Math Is Fun Forum

oem7110

Member

Registered: 2010-12-08

Posts: 24

Normal and biased dice

I find an interesting topic about dice on the web, but don't know the answer, does anyone have any suggestions on how to work out this math?

You are given 3 dice. One dice is normal unbiased, one dice is biased for even numbers and the third is biased for odd number.

Biased for even(odd) numbers means that when the dice is rolled, the probability that the number facing up will be even(odd) is 4/5. Also, each of the even(odd) numbers have equal probability of being on top and so do the odd(even).

That is, if you roll the dice biased for even numbers, the probability of having 2 on top is 4/5 x 1/3 = 4/15, while the probability of having 1 on top is 1/5*1/3 = 1/15
In the unbiased dice, each number (1 to 6) has equal probability of being on the top face.

The three dice are rolled together over and over, and their sum (of numbers on the top faces of the three dice) "S" noted each time.  What value of "S" is expected to be most frequent?

Thanks in advance for any suggestions

Bob

Administrator

Registered: 2010-06-20

Posts: 10,184

Re: Normal and biased dice

hi rcwitt,

Did you understand my other post about normal distributions?

I did these tables in MS WORD so hopefully they will display properly.

unbiased(X)    1       2       3      4       5       6
Probability    1/6    1/6    1/6    1/6    1/6    1/6

Biased to even
even bias(X)    1        2        3         4        5          6
Probability    1/15    4/15    1/15    4/15    1/15    4/15

Biased to odd
odd bias(X)     1         2         3         4         5         6
Probability    4/15    1/15    4/15    1/15    4/15    1/15

To calculate the ‘expected’ value you do Σ X.P
For table 1         E = 1/6 + 2/6 + 3/6 + 4/6 + 5/6 + 6/6 = 21/6 = 3.5
For table 2         E = 1/15 + 2x4/15 + 3/15 + 4x4/15 + 5/15 + 6x4/15 = 57/15 = 3.8
For table 3         E = 1x4/15 + 2/6 + 3x4/15 + 4/15 + 5x4/15 + 6/15 = 48/15 = 3.2

The (expected value for the sum) is just the (sum of the expected values) = 3.5 + 3.8 + 3.2 = 10.5
But, take care because you cannot just add the variances.

Last edited by Bob (2010-12-09 05:16:12)

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

oem7110

Member

Registered: 2010-12-08

Posts: 24

Re: Normal and biased dice

I understand your approach on determining the expected value, but I don't understand what I need to take care because I cannot just add the variances.
Could you please give me any further suggestions?
Thank you very much for your suggestions

Bob

Administrator

Registered: 2010-06-20

Posts: 10,184

Re: Normal and biased dice

hi

In statistical theory it is right to add expected values to get the combined value.  This is not true for variances; the calculation is more complex.  You didn't want to know variances so I should have kept quiet about them, but I was just trying to help you avoid a problem in the future. 

So, no need to worry.

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob