De Mere's problem
April 26th, 2011 at 11:53:11 PM
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I know the normal way to calculate the probability of rolling at least one six in four rolls: 1 - (5/6)^4 = 0, 517747, but I would like to know how to get the same result by calculating the probability of 1 six + 2 sixes + ...... + 6 sixes.
By using Excel I can easily generate a list of all possible dice rolls and find that there are 500 combinations with 1 six, 150 combinations with 2 sixes, 20 combinations with 3 sixes and 1 with four sixes.
I hope someone will help me in understading the combinatorial reasoning that leads to the above numbers, 500, 150 , 20 and 1.
By using Excel I can easily generate a list of all possible dice rolls and find that there are 500 combinations with 1 six, 150 combinations with 2 sixes, 20 combinations with 3 sixes and 1 with four sixes.
I hope someone will help me in understading the combinatorial reasoning that leads to the above numbers, 500, 150 , 20 and 1.
April 27th, 2011 at 4:28:30 AM
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1 way to get 4 sixes, 1 way to arrange them = 1
5 ways to get 3 sixes (1*1*1*5), 4 ways to arrange them. 5*4 = 20
25 ways to get 2 sixes (1*1*5*5), 4 choose 2 = 6 ways to arrange them. 25*6 = 150
125 ways to get 1 six (1*5*5*5), 4 choose 3 = 4 ways to arrange them. 125*4 = 500
5 ways to get 3 sixes (1*1*1*5), 4 ways to arrange them. 5*4 = 20
25 ways to get 2 sixes (1*1*5*5), 4 choose 2 = 6 ways to arrange them. 25*6 = 150
125 ways to get 1 six (1*5*5*5), 4 choose 3 = 4 ways to arrange them. 125*4 = 500
Wisdom is the quality that keeps you out of situations where you would otherwise need it
April 27th, 2011 at 5:14:22 AM
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It's a binomial distribution.
Sum (1 six + 2 sixes + 3 sixes + 4 sixes)
= (1/6)(5/6)^3(4 choose 1) + (1/6)^2(5/6)^2(4 choose 2) + (1/6)^3(5/6)(4 choose 3) + (1/6)^4(5/6)^0(4 choose 4)
= (1/6)(125/216)(4) + (1/36)(25/36)(6) + (1/216)(5/6)(4) + (1/1296)(1)(1)
= 500/1296 + 150/1296 + 20/1296 + 1/1296
= 671/1296 = .51775
For s number of sixes, each term is [(1/6)^s][(5/6)^(4-s)][4 choose s].
Sum (1 six + 2 sixes + 3 sixes + 4 sixes)
= (1/6)(5/6)^3(4 choose 1) + (1/6)^2(5/6)^2(4 choose 2) + (1/6)^3(5/6)(4 choose 3) + (1/6)^4(5/6)^0(4 choose 4)
= (1/6)(125/216)(4) + (1/36)(25/36)(6) + (1/216)(5/6)(4) + (1/1296)(1)(1)
= 500/1296 + 150/1296 + 20/1296 + 1/1296
= 671/1296 = .51775
For s number of sixes, each term is [(1/6)^s][(5/6)^(4-s)][4 choose s].
April 27th, 2011 at 9:48:26 AM
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Thank you very much.
April 27th, 2011 at 11:20:21 AM
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Quote: chapmanBy using Excel I can easily generate a list of all possible dice rolls and find that there are 500 combinations with 1 six, 150 combinations with 2 sixes, 20 combinations with 3 sixes and 1 with four sixes.
I hope someone will help me in understading the combinatorial reasoning that leads to the above numbers, 500, 150 , 20 and 1.
For these type of simple problems, using Excel, one does not have to generate all possible rolls and then count the ways. There is an easier and faster way.
The combinatorial way can be a difficult way to arrive at the answer, unless you really understand it, and the very easiest way to introduce any number of errors into the final answers. At least this speaks for me. I understand the combinatorial reasoning, but my brain most times just goes blank.
Since you now understand the combinatorial way, let us use Excel to verify the answers using Excel's BINOMDIST function.
Col A
# of 6s Prob ways Prob formula
0 0.482253086 625 =BINOMDIST(A2,4,1/6,0)
1 0.385802469 500 =BINOMDIST(A3,4,1/6,0)
2 0.115740741 150 =BINOMDIST(A4,4,1/6,0)
3 0.015432099 20 =BINOMDIST(A5,4,1/6,0)
4 0.000771605 1 =BINOMDIST(A6,4,1/6,0)
total possible sequences: 6^4 = 1296
ways = Prob * 1296
April 27th, 2011 at 10:35:01 PM
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Yes, I fully understand what you write.
Until I saw 7outlineaway's answer I did not realize that it is a binomial distribution. I knew that I did not have to generate all possible rolls (by Excel) by that was my attempt (before asking here) in trying to understand the combinatorics.
Until I saw 7outlineaway's answer I did not realize that it is a binomial distribution. I knew that I did not have to generate all possible rolls (by Excel) by that was my attempt (before asking here) in trying to understand the combinatorics.
April 27th, 2011 at 11:21:08 PM
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Very good.Quote: chapmanYes, I fully understand what you write.
Until I saw 7outlineaway's answer I did not realize that it is a binomial distribution. I knew that I did not have to generate all possible rolls (by Excel) by that was my attempt (before asking here) in trying to understand the combinatorics.
For an interesting read on De Mere's actual question that it was not advantageous to wager
on a double six being rolled at least one time in 24 rolls of a pair of dice.
Here is a link to a free sample pdf of the book "Understanding Probability". worth reading.
http://personal.vu.nl/h.c.tijms/sample.pdf
Page 2 and 3 in the Introduction.
Understanding Probability
Henk Tijms
