MHB Find Probability of Double Throwing 6-Sided Die 5 Times

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The discussion focuses on calculating the probabilities of different outcomes when throwing a 6-sided die five times. Initially, the user calculated probabilities for outcomes where all throws are different, doubles, trebles, quartets, and quintets, summing to 1527/1296. Upon realizing double counting issues, they attempted to adjust for overlaps between doubles and trebles but still did not reach a total of 1. A revised approach was suggested, introducing additional outcomes like combinations of doubles and trebles, which successfully summed to 1. The final probabilities for the new outcomes were calculated, confirming the correctness of the revised method.
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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?
 
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Josefk said:
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.
 
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