Laws of Probability - Statistics

[martinh]

If I have 7 six sided dice what is the probability of rolling all '6s'; 6x6's; 5x6's 4x6's; 3x6's; 2x6's, 1x6's - I assume that the probability of rolling 1x6 must be more than 1in1 as each dice has a 1in6 chance of being a six - help please.

[rja211077]

well firstly you can't get a probability of more than "1in1" as that is a certainty. Probabilities should also be expressed as fractions although I appreciate they're hard to express in typing on this site. The probability of rolling 7x6 is 1/6 x 1/6 x 1/6 x 1/6 x 1/6 x 1/6 x 1/6 = 1/279936. I'll get back to you on the others - unless someone beats me to it!

[Hamish]

The calculation you need to do is number of possible outcomes to the power of number of dice. So for 7 6's it is one in 6^7 or one in 279936 as rja211077 says. Assuming you don't mind what the other dice show, the probability for the others will be 6^6, 6^5, 6^4, 6^3, 6^2 and 6^1 (which is 6). If the other dice must not show a six, I think the solution is 5x6^6, 5^2x6^5 etc but I'm not sure..... I need to think about this a bit more too.... Hamish

[rja211077]

i'm ashamed to say this has defeated me. Probability is expressed as a fraction: number of favourable outcomes/no of possible outcomes. The denominator is easy. That is x^y where x is the possible outcomes per die (=6 for a six sided die) and y=no of dice. I can't work out a pattern for the numerator. For six dice the denominator is 6^6 = 46656. I have worked out using excel that the numerator (for six dice) is 15625, 18750, 9375, 2500, 375, 30 and 1 for throwing exactly 0,1,2,3,4,5 and 6 sixes respectively. Help! can anybody spot the pattern that enables derivation of a generic formula!?!?

[(2^2)n3!]

There are 6 possible faces 7 times so the denominator is 6^7. The numerator combinations or combinatorics or combinatorial probability for the number of 6's uppermost are as follows:-  there are 7 faces uppermost showing the number 6 and you must choose all 7 (7C7=7!/7!)=1, there are 6 faces uppermost showing the number 6 (7C6= 7/1)=7, 5 faces ... (7C5=7X6/1X2)=21, 4 faces (7C4=7.6.5/1.2.3)=35, 3 faces (7C3=7.6.5.4/1.2.3.4)=35, 2 faces (7.6.5.4.3/1.2.3.4.5)=21, 1 face

[(2^2)n3!]

There are 6 possible faces 7 times so the denominator is 6^7. The numerator combinations or combinatorics or combinatorial probability for the number of 6's uppermost are as follows:-  there are 7 faces uppermost showing the number 6 and you must choose all 7 (7C7=7!/7!)=1, there are 6 faces uppermost showing the number 6 (7C6= 7/1)=7, 5 faces ... (7C5=7X6/1X2)=21, 4 faces (7C4=7.6.5/1.2.3)=35, 3 faces (7C3=7.6.5.4/1.2.3.4)=35, 2 faces (7.6.5.4.3/1.2.3.4.5)=21, 1 face with a

[(2^2)n3!]

There are 6 possible faces 7 times so the denominator is 6^7. The numerator combinations or combinatorics or combinatorial probability for the number of 6's uppermost are as follows:-  there are 7 faces uppermost showing the number 6 and you must choose all 7 (7C7=7!/7!)=1, there are 6 faces uppermost showing the number 6 (7C6= 7/1)=7, 5 faces ... (7C5=7X6/1X2)=21, 4 faces (7C4=7.6.5/1.2.3)=35, 3 faces (7C3=7.6.5.4/1.2.3.4)=35, 2 faces (7.6.5.4.3/1.2.3.4.5)=21, 1 face with a 6