120/216 * 6/216

Prob (case 2 roll2)= .0154

note that Prob (A and B)=P(A)*P(B)

----------------------------------------------------------------------

Conclusion:

We add the prob(all 3 same on 1st roll) + prob( 2same on 1st

roll+match on second) + prob (all diff on 1st roll +3 same on second

roll)

.027777+.06944+.0154=.112617 or 11.26% prob. of success within 2 rolls.

The Computer simulation Yahtzee and Yahtzee2 matched perfectly.

The main math models are :

Prob (A and B)=P(A)*P(B)

P(A or B) = P(A U B) = P(A) + P(B) for mutually exclusive events

logic error txt
Logic error.txt

Aside

Programming Note

When we ran Yahtzee2, which finds the prob. of getting all three die the same within

two rolls (where you could keep two from the first roll if they were the same) the program

indicated a 15.8% prob. When I used the formulas to calculate the prob. I came up with

11.261% prob. At first I thought I had made a mistake with the formulas and trusted the computer

but after reworking the problem, I went back to the code. I found a logical error when finding

the case where two die were the same on the first roll and the second roll gave a third die that

matches. I had if(a==d or b=d) then add 1 to the success counter. D is the 4th random number

representing the new roll on try 2. The problem is that a or b could have been the third die that

was not the same as the other two. I corrected the line to read if(a==b && a=d) then success and

listed all three cases. [if(a==c && a=d) and if(b=c && b=d)]

The interesting point to me is that we used a formula to predict what the computer simulation

should give us and when it didn't we could recognize that we had a problem either in the program

or the math model (or solving the math model).