[Guest]

Well, I made up this problem. So if you need any clarification, just ask. I would be glad to explain.

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?

[soop]

It should be either 10 or 11. I think.

[Guest]

I think the average for 1 die (Normal) over a period of throws is 3.5. (The sum of all sides divided by the number of sides). So for the even weighted die you would multiply the even numbers by 4 and the odd by 1 and divide the sum of these numbers by 15 (57/15=3.8) and for the odd die you would multiply the odd numbers by 4 and the even by 1 and divide the sum of these numbers by 15 (48/15 = 3.2). Add the three averages (3.5 + 3.8 + 3.2 = 10.5). I think the answer is 10 or 11.

[Guest]

I used the exact same method as Zeron, so I came to same conclusion.

[Guest]

Unless we are all wrong, that answer is no different than if all 3 dice were unbiased.

Here are some interesting follow-up questions:

A: what is the 2nd most likely number(s) if you only roll the 2 biased dice.

B: how could you make a die that would be biased toward even/odd numbers?

[Guest]

The most likely sums are 10 and 11. The chance is equal for both. This is also true for normal dice. Below are the chances for every combination of both loaded and normal dice.

SUM LOADED NORMAL

3 0,0029630 0,0046296

4 0,0155556 0,0138889

5 0,0244444 0,0277778

6 0,0496296 0,0462963

7 0,0644444 0,0694444

8 0,1022222 0,0972222

9 0,1140741 0,1157407

10 0,1266667 0,1250000

11 0,1266667 0,1250000

12 0,1140741 0,1157407

13 0,1022222 0,0972222

14 0,0644444 0,0694444

15 0,0496296 0,0462963

16 0,0244444 0,0277778

17 0,0155556 0,0138889

18 0,0029630 0,0046296

[Guest]

the answer is not unique

open the foil

read the spoil

yes the answer is 10 or 11.

only that i found my way slightly different and tedious

dice e=even, o=odd, n=neutral

4/15 times; e=2, and 12/15times; o=1,3,5 hence 4 x 12/ 225; e+o=3,5,7

4/15 times; e=4, and 12/15 times; o=1,3,5 hence 4 x 12/ 225; e+o=5,7,9

4/15 times; e=6, and 12/15 times; o=1,3,5 hence 4 x 12/ 225; e+o=7,9,11

now that n can have any value from 1 to 6... the range becomes as follows:

48/225 times; S=4 to 9, 6 to 11, or 8 to 13

48/225 times; S=6 to 11, 8 to 13 or 10 to 15

48/225 times; S=8 to 13, 10 to 15, or 12 to 17..

any number has 16/225 x 1/6 (for the neutral dice) prob. for each set that it appears in

both 10 and 11 appear equally in 7 sets.

now..

the other condition...

one biased dias responds probably and the other biased dias responds uprobably

i.e.

you match.. 2,4,6 with 2,4,6 and 1,3,5, with 1,3,5.

for the total of the three dice to be 10,

e+o should be 4 to 9 (as n ranges 1 to 6) thus,

4 and 6

6 and 4

6 and 6

1 and 1

5 and 5 don't count.. (five out of eighteen possibilities don't count)

for the total to be 11, e+o should be 5 to 10 thus,

2 and 2

6 and 6

1 and 1

1 and 3

3 and 1 don't count.. (five out of eighteen possibilities don't count)

again both 10 and 11 have equal chance. (both biased dice unfavourable is same as both bias dice favourable, only that the probabability is reduced by 16)

Edited September 15, 2009 by cmgogo00

[Guest]

Unless we are all wrong, that answer is no different than if all 3 dice were unbiased.

Here are some interesting follow-up questions:

A: what is the 2nd most likely number(s) if you only roll the 2 biased dice.

B: how could you make a die that would be biased toward even/odd numbers?

paint the numbers 1 3 and 5 with lead and roll this dice into a cube.. you shall have your biased dice

. 3

1 5 6 2

. 4

the opposites add up to 7, so its a true dice as well

Edited September 15, 2009 by cmgogo00

[Guest]

paint the numbers 1 3 and 5 with lead and roll this dice into a cube.. you shall have your biased dice

. 3

1 5 6 2

. 4

the opposites add up to 7, so its a true dice as well

I'm no expert on weighting dice but I don't think you could get such a change in odds (4/5 1/5) from simple lead paint. I have another idea...

Also, nobody answered my 1st question of "A: what is the 2nd most likely number(s) if you only roll the 2 biased dice.", maybe it is too easy.

[Guest]

Unless we are all wrong, that answer is no different than if all 3 dice were unbiased.

Here are some interesting follow-up questions:

A: what is the 2nd most likely number(s) if you only roll the 2 biased dice.

B: how could you make a die that would be biased toward even/odd numbers?

Fill mercury near the even (or odd as desired) numbered sides to "load" the dice.

In fact, casino dice are always transparent to show that the dice are not loaded with mercury or any other material!

Edited September 17, 2009 by DeeGee

[soop]

Geeze, all this maths.

I just looked at it that there are 3 even numbers and 3 odd. If you're more likely to get an odd and an even, the average score between the two loaded dice is 7. That makes it the same as two normal dice. Then with the last dice added, you're likely to get another 3/4 on top.