How to solve probability problems with dice

Probability theory is the branch of mathematics concerned with probability. Although there are several different probability interpretations, probability theory treats the concept in a rigorous mathematical manner by expressing it through a set of axioms.

[eimqn]

how do u test / check if a dice is fair or biased?

[岡安一壽-gy]

Thank you sir.
Many people get the following question wrong, so please explain.
[Question]
Find the probabilities that the following events will occur when two dice are rolled.
①Both show odd numbers
②One shows odd number and another shows evenn number
③Both show even number

P(E):the probability that event E occurs.

【Distinguishable dice A and B】
(the number that dice A shows, the number dice B shows):event
If we only judge whether the numbers indicated by the dice are odd or even, the following four events will occur.
　　(odd, odd) (odd, even)

　　(even, odd) (even, even)
If each event occurs with equal probability, the probability is 1/4.
Therefore, P(①)=1/4, P(②)=1/2, P(③)=1/4 (1)
If we judge the number(1, 2, ···, 6) indicated by the dice, the following 36 events will occur.
(1, 1), (1, 3), (1, 5) (1, 2), (1, 4), (1, 6)
(3, 1), (3, 3), (3, 5) (3, 2), (3, 4), (3, 6)
(5, 1), (5, 3), (5, 5) (5, 2), (5, 4), (5, 6)

(2, 1), (2, 3), (2, 5) (2, 2), (2, 4), (2, 6)
(4, 1), (4, 3), (4, 5) (4, 2), (4, 4), (4, 6)
(6, 1), (6, 3), (6, 5) (6, 2), (6, 4), (6, 6)
If each event occurs with equal probability, the probability is 1/36.
Therfore
P(①)=9×(1/36)=1/4, P(②)=18×(1/36)=1/2, P(③)=9×(1/36)=1/4 (2)
(2) matches (1).

【Indistinguishable two dice】
(odd, even)is the same event as(even, odd).
Thererore, if we only judge whether the numbers indicated by the dice are odd or even, the following three events will occur.
　　(odd, odd)

　　(even, odd) (even, even)
If each event occurs with equal probability, the probability is 1/3.
Therefore
P(①)=1/3, P(②)=1/3, P(③)=1/3 (3)
(1, 3) is the same event as (3, 1).
If we judge the number(1, 2, ···, 6) indicated by the dice, the following 21 events will occur.
(1, 1)
(3, 1), (3, 3)
(5, 1), (5, 3), (5, 5)

(2, 1), (2, 3), (2, 5) (2, 2)
(4, 1), (4, 3), (4, 5) (4, 2), (4, 4)
(6, 1), (6, 3), (6, 5) (6, 2), (6, 4), (6, 6)
If each event occurs with equal probability, the probability is 1/21.
Therefore
P(①)=6×(1/21)=2/7, P(②)=9×(1/21)=3/7, P(③)=6×(1/21)=2/7 (4)
(4) contradicts (3).

[obiwonton]

how to figure long odds of repeater rolls, 2 die total 7 x 7 consecutive rolls. Yes each roll is random event but we rarely see 2/2 x 4 consecutive rolls or 7x7. Does math explain this phenom?