Simple Statistics problem, just need verification.

Jun 2010
16
4
Ok, so I was playing a dice game with my brother, and he had 3 dice.

On 1 roll he rolled three 1s, and on the next roll three 4s.

I did some quick calculations in my head, and I got the probability of that happening as 1 in 46656... Then realised I may be completely off the mark.

My mental calculations took the root \(\displaystyle 6^3\)x\(\displaystyle 6^3\), or \(\displaystyle 216\)x\(\displaystyle 216\)... I got this because each dice has a 1 in 6 chance of being a certain number (1 through 6), and there were 3 dice each time. I then multiplied the \(\displaystyle 6^3\)s together (obviously also known as \(\displaystyle 6^6\)) to get, after a little thought, 46,656.

This just seems wrong for some reason, please correct me or verify that my reasoning is correct (Rofl)



((Yes, statistics is my weak point, and I'm only 15 anyway (Giggle)))





((Also feel free to be as complicated as you like in any replies, I'm not stupid (Wink)))
 
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mr fantastic

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Ok, so I was playing a dice game with my brother, and he had 3 dice.

On 1 roll he rolled three 1s, and on the next roll three 4s.

I did some quick calculations in my head, and I got the probability of that happening as 1 in 46656... Then realised I may be completely off the mark.

My mental calculations took the root \(\displaystyle 6^3x6^3\), or \(\displaystyle 216x216\)... I got this because each dice has a 1 in 6 chance of being a certain number (1 through 6), and there were 3 dice each time. I then multiplied the \(\displaystyle 6^3\)s together (obviously also known as \(\displaystyle 6^6\)) to get, after a little thought, 46,656.

This just seems wrong for some reason, please correct me or verify that my reasoning is correct (Rofl)



((Yes, statistics is my weak point, and I'm only 15 anyway (Giggle)))





((Also feel free to be as complicated as you like in any replies, I'm not stupid (Wink)))
The probability of rolling three of anything with three rolls is p = (1/6)^3. Therefore the required probability is p^2 and you can check whether this gives the same number that you got.
 
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Jun 2010
16
4
The probability of rolling three of anything with three rolls is p = (1/6)^3. Therefore the required probability is p^2 and you can check whether this gives the same number that you got.
I'll see if I can do this without having to go find my calculator (Wink)

To simplify the (1/6)^3, I assume I just have to know 6^3 which is 216, and therefore (1/6)^3 = 1/216

So p = 1/216 and to find the final probability I simply need to do p^2 which is (1/216)^2 = 1/46656 or 1 in 46,656.

YAY. Thank you ^_^