20-Sided Dice

1. The "house" rolls two 20-sided dice (each numbered 1-20) and the "player" rolls an identical 20-sided die. If the player rolls a number on his die between the two numbers the house rolled, then the player wins. Otherwise, the house wins (including ties). What is the probability of the player winning?
2. The game is changed slightly: two of the 20-sided dice are rolled, and the player must choose one of them for his number. Then the third die is rolled to determine the house's second number. What should the player's method be for making his choice, and what is the probability of winning now?
3. If the player is given an option of either choosing a house die or rolling his own, what is his best strategy, and what is the probability of winning?

Source: rec.puzzles, with original extensions.

1. The odds for winning the first game are 2280/8000 = 0.285.
2. The odds for winning the second game are 3510/8000 = 0.43875.
3. The odds for winning the third game are 4332/8000 = 0.5415.

Denis Borris' solutions:
CASE#1
Winning Probability : 57 / 200
N = Number of sides on dice:
General Formula: [N^2 - 3N + 2] / [3N^2]      (N>0)

CASE2
Winning Probability : 351 / 800 (strategy: pick best option!)

General formula: [2N^2 - 5N       + 2] / [4N^2]   (N is even)
                 [2N^2 - 5N + 1/N + 2] / [4N^2]   (N is odd)

CASE#3
Winning Probability : 1083 / 2000  (same strategy!)
(Vegas will NEVER allow this!)

General Formula: combination of the "best" of above 2 Cases;
(whoever comes up with one deserves top spot in the 
 "Duisenberg Hall Of Fame", Weird Formulae Section!!)

Anyhoo, these are the "chances" per roll possibility (20 sides case):
(the "player" assumed to soberly pick properly!)
   ROLL 'EM     CHANCES
   1   2-20       261
   2   3-20       234
   3   4-20       208
   4   5-20       184
   5   6-20       161
   6   7-20       140
   7   8-20       120
   8   9-20       104
   9  10-20        94
  10  11-20        90
  11  12-20        90
  12  13-20        88
  13  14-20        84
  14  15-20        78
  15  16-20        70
  16  17-20        60
  17  18-20        48
  18  19-20        34
  19     20        18
                 ====
                 2166
 2166 / 4000 = 1083 / 2000 = winning prob. as reduced fraction.
   
Putting it all as a winning percentage makes for
this interesting(?) table:

   SIDES     CASE#1      CASE#2      CASE#3
      1        0.00        0.00        0.00
      2        0.00        0.00        0.00
      3        7.41       14.81       22.22
      4       12.50       21.88       28.13
      5       16.00       27.20       35.20
      6       18.52       30.56       38.89
      7       20.41       33.24       41.98
      8       21.88       35.16       44.14
      9       23.05       36.76       46.09
     10       24.00       38.00       47.40
     11       24.79       39.07       48.69
     12       25.46       39.93       49.65
     13       26.04       40.69       50.52
     14       26.53       41.33       51.24
     15       26.96       41.90       51.91
     16       27.34       42.38       52.44
     17       27.68       42.83       52.96
     18       27.98       43.21       53.40
     19       28.25       43.56       53.80
     20       28.50       43.88       54.15

Al Zimmermann's solutions:
Assume we are working with n-sided dice, rather than 20-sided dice.

(1)  When one rolls three n-sided dice, there are n ^ 3 possible outcomes.  We
must count how many of these are good outcomes, that is, in how many of these
outcomes the player's die is the middle-valued die.

If any two dice have the same number, it is impossible for the player to have
won.  The number of outcomes in which all three numbers are different is n *
(n - 1) * (n - 2).  By symmetry, the middle-valued die is the player's die in
exaclty one-third of the cases.  So the number of desirable outcomes is n * (n
- 1) * (n - 2) / 3.

Thus, the probability of the player's winning is n * (n - 1) * (n - 2) / (3 *
n ^ 3).  For n = 20, this is 2280 / 8000 = .285.

(2) The player should choose whichever of the two dice has a number closer to
(n+1)/2.  If the two numbers are equidistant from (n+1)/2, he can choose
either die.

The analysis of the exact probability is too complicated for this email, but
the general solution for n-sided dice is

n * (2 * n - 1) * (n - 2) / (4 * n ^ 3).

For n = 20, this is 3510 / 8000 = .43875.

(3) After the first two dice are rolled, the values on those two dice create
three sets of numbers:  numbers less than the lower value, numbers between the
two values, and numbers greater than the higher value.  The player's should
determine which of these three sets of numbers is the largest (that is, has
the most members) and then act as follows:  (1) if the largest of the three
sets of numbers is the set of numbers less than the lower value, the player
should take the lower die for his own, (2) if the largest of the three sets of
numbers is the set of number greater than the higher value, the player should
take the higher die for his own, and (3) if the largest set is the set of
numbers between the two values, the player should choose the value on the
third die as his own.

The probability of the player's winning is given by six different formulae,
one of which should be chosen according to the value of n modulo 6:

Case n mod 6 = 0:  (22 * N ^ 3 - 51 * N ^ 2 + 18 * N + 0) / (36 * n ^ 3)
Case n mod 6 = 1:  (22 * N ^ 3 - 51 * N ^ 2 + 18 * N + 11) / (36 * n ^ 3)
Case n mod 6 = 2:  (22 * N ^ 3 - 51 * N ^ 2 + 18 * N - 8) / (36 * n ^ 3)
Case n mod 6 = 3:  (22 * N ^ 3 - 51 * N ^ 2 + 18 * N + 27) / (36 * n ^ 3)
Case n mod 6 = 4:  (22 * N ^ 3 - 51 * N ^ 2 + 18 * N - 16) / (36 * n ^ 3)
Case n mod 6 = 5:  (22 * N ^ 3 - 51 * N ^ 2 + 18 * N + 19) / (36 * n ^ 3)

For n = 20, the probability is 4332 / 8000 = .5415.

Regards,
Al Zimmermann
New York, New York