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April 3rd, 2017, 08:53 PM   #1
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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.
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April 3rd, 2017, 09:12 PM   #2
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have you made any attempts so far?

you should at least be able to determine what the probability of each number being rolled is.
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April 4th, 2017, 09:47 PM   #3
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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
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April 4th, 2017, 10:00 PM   #4
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Quote:
Originally Posted by Posher View Post
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.
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April 4th, 2017, 10:25 PM   #5
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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.
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April 4th, 2017, 10:49 PM   #6
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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
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April 4th, 2017, 10:52 PM   #7
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Quote:
Originally Posted by Posher View Post
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 $k=1,2,3$

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

$P[k=6|unfair]>P[k=6|fair]$
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Last edited by skipjack; April 6th, 2017 at 02:07 PM.
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