The object of the dice game is to be the first player to reach a score of at least 100 points.
Each player’s turn consists of repeatedly rolling a die.
After each roll, the player has two choices: roll again, or stop.
 If the player rolls 1, nothing is scored in that turn and it becomes the opponent’s
turn.
 If the player rolls a number other than 1, the number is added to the player’s turn total and the player’s turn continues.
 If the player stops, the turn total (the sum of the rolls during the turn), is added to the player’s score, and it becomes the opponent’s turn.
What's your strategy?
Strategy inside a concrete game will vary with the livingscore of the other players.
But we can think we are playing to scores high most of the times(in the long run, maximum score with minimum number of turns). We have to maximize the ratio 100/n, con n = number of turns.
One approach could be assume that the statistical sequence of rolls, centred in value 1, would have the general form: v1, v2, v3, 1, v4, v5, (or v1, v2, 1, v3, v4, v5) where v1+v2+v3+v4+v5 = 20. (p. ej: 3+5+4+6+2)
We can distribute (v1, v2) (v3, 1) (v4, v5). If v3 is the medium value of the series excluding 1, we have (4, 1). These score will be 0. It means that a set of three turns fruits a v1+v2+v5+v6 = 16 (medium value of three turns are 16 points). To score 100, we have to plan then 100/16 = 6,25 threesetturns (<19 turns = 18 turns). So we can stop the first turn when we are up 8. We still playing like that, assuming that in each threeturnsset we can accept one unproductive score (score 0) and two scores up 8 (for a medium score > 16/3 = 6 each turn).
If during the game, numbers aren't going like that (f. es. we have had two scores 0 in the three first set) we have to recover points, taking more risk and increasing the quote for next three turns. If f. ex., we have just score 6 in the first three turns, we have to recover 10 points. We have other 5 series of three turns. It means in each series we have to score more than 2 more points, then 18 instead of 16, and so we will stop only if we rolls 9 or up.
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Other form to calculate medium value would be this. Considering all the possible 6! forms of the sequence 1, 2, 3, 4, 5, 6, we can reduce to value 0 what is on the left of value 1 and conserve instead the sum of what is on the right (f. ex. in 342165, only 6+5 =11). If we do this for all the possible forms, we have a medium value of 6,86 points/form).
This would lead to other strategy. For example: to try to be always up a 7 medium value referred to the number of turns we have done. In this case if we are down after rolling n times (< 7*n) we have to take more risks (not stopping) to try to get that level.
Edited on April 22, 2005, 9:13 pm
Edited on April 22, 2005, 9:15 pm

Posted by armando
on 20050422 15:42:51 