The Parcheesi is played with 1 die and 4 chips for each player (Two to four, although there are also boards for 6 or 8). The object of the game is that players take all the chips from start to finish. The first to get wins.
But let's what matters (or, if someone offends you, to what matters to me.) What is the minimum number of shifts required to win a game of Parcheesi? This question came to me after riding a number of problems and random chance, as the ability to finish a game taking only five (not eat or be eaten, and without encountering walls, (1 / 6) ^ 60, but they would all tabs in the last box.)
Given that the rules of this game are so ambiguous, I will set one over which guided me:
- No more than two cards in one box.
- If all your chips are in play, always bring out 6 in the die, advance 7 squares instead of 6.
- If you have two tabs in the output box (creating a wall) and you roll a 6, you must move one of those two chips.
- If you roll a 6, given strips again after moving the tab. Should take three 6 in a row in the same turn, you must return the last tab you've moved (the second 6) home and it can not shoot again.
- For put a chip in the goal, you have to get the right number, that is, if you're in the last box, you can only enter with a 1.
- If you put a chip in the goal, go along 10 blocks with any other card (ie it going 10).
- Moreover, in the hypothetical game would neither eat or you eat chips, or you'd find walls / barriers along the road.
Notably, if we exclude the first rule, we can find another item that requires more turns but less runs.
We presume that the board has 68 squares, plus 8 for each aisle end of each color (including the target). Thus, each card would have to travel 71 squares to complete travel (76-5 for which never happens, between the entrance hall and box out, including these 5 outlet box itself, since it is not necessary to remove 1 to reach it.) After these clarifications, let us turn to the development of the game.
is clear that the first printing would be 5. This would pull the first tab (tab 1). Each and every one of these shifts would be 3 rolls, taking 6 in the first two. In the second round , would we get June 2 and 5. With the June 2 would move the information 1 and with 5'd get the 2. The rolls of the next turn ( third ) would be equal (the 6-6-5 is usually with some exceptions). The June 2 would go back to tab 1 and 5 would give the output to the 3. In the fourth turn again 6-6-5, the first 6 would be to move one of the two chips in the box would output (Chart 2, for example), the second 6 would be for 1 tab and with 5'd get the last tab, tab 4. Now all the 6 sacks count as 7. The
fifth turn, the first 6 (good, and 7) go to tab 3, which would be next to the page 4 in the output. The second 7 and 5 would be for the 1. The three runs the sixth turn (7-7-5) would add to the 1. Now, the 1 tab would be missing 10 squares to reach the goal (the total length is 71 slots). Therefore, the seven would turn 7-7-3: The first 7 and 3 would Schedule 1 of the goal, and the second 7 would add to the 2. Sheet 2 would advance the 10 squares obtained by putting the tab 1 on the goal. The next two shifts ( eighth and ninth), 7-7-5, count toward the 2. In the tenth shift happen as in the seventh, 7-7-3, 7 +3 for the record 2, reaching the goal, and 7 for the information 3 (10 per meter and the card 2) . In turn eleven twelve , 7-7-5 again, would move only the 3. The thirteenth turn would be similar to the seventh and tenth: 7-7-2, 7 +2 for the 3 tab tab 7 for 4 (and, obviously, put the tab 10 for 3). The next three turns ( fourteen, fifteen and sixteen ) serve to finish the game finally getting the card 4 at the finish. Would be 7-7-5, 7-7-5 and 7-7-2.
This chart, tab 1, tab 2 , sheet 3 and sheet 4 .
1 - 5
2 - 6 - 6 - 5
3º - 6 - 6 - 5
4º - 6 - 6 - 55º - 7 - 7 - 56º - 7 - 7 - 57º - 7 - 7 - 3 ( +10 )8º - 7 - 7 - 59º - 7 - 7 - 510º - 7 - 7 - 3 ( +10 )11º - 7 - 7 - 512º - 7 - 7 - 513º - 7 - 7 - 2 ( +10 )14º - 7 - 7 - 515º - 7 - 7 - 516º - 7 - 7 - 2
So, the game would end after 16 turns and 46 rolls of winning player. The game I said allowing 4 cards are in the same space would require 18 turns but "only" 45 runs. How to distribute the runs for each tab is very variable, this way is just more organized.
And all this if you do not eat or eat any chips, and that you are no wall along the road. Well, if someone manages to link this sequence of tosses, the probability is (1 / 6) ^ 45, it will not cost much to avoid being eaten and stuff (although doing so would be more by chance than by virtue).
PS: Yesterday I lost a game of Parcheesi of three players, I was last.