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# Rolling a 1 or 5

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.
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Type: Question • Score: 2 • Views: 9,516 • Replies: 24
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1
Tue 10 Mar, 2009 04:15 pm
114/216 (I think)
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Robert Gentel

1
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.

1
Tue 10 Mar, 2009 04:26 pm
@Robert Gentel,
Ah....

Wanna play "Yahtzee"?
joefromchicago

2
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.
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fresco

1
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
Tue 10 Mar, 2009 05:01 pm
@fresco,
...though I quite like Joe's answer !
0 Replies

Robert Gentel

1
Tue 10 Mar, 2009 05:04 pm
Eh, brainfart on my first step. I needed to reverse the individual die odds and multiply.
fresco

1
Tue 10 Mar, 2009 05:14 pm
@Robert Gentel,
And I have a mistake in visualizing intersecting planes of the cube.
fresco

1
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
Tue 10 Mar, 2009 10:06 pm
@fresco,
fresco wrote:
(in agreement with Joe.)

There's a first time for everything.
markr

1
Wed 11 Mar, 2009 12:38 am
@fresco,
"=148/216 (in agreement with Joe.) "

148 != 150
markr

1
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
Wed 11 Mar, 2009 01:20 am
@markr,
Whoops !
oolongteasup

1
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
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

1
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....

1
Wed 11 Mar, 2009 07:05 am
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
Wed 11 Mar, 2009 07:52 am
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.

1
Wed 11 Mar, 2009 08:01 am
@FreeDuck,
Yes. I will remember that approach in the future.

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