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April 3rd, 2017, 08:53 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, 09:12 PM  #2 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,493 Thanks: 752 
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, 09:47 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, 10:00 PM  #4  
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,493 Thanks: 752  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 02:03 PM.  
April 4th, 2017, 10:25 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 02:04 PM. 
April 4th, 2017, 10:49 PM  #6 
Senior Member Joined: Jul 2008 From: Western Canada Posts: 3,193 Thanks: 33 
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, 10:52 PM  #7  
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,493 Thanks: 752  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 02:07 PM.  

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