
Probability and Statistics Basic Probability and Statistics Math Forum 
 LinkBack  Thread Tools  Display Modes 
April 3rd, 2017, 09:53 PM  #1 
Member Joined: Mar 2017 From: Tasmania Posts: 36 Thanks: 2  Dice Prob
A die is 'fixed' so that a certain number appears more often. The probability that a 6 appears is twice the probability of a 5 and three times the probability of a 4. The probabilities of 3,2 and 1 are unchanged from a normal die The probability of getting a 'double with two of these dice. Compare with the probability of getting a double with a normal die. 
April 3rd, 2017, 10:12 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 1,646 Thanks: 836 
have you made any attempts so far? you should at least be able to determine what the probability of each number being rolled is. 
April 4th, 2017, 10:47 PM  #3 
Member Joined: Mar 2017 From: Tasmania Posts: 36 Thanks: 2 
I have done this in a prob distribution table x 1 2 3 4 5 6 P(X=x) 1/6 1/6 1/6 2/22 3/22 6/22 
April 4th, 2017, 11:00 PM  #4  
Senior Member Joined: Sep 2015 From: USA Posts: 1,646 Thanks: 836  Quote:
For "doubles", there is only a single way the roll can occur. So just take the product of the individual probabilities. Last edited by skipjack; April 6th, 2017 at 03:03 PM.  
April 4th, 2017, 11:25 PM  #5 
Member Joined: Mar 2017 From: Tasmania Posts: 36 Thanks: 2 
When you take the product of the individual prob the number is a lot smaller than if the dice is not fixed; surely this is not right.
Last edited by skipjack; April 6th, 2017 at 03:04 PM. 
April 4th, 2017, 11:49 PM  #6 
Senior Member Joined: Jul 2008 From: Western Canada Posts: 3,307 Thanks: 39 
For a fair die, the probability of getting a double is the probability of getting a one on the first roll and a one on the second roll (1/6*1/6), plus the probability of getting a two on the first roll times the probability of getting a two on the second roll (1/6*1/6), and so on... Or 6*(1/6)^2 = 1/6 For the loaded die, the calculation is the same except that you substitute the new probabilities: (1/6)^2 + (1/6)^2 + (1/6)^2 + (2/22)^2 + (3/22)^2 + (6/22)^2 
April 4th, 2017, 11:52 PM  #7  
Senior Member Joined: Sep 2015 From: USA Posts: 1,646 Thanks: 836  Quote:
The probabilities with your unfair dice are $\Large \left\{\frac{1}{36},\frac{1}{36},\frac{1}{36}, \frac{1}{121},\frac{9}{484},\frac{9}{121}\right\}$ These two distributions are identical for $k=1,2,3$ $P[k=3,4unfair]< P[k=3,4fair]$ $P[k=6unfair]>P[k=6fair]$ Last edited by skipjack; April 6th, 2017 at 03:07 PM.  

Tags 
dice, prob 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
If you roll 10sided dice, what is the prob that two dice add up to either 19 or 3?  mepoom  Probability and Statistics  2  September 12th, 2014 06:05 PM 
Dice prob  tiba  Algebra  4  June 12th, 2012 03:23 PM 
Dice prob  daemonlies  Advanced Statistics  3  September 3rd, 2011 07:31 PM 
Prob  jpatche  Advanced Statistics  1  February 26th, 2010 02:45 PM 
Dice probability with multicolored dice pips  Brimstone  Algebra  10  May 30th, 2008 08:36 PM 