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

  • Thread starter Thread starter Josefk
  • Start date Start date
  • Tags Tags
    Probability
AI Thread Summary
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.
Josefk
Messages
2
Reaction score
0
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?
 
Mathematics news on Phys.org
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.
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top