2
   

Rolling a 1 or 5

 
 
Reply Tue 10 Mar, 2009 04:01 pm
What is the probability that out of three dice rolled, any of the dice will be either a 1 or 5, excluding three ones or three fives.
  • Topic Stats
  • Top Replies
  • Link to this Topic
Type: Question • Score: 2 • Views: 9,355 • Replies: 24
No top replies

 
DrewDad
 
  1  
Reply Tue 10 Mar, 2009 04:15 pm
114/216 (I think)
0 Replies
 
Robert Gentel
 
  1  
Reply Tue 10 Mar, 2009 04:21 pm
@babemomlover,
Doing this quickly so I may have mistakes and I'm assuming 6 sided dice:

There is a 2/6 chance each die will roll either 1 or 5. So there is a 1/1 chance that one will be a 1 or 5.

There is a 1/108 chance of rolling 3 1s or 3 5s. So there's a 107/108 chance of the situation you describe.
DrewDad
 
  1  
Reply Tue 10 Mar, 2009 04:26 pm
@Robert Gentel,
Ah....

Wanna play "Yahtzee"?
joefromchicago
 
  2  
Reply Tue 10 Mar, 2009 04:36 pm
@babemomlover,
babemomlover wrote:

What is the probability that out of three dice rolled, any of the dice will be either a 1 or 5, excluding three ones or three fives.

The easiest way to figure this out is by determining how many results do not contain a 1 or a 5. There are 216 possible combinations of three six-sided dice (6x6x6). There are four numbers that do not yield either a 1 or a 5, so there are 64 results that contain neither a 1 nor a 5 (4x4x4). So 216 total results minus 64 results equals 152. The 1-1-1 and 5-5-5 combinations come up once each out of 216 results, so subtract 2 from 152 yields 150 total results where either a 1 or a 5 is rolled but not triples. The probability, then, of that result is 150/216, or roughly 69% of the time.
0 Replies
 
fresco
 
  1  
Reply Tue 10 Mar, 2009 04:52 pm
@babemomlover,
Assuming 6 face dice, these would generate a 3D sample space of 216 cubes of 2 would be the excluded results (1,1,1) and (5,5,5).
Trying to visualize it
I think p can be calculated by (36+36+30+30+24+24 -2)/216

P=178/216=0.824
fresco
 
  1  
Reply Tue 10 Mar, 2009 05:01 pm
@fresco,
...though I quite like Joe's answer !
0 Replies
 
Robert Gentel
 
  1  
Reply Tue 10 Mar, 2009 05:04 pm
@DrewDad,
Eh, brainfart on my first step. I needed to reverse the individual die odds and multiply.
fresco
 
  1  
Reply Tue 10 Mar, 2009 05:14 pm
@Robert Gentel,
And I have a mistake in visualizing intersecting planes of the cube.
fresco
 
  1  
Reply Tue 10 Mar, 2009 05:24 pm
@fresco,
.....p=(36+36+24+24+15+15-2)/216

=148/216 (in agreement with Joe.)
joefromchicago
 
  1  
Reply Tue 10 Mar, 2009 10:06 pm
@fresco,
fresco wrote:
(in agreement with Joe.)

There's a first time for everything.
markr
 
  1  
Reply Wed 11 Mar, 2009 12:38 am
@fresco,
"=148/216 (in agreement with Joe.) "

148 != 150
markr
 
  1  
Reply Wed 11 Mar, 2009 12:41 am
@babemomlover,
"What is the probability that out of three dice rolled, any of the dice will be either a 1 or 5, excluding three ones or three fives."

Are you playing 10,000 (AKA Farkle or Dix Mille)?
0 Replies
 
fresco
 
  1  
Reply Wed 11 Mar, 2009 01:20 am
@markr,
Whoops !
oolongteasup
 
  1  
Reply Wed 11 Mar, 2009 01:22 am
@joefromchicago,
P(none of 1 or5) is 3C0*(2/6)^0* (4/6)^3 = 64/216
P(exactly one of 1 or 5) is 3C1*(2/6)^1*(4/6)^2=96/216
P(exactly two of 1 or 5) is 3C2*(2/6^2*(4/6)^1=48/216
P(exactly three of 1 or 5) is 3C3*(2/6)^3*(4/6)0=8/216

ergo 144/216 , funny how the probabilities add to 1
fresco
 
  1  
Reply Wed 11 Mar, 2009 01:44 am
@fresco,
The problem of visualizing the allowable cells in the 3D sample space is indeed simplified by a subtraction method by placing the items 1 and 5 next to each other on the 6x6x6 cube, This gives a 4x4x4 group to take out of 216 together with the disallowed (1,1,1) and (5,5,5) cells . I concur p=150/216
0 Replies
 
DrewDad
 
  1  
Reply Wed 11 Mar, 2009 06:54 am
@oolongteasup,
oolongteasup wrote:

P(none of 1 or5) is 3C0*(2/6)^0* (4/6)^3 = 64/216
P(exactly one of 1 or 5) is 3C1*(2/6)^1*(4/6)^2=96/216
P(exactly two of 1 or 5) is 3C2*(2/6^2*(4/6)^1=48/216
P(exactly three of 1 or 5) is 3C3*(2/6)^3*(4/6)0=8/216

ergo 144/216 , funny how the probabilities add to 1

OK, explain to me how there are eight ways to roll three ones or three fives....
DrewDad
 
  1  
Reply Wed 11 Mar, 2009 07:05 am
@DrewDad,
To explain my methodology:

There are 216 possible combinations.

1 1 1
1 1 2
1 1 3

etc. through

6 6 4
6 6 5
6 6 6

There are 36 combinations for each beginning number (1 1 1 through 1 6 6).

So:

35 combinations starting with "1" that match the criteria. 35 combinations starting with "5" that match the criteria. 20 combinations for each of the other starting numbers (2,3,4,6)

For total odds of 150/216. Which doesn't match my original probability, because I originally made a fatal flaw in my math.
FreeDuck
 
  1  
Reply Wed 11 Mar, 2009 07:52 am
@DrewDad,
I get the same probability by taking all possible combinations (216) and subtracting out all cases of no 1s or 5s (4^3) and subtracting out the two cases of all 1s and all 5s. 216-64-2 = 150 for 150/216.
DrewDad
 
  1  
Reply Wed 11 Mar, 2009 08:01 am
@FreeDuck,
Yes. I will remember that approach in the future.
 

Related Topics

Poker odds - Question by boomerang
what is the probability of... #4 - Question by w0lfshad3
Texas Holdem - Calculating your odds quickly - Discussion by Robert Gentel
How did bookmaker calculate odds ? - Question by CAPABLE2KNOW
Calculate odds of 16 sided dice - Question by gamblingdude1234
The birthday paradox with a twist - Question by Justonequestion
Fancy a punt on the Royal baby's name? - Discussion by izzythepush
Even money - Question by ascribbler
What are the odds? - Question by Nerms
Twelve to one agin the field - Question by Maggie Tong
 
  1. Forums
  2. » Rolling a 1 or 5
Copyright © 2021 MadLab, LLC :: Terms of Service :: Privacy Policy :: Page generated in 0.03 seconds on 09/24/2021 at 12:44:08