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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: CA Posts: 1,303 Thanks: 666 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
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Quote:
 Originally Posted by Posher 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
Assuming independent dice, the probability of a given roll is the product of the individual probabilities times the number of indistinguishable ways that roll can occur.

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: 2,976 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 Thanks from Posher
April 4th, 2017, 10:52 PM   #7
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Quote:
 Originally Posted by Posher 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.
The probability for doubles with 2 fair dice is $\dfrac {1}{36}$ for all of them.

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 $k=1,2,3$

$P[k=3,4|unfair]< P[k=3,4|fair]$

$P[k=6|unfair]>P[k=6|fair]$

Last edited by skipjack; April 6th, 2017 at 02:07 PM.

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