A question about probability.?

Question

A biased die, marked 1 to 6, is such that:
Pr(6)=Pr(2)=Pr(1)
Pr(not 6)=5×Pr(5)+Pr(1)
Pr(3)=Pr(4)+Pr(2)
Find the probability of rolling each number.

Answer 1

Not enough information. There are infinite solutions.
There are basically 5 equations and 6 unknowns:
p(6) = p(2)
p(2) = p(1)
p(not 6) = p(1)+p(2)+p(3)+p(4)+p(5) = 5p(5) + p(1)
p(3) = p(4) + p(2)
p(1)+p(2)+p(3)+p(4)+p(5)+p(6) = 1

5 independent degree-1 equations with 6 unknowns is always linear in any 2 of the variables, so a lengthy computational derivation is unnecessary here: the solution is sufficient.
It is enough that 5x + 2y = 1 (and 2x≥y≥0) and the probabilities of the six dice are:
p(1) = y
p(2) = y
p(3) = 2x
p(4) = 2x - y
p(5) = x
p(6) = y

so for instance, {1/7, 1/7, 2/7, 1/7, 1/7, 1/7} works,
but so does {0, 0, 2/5, 2/5, 1/5, 0}
and also {2/9, 2/9, 2/9, 0, 1/9, 2/9}, etc.
(these last 2 solutions represent the endpoints of the line segment of 5x + 2y = 1 within the boundary constraints 2x≥y≥0
Therefore, we see that 0≤p(1)≤2/9, 0≤p(2)≤2/9, 2/9≤p(3)≤2/5, etc.)

NOTE: Nothing in the question prohibits p(k)=0 (it is certainly possible to construct a biased die which could never come up on a certain face), so I cannot agree with the completeness of the answers which follow that include that supposition.

Answer 2

P(1) = P(2) = P(6) ............. (1)
P(not 6) = 5 P(5) + P(1) .... (2)
P(3) = P(2) + P(4) ............. (3)

Given that there are actually 6 faces, assume that p(n) ≠0

P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1 ... (4)
P(not 6) = P(1) + P(2) + P(3) + P(4) + P(5) = P(1) + 5 P(5) ... (5)
So from (5) P(2) + P(3) + P(4) = 4P(5) ................ (6)
But P(2) + P(4) = P(3) .... (3)
Thus from (3) and (6) 2P(3) = 4P(5) ......................(7)
ie P(3) = 2P(5)

So let P(1) = a
So P(2) = a and P(6) = a
let P(5) = b
and from (7) P(3) = 2b
Thus from (3) P(4) = 2b - a
Thus a + a + b + 2b - a + b + a = 1
ie 2a + 5b = 1
So if a = 1/7 b = 1/7
and P(1) = 1/7
P(2) = 1/7
P(3) = 2/7
P(4) = 1/7
P(5) = 1/7
P(6) = 1/7

So if a = 1/12 b = 1/6
and P(1) = 1/12
P(2) = 1/12
P(3) = 1/3
P(4) = 1/4
P(5) = 1/6
P(6) = 1/12

So , in general, if a = 1/(5n + 2) b = n/(5n + 2); and P(1) = 1/(5n + 2)
P(2) = 1/(5n + 2)
P(3) = 2n/(5n + 2)
P(4) = (2n - 1)/(5n + 2)
P(5) = n/(5n + 2)
P(6) = 1/(5n + 2)
and n is any number such that n > 0.5 (as 2n -1 >0 since P(4) > 0)

Answer 3

P(1) = P(2) = P(6) ............. (1)
P(not 6) = 5 P(5) + P(1) .... (2)
P(3) = P(2) + P(4) ............. (3)

Given that there are actually 6 faces, assume that p(n) ≠0

P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1 ... (4)
P(not 6) = P(1) + P(2) + P(3) + P(4) + P(5) = P(1) + 5 P(5) ... (5)
So from (5) P(2) + P(3) + P(4) = 4P(5) ................ (6)
But P(2) + P(4) = P(3) .... (3)
Thus from (3) and (6) 2P(3) = 4P(5) ......................(7)
ie P(3) = 2P(5)

So let P(1) = a
So P(2) = a and P(6) = a
let P(5) = b
and from (7) P(3) = 2b
Thus from (3) P(4) = 2b - a
Thus a + a + b + 2b - a + b + a = 1
ie 2a + 5b = 1
So if a = 1/7 b = 1/7
and P(1) = 1/7
P(2) = 1/7
P(3) = 2/7
P(4) = 1/7
P(5) = 1/7
P(6) = 1/7

So if a = 1/12 b = 1/6
and P(1) = 1/12
P(2) = 1/12
P(3) = 1/3
P(4) = 1/4
P(5) = 1/6
P(6) = 1/12

So , in general, if a = 1/(5n + 2) b = n/(5n + 2); and P(1) = 1/(5n + 2)
P(2) = 1/(5n + 2)
P(3) = 2n/(5n + 2)
P(4) = (2n - 1)/(5n + 2)
P(5) = n/(5n + 2)
P(6) = 1/(5n + 2)
and n is any number such that n > 0.5 (as 2n -1 >0 since P(4) > 0)

Answer 4

There are basically 5 equations and 6 unknowns:
p(6) = p(2)
p(2) = p(1)
p(6) = p(1)
p(not 6) = p(1)+p(2)+p(3)+p(4)+p(5) = 5p(5) + p(1)
p(3) = p(4) + p(2)

p(1)+p(2)+p(3)+p(4)+p(5)+p(6) = 1

let,
p(1) = y
p(2) = y
p(6) = y

let, p(5) = x
p(4) = z
p(3) = p(4) + p(2) = x + z

p(1)+p(2)+p(3)+p(4)+p(5) = 5p(5) + p(1) = 5x + y
or, y + y + x+z + z + x = 5x + y
or, 3x -y - 2z = 0 .. .(i)

again,
p(1)+p(2)+p(3)+p(4)+p(5)+p(6) = 1
or, y+y+x+z+x+z+y = 1
or, 2x + 3y + 2z = 1 .....(ii)

So we have 3 variables and 2 independent equations.

There are infinitely many solutions

Answer 5

I can only help you with the formula for probability which is :
Probability= no of favorable cases over no or possible cases...
:)Hope i was helpful