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You must create a set of 6-sided dice and play a game with an opponent. The dice are like regular dice, but you can change the numbers on the dice. For example, regular dice have sides 123456. You could create one with sides 244678 if you want.

The game starts with the opponent selecting any one die. After this, you also must select one (also from the same set). If both of you roll the dice, the probability of you rolling a higher number should be at least 60%.

What is the largest set of such dice that can be created?

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    $\begingroup$ I feel like this would be a good question of the math forum. Since it is a probability question and not a puzzle. $\endgroup$ – Steven Walton May 13 '16 at 4:23
  • $\begingroup$ This is indeed a puzzle, one that involves probability. How many pure probability questions begin with "you must create"? Admittedly, this would probably find additional interest at both mathematics.stackexchange and creationism.stackexchange . . . $\endgroup$ – humn May 13 '16 at 8:22
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    $\begingroup$ I fail to see how this is a real puzzle. If any such set exists (and it does), then you can obviously have an arbitrarily large set of such dice. Just add dice with only 0 faces. If your opponent chooses one of these, you win! If your opponent chooses a die from the original set, you win! $\endgroup$ – Édouard May 13 '16 at 8:43
  • $\begingroup$ @Édouard, your comment would be a solution to this puzzle if you gave an example of such a set. $\endgroup$ – humn May 13 '16 at 13:09
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    $\begingroup$ I suggest a modification of the puzzle as follows: After the opponent selects a die, you must select two dice (also from the same set). One of the dice you selected should beat the opponent's die at least 60% of the time, the other die should lose against you opponent's die at least 60% of the time. $\endgroup$ – Johannes May 13 '16 at 17:42
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You can make arbitrarily large sets of dice with this property.

Start with Efron's dice:

  • A: 4, 4, 4, 4, 0, 0
  • B: 3, 3, 3, 3, 3, 3
  • C: 6, 6, 2, 2, 2, 2
  • D: 5, 5, 5, 1, 1, 1

A beats B, B beats C, C beats D, and D beats A with probability $\frac{2}{3}>60\%$.

Now add many copies of die B, each using a different value between 2 and 4. For example, one of these new dice could have 3.5 on every side. (If the numbers need to be integers, multiply everything by a large amount first.) A will still beat all of the B dice with the same probability, and any of the B dice will still beat C. So the strategy is:

  • If the opponent chooses A, you choose D.
  • If the opponent chooses any of the B's, you choose A.
  • If the opponent chooses C, you choose any of the B's.
  • If the opponent chooses D, you choose C.
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  • $\begingroup$ How many sets can you make given: numbers must be 1..6 and integer? $\endgroup$ – Ole Tange May 16 '16 at 6:27

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