I do not generally look at a question another tutor has answered, but since I thought a long time about it over night and came to another way to attack the problem before I knew Andy had answered, I will add my thoughts.
There are 81 possible outcomes and the effect of rolling the 6 needs to be successively eliminated. The table below shows how this happens.
81 xxx6 27 0 => 27 outcomes end in 6 which moves point finally to origin
54 xx6x 18 => of the remaining 54 outcomes 18 have a 6 in the next to last place...can't go to origin
36 6xxx 12 => of the remaining 36 outcomes 12 have an initial 6...can't go to origin
24 x6xx 4 4 => of the remaining 24 outcomes 8 have a 6 in the 2nd position of which 1/2 go to origin
16 xxxx 6 10 => of the last 16 with no 6, 10 cannot go to origin and 6 do
37 44
Therefore a total of 37 of the 81 outcomes go to the origin, i.e. probability is 37/81, approximately .46
Paul M.
tutor
I now think I have the correct answer (not 100% certain), but I think this is correct. I certainly could not have gotten it in the framework of a test unless there was no time limit on the test!
There are 64 = 1296 equally probable outcomes of 4 tosses with a single die.
Of these 216 have a 6 at the last toss, which returns the point to the origin.
Of the remaining 1080 outcomes 180 show a 6 at the 3rd toss, which means the point cannot return to the origin.
Of the remaining 900 outcomes 150 show a 6 at the first toss, which also means the point cannot return to the origin.
Of the 750 remaining outcomes 125 show a 6 at the 2nd toss. In the last 2 tosses there are 3 ways to get a +1 and 2 ways to get a -1 and 2 permutations, that is 12 ways to get a +1 and a -1 in the last 2 tosses, which causes a return to the origin. With the 5 ways to get the 1st toss, there are 60 outcomes which cause a return to the origin.
Now there are 625 remaining possible outcomes and each of these outcomes has no 6. In order to return to the origin 2 +'s and 2 -'s need to occur. There are 4 pairs which cause a -, namely 4,4 4,5 5,4 and 5,5 and these can be placed in the 4 outcome boxes in 4C2 = 6 ways. There are 9 pairs which cause a +, namely 1,1 1,2 13, 2,1 2,2 2,3 3,1 3,2 and 3,3 and these can be placed in the remaining 2 boxes. This means there are 4 * 6 * 9 = 216 outcomes which result in a return to the origin.
In total then there are 216 + 60 + 216 = 492 outcomes which cause a return to the origin for a probability of 492/1296 = 41/108 = .37963.
Report
07/22/18
Paul M.
tutor
Still didn't quite get it right...but now I have it and the answer I confirmed independently by a method for which I used a spreadsheet. I am going to repeat the whole method with the small change which makes the last bit of difference in the result.
The are 64 = 1296 equally probable outcomes of 4 tosses witha single die.
Of these 216 have a 6 at the last toss, which returns the point to the origin at the end.
Of the remaining 1080 outcomes 180 show a 6 at the third toss, which means the point cannot return to the origin at the end.
Of the remaining 900 outcomes 150 show a six at the 2nd toss. In the last 2 tosses there are 3 ways to get a +1 and 2 ways to get a -1 and 2 permutations, that is 12 ways to get a +1 and -1 in the last two tosses which causes a final return to the origin. With 6 ways to get the first toss, there are 72 outcomes which cause the return to the origin.
Of the 750 remaining outcomes 125 show a six at the first toss, which also means there is no terminal return to the origin.
Now there are 625 remaining possible outcomes and each of these outcomes contains no 6. In order to have a return to the origin 2 +'s and 2 -'s need to occur. There are 4 pairs which cause a -, namely 4,4; 4,5; 5,4 and 5,5 and these can be placed in the 4 outcome boxes in 4C2 = 6 ways. There are 9 pairs which cause a +, namely 1,2; 1,2; 1,3; 2,1; 2,2; 2,3; 3,1; 3,2 and 3,3 and these can be placed in the remaining 2 boxes. This means there are 4 * 6 * 9 = 216 outcomes which result in a terminal return to the origin.
In total there are 216 +72 + 216 = 504 outcomes which cause a terminal return to the origin for a probability of 50//1296 = 42/108 = .38889.
This agrees with the value determined independently on a spreadsheet. At last, I am pretty sure this is correct!
Report
07/22/18
Paul M.
07/21/18