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JayBee
JayBee
JayBee
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December 29th, 2014 at 8:55:43 AM permalink
There is a dice game called "ForGet It" where you roll six dice, each die is numbered 1-5 and the 6th spot is labeled "FOR" on 2 of the dice, "Get" on 2 of the dice and "IT" on 2 of the dice.

Trying to figure out what the odds would be in rolling the complete phrase "FOR" "GET" "IT" on the initial roll out using the 6 dice?

Trying to settle an argument of answers....thanks
SOOPOO
SOOPOO
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December 29th, 2014 at 9:18:39 AM permalink
Quote: JayBee

There is a dice game called "ForGet It" where you roll six dice, each die is numbered 1-5 and the 6th spot is labeled "FOR" on 2 of the dice, "Get" on 2 of the dice and "IT" on 2 of the dice.

Trying to figure out what the odds would be in rolling the complete phrase "FOR" "GET" "IT" on the initial roll out using the 6 dice?

Trying to settle an argument of answers....thanks



11/36 x 11/36 x 11/36 = 1331/46656, or about 1 in 35 initial spins.
wudged
wudged
wudged
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December 29th, 2014 at 9:21:25 AM permalink
(1-(5/6)^2)^3 = 0.02852794924554183813443072702332 about 1 in 35

Edit: soopoo's answer is obviously much simpler!
JayBee
JayBee
JayBee
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December 29th, 2014 at 12:14:49 PM permalink
Thank you, let me just clarify again to make sure everything is correct as described before.

Out of the 36 dice sides, 6 sides are taken up by a word, 1 words per die. Two dice have the word "FOR".....two dice have "GET"...and two have "IT".

How do you get the 11 in the numerator?
PeeMcGee
PeeMcGee
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December 29th, 2014 at 12:44:51 PM permalink
Quote: JayBee

Thank you, let me just clarify again to make sure everything is correct as described before.

Out of the 36 dice sides, 6 sides are taken up by a word, 1 words per die. Two dice have the word "FOR".....two dice have "GET"...and two have "IT".

How do you get the 11 in the numerator?


Since one side of a die (out of six) has a word, the probability to not roll a word with one die is 5/6.

Therefore, probability to not roll the word with two dice is 5/6 * 5/6 = 25/36

Therefore, the probability to roll at least one word is the above number subtracted from 1.
1 – 25/36 = 11/36.
wudged
wudged
wudged
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December 29th, 2014 at 12:54:08 PM permalink
You can also look at it as 6/36 on the first die + 6/36 on the second die = 12/36. However, the combination of both dice showing the word is being counted twice, so you have to subtract one of those duplicates, 1/36, which gives 11/36.
Beardgoat
Beardgoat
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December 29th, 2014 at 1:19:11 PM permalink
I got this game for Christmas except it came with 9 die
JayBee
JayBee
JayBee
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December 29th, 2014 at 2:02:47 PM permalink
Quote: Beardgoat

I got this game for Christmas except it came with 9 die



You must be correct as I was watching and not playing.

Now with the 9 die in action, 3 would have the word "FOR", 3 would have "GET" and 3 would have "IT" (correct me if I'm wrong on the # of words)

so what are the odds now throwing "FOR" "GET" "IT" using the 9 die?


Thanks again for all your replies and answers
PeeMcGee
PeeMcGee
PeeMcGee
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December 29th, 2014 at 2:54:05 PM permalink
Quote: JayBee

You must be correct as I was watching and not playing.

Now with the 9 die in action, 3 would have the word "FOR", 3 would have "GET" and 3 would have "IT" (correct me if I'm wrong on the # of words)

so what are the odds now throwing "FOR" "GET" "IT" using the 9 die?


Thanks again for all your replies and answers


Change the 2 to a 3 in wudged’s equation above.
(1 – (5/6)3)3
= 0.07478

Or about 1 in 13
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