bonanova Posted March 30, 2008 Report Share Posted March 30, 2008 A standard cubic fair die has 6 faces with equal probability of 1/6 for each face to show. Let die A be numbered 1 2 3 4 5 6. Let die B be numbered 1 1 2 4 6 7. Let die C be numbered 1 1 3 4 6 7. If A is rolled against B, in 16 cases A is larger; in 15 cases B is larger; in 5 cases it's a tie. If C is rolled against A, in 16 cases C is larger; in 15 cases A is larger; in 5 cases it's a tie. If C is rolled against B, in 15 cases C is larger; in 14 cases B is larger; in 7 cases it's a tie. Thus, A is said to beat B; C is said to beat A; and C is said to beat B. Now, Is it possible for the faces of three dice D, E,and F to be numbered in a way that: D beats E; E beats F; and F beats D? [Edited to make it clear we're talking about three new dice.] Quote Link to comment Share on other sites More sharing options...
0 Guest Posted March 30, 2008 Report Share Posted March 30, 2008 (edited) But I'd have to say no. Oh wait! I completely misunderstood the question. I thought you were asking if the dice you proposed could fit the problem. Hmmm... Now I need to think Edited March 30, 2008 by Noct Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted April 3, 2008 Author Report Share Posted April 3, 2008 But I'd have to say no.Oh wait! I completely misunderstood the question. I thought you were asking if the dice you proposed could fit the problem. Hmmm... Now I need to think OP edited for clarity. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted April 8, 2008 Author Report Share Posted April 8, 2008 Helpful idea. Find sets of three numbers a1 a2 a3 that beat b1 b2 b3 but are beaten by c1 c2 c3. Then label the dice a1 a1 a2 a2 a3 a3 etc. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 9, 2008 Report Share Posted April 9, 2008 I have an answer that works, I believe. Die D = 1,1,5,5,5,5 Die E = 3,3,3,3,3,6 Die F = 2,2,2,2,6,6 Die D is higher than Die E 20/36 possibilities. Die E is higher than Die F 20/36 possibilities. Die F is higher than Die D 20/36 possibilies (12 rolls when die D is a 1 + 8 rolls when die D is a 5 and die F is a 6) Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted April 10, 2008 Author Report Share Posted April 10, 2008 I have an answer that works, I believe.Die D = 1,1,5,5,5,5 Die E = 3,3,3,3,3,6 Die F = 2,2,2,2,6,6 Die D is higher than Die E 20/36 possibilities. Die E is higher than Die F 20/36 possibilities. Die F is higher than Die D 20/36 possibilies (12 rolls when die D is a 1 + 8 rolls when die D is a 5 and die F is a 6) You've got it. The winning totals actually are 20-16, 24-10 [and two ties, with the 6's] and 20-16. Quote Link to comment Share on other sites More sharing options...
Question
bonanova
A standard cubic fair die has 6 faces with equal probability of 1/6 for each face to show.
Let die A be numbered 1 2 3 4 5 6.
Let die B be numbered 1 1 2 4 6 7.
Let die C be numbered 1 1 3 4 6 7.
If A is rolled against B, in 16 cases A is larger; in 15 cases B is larger; in 5 cases it's a tie.
If C is rolled against A, in 16 cases C is larger; in 15 cases A is larger; in 5 cases it's a tie.
If C is rolled against B, in 15 cases C is larger; in 14 cases B is larger; in 7 cases it's a tie.
Thus, A is said to beat B; C is said to beat A; and C is said to beat B.
Now,
Is it possible for the faces of three dice D, E,and F to be numbered in a way that:
D beats E; E beats F; and F beats D?
[Edited to make it clear we're talking about three new dice.]
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