Find Probability of Double Throwing 6-Sided Die 5 Times

[Josefk]

For throwing a 6-sided die 5 times, i’m trying to summate the following (only) possible outcomes to make 1.
a) all throws are different
b) two throws are the same (a double)
c) three throws are the same (a treble)
d) four throws are the same (a quartet)
e) all five throws are the same (quintet).

So far i have worked out that:

a) (1/6)^5 . 6!/5!1! = 1/1296
b) 6 . (1/6)^2 . (5/6)^3 . 5!/3!2! = 1250/1296
c) 6 . (1/6)^3 . (5/6)^2 . 5!/2!3! = 250/1296
d) 6. (1/6)^4 . (5/6)^1 . 5!/4!1! = 25/1296
e) 6 . (1/6)^5 = 1/1296

I summated the above, expecting them to come to 1, but they come to 1527/1296

I then realized that i was double counting some of the outcomes because when a treble is thrown there is a chance a double is ALSO thrown and when a double is thrown there is a chance that a treble is thrown, so to deduct these i did

i) chance of throwing a double given that a treble is thrown AND not throwing a quintet=
5 . (1/6)^2 = 5/36 = 180/1296
ii) chance if throwing a treble given that a double is thrown AND not throwing a quintet=
5 . (1/6)^3 = 5/216 = 30/1296

1527/1296 - 180/1296 - 30/1296 = 1317/1296

Can anyone tell me why this still doesn't come to 1?

 

[Opalg]

For throwing a 6-sided die 5 times, i’m trying to summate the following (only) possible outcomes to make 1.
a) all throws are different
b) two throws are the same (a double)
c) three throws are the same (a treble)
d) four throws are the same (a quartet)
e) all five throws are the same (quintet).

So far i have worked out that:

a) (1/6)^5 . 6!/5!1! = 1/1296
b) 6 . (1/6)^2 . (5/6)^3 . 5!/3!2! = 1250/1296
c) 6 . (1/6)^3 . (5/6)^2 . 5!/2!3! = 250/1296
d) 6. (1/6)^4 . (5/6)^1 . 5!/4!1! = 25/1296
e) 6 . (1/6)^5 = 1/1296

I summated the above, expecting them to come to 1, but they come to 1527/1296

I then realized that i was double counting some of the outcomes because when a treble is thrown there is a chance a double is ALSO thrown and when a double is thrown there is a chance that a treble is thrown, so to deduct these i did

i) chance of throwing a double given that a treble is thrown AND not throwing a quintet=
5 . (1/6)^2 = 5/36 = 180/1296
ii) chance if throwing a treble given that a double is thrown AND not throwing a quintet=
5 . (1/6)^3 = 5/216 = 30/1296

1527/1296 - 180/1296 - 30/1296 = 1317/1296

Can anyone tell me why this still doesn't come to 1?

I would use these possible outcomes:

a) all throws are different
b) two throws are the same (a double) and all the others are different
c) three throws are the same (a treble) and both the others are different
d) four throws are the same (a quartet) and the other one is different
e) all five throws are the same (quintet)
f) a double and a treble
g) two doubles and a single.

The probabilities are then

a) 120/1296
b) 600/1296
c) 200/1296
d) 25/1296
e) 1/1296
f) 50/1296
g) 300/1296.

Those add up to 1, as required.