I recently purchased David Parlett's "The Oxford History of Board Games." I did not realize that David created the "Hare & Tortoise" board game that has sold millions over the years.
In chapter 6, he explains that the
cost of movement [in terms of carrots] increases geometrically with the number of spaces moved. To move
2 forward costs 3 units (1+2)
3 forward costs 6 units (1+2+3)
4 forward costst 10 units (1+2+3+4)
. . . and so on.
The beauty of this formula is that there is a total of 63 squares on the board and 65 units of energy to spend. He goes on to say:
From the Start, therefore you can get Home in 64 moves by the expenditure of 1 unit at a time. On the other hand, you could get nearly a quarter of the way round on the first two turns, leaving you well in advance but utterly played out, have spent either all 65 units on moves of ten or four, or 64 moves of seven and eight.
What this amounts to is the player having to find a balance of going to fast or going to slow:
It was at this point in the originally abstract development of the game that the title of Hare & Tortoise naturally suggested itself, since skill consists in not haring ahead so fast as to run out of energy too soon, nor lagging tortoise-like behind so as to be unable to catch up when necessary.
I was thinking of using this geometric formula of moving n forward: (n^2 + n)/2. However, instead of having a large amount of energy units (like 65), I was thinking of having a smaller amount where you would have the probability of loosing an energy unit dependant on the speed you were traveling at by using this geometric formula. For example, if you travel at speed 1 (the max is speed 10 for a total amount of 55 in the equation above) the probability of loosing an energy unit is 1/55. If traveling at speed 2, the probability of loosing an energy unit is 3/55 and so on.
Does anyone have an idea to reflect a geometric distribution of probability using dice? If I had a 55 sided die I could do it, but I don't they exist out there. Anyone have any ideas besides creating 55 cards with different distributions of 1-10 on it?
Thanks,
--DarkDream
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If you don't care to be 100% accurate, you could use percentile dice (2d10) and come reasonably close:
1 = 2%
2 = 5%
3 = 11%
4 = 18%
5 = 27%
6 = 38%
7 = 51%
8 = 65%
9 = 82%
10 = 100%