EXERT FROM APPENDIX 1 from Don Featherstone's Battles With Model Soldiers
(The book that got me started.)

"Nothing in these pages is a dictate, no word says you must or you shall do it this way. On the contrary, the book sets out from the very beginning to stimulate the reader to think for himself, and to use what he has read merely as a foundation for efforts and ideas which reflect his own temperament and character. Only in this way will he obtain maximum satisfaction from the hobby of battling with model soldiers."

-Don Featherstone 1918 - 2013

Wednesday, January 20, 2010

What are the odds?

It occured to me out of the Blue today  that I may be off track. If my 20 sided die is marked 0-19 then I'm ok but if its 1-20 then I'm off.  (I'd look but the die in quest lives 75km away, where one day, with the favour of the powers that be, I will also live full time at last)

Take the Greeks, assuming a die number 1 to 20, if a score equal to or greater that 18 will cause a hit then there are 3 scores out of 20 that will hit, 18, 19 & 20 (15% if my math is right) that miss, on the other hand, if the die is number 0-19 then there are only 2 chances. If I half the score to 9 and use a d10 then there are  2 rolls out of 10 that hit or 20% and so on down the line, only the odd factor, cavalry, 15 to hit giving 6 chance out of 20  equal to 3 out of 10.

I now see 3 options (apart from trying harder to buy/borrow/make dice):
a)alter the odds and live with the small relative difference,

b) use d10 and combine with a d6 die roll to indictate whether the result is high (11-20) or low (1-10). In theory this would be done by rolling teh pair of dice at the same time but then I might as well use my 1 20 siders. Since the lowest score that can possible score a hit is 12, I could in effect just roll the d6. 1,2,3 = low, no hit regardless of what the d10 rolls, 4,5,6 is high, roll again and add 10 to the score.

c) Roll a d10 and double the score. (giving possible scores of 2,4,6 etc). Each face of the 10 sided die represents 2 x the number shown and 1 less than that eg 2 = 3 & 4, 9=17,18, 0=19 & 20.    If the score required to hit is exceeded then the hit is scored, if the number is met then you need to roll again 1,2,3 it is the lower number and the hit is missed, 4,5,6 then it is the higher number and the hit counts. So, against hoplites a score of 0 on the d10 would hit, a score of 8 or less would miss and a score of 9 would have to be confirmed witha  50% chance each of missing or hitting.

Perhaps I should have called this post, "Storm in a Teacup"


       

4 comments:

  1. My d20 are numbered 0-19. I've always assumed the 0 stood for 20. What was the problem again?
    Ah yes, teacups.
    Don't lose any sleep over this...

    Pjotr

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  2. Yes Pjotr, this is a perfect example about the advantages of playing games rather than thinking about them!
    -Ross

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  3. Of course, the die might not be where the problem lies. The suggestions made in these comments come quickly to mind, but what was the context of the question? Suppose you got a commercial set of rules which used an icosahedron as a die. I have couple marked 0-10 twice (I painted half the numbers black so I could use it as a D20, but, mark you, it has 2 zeros. But I digress).
    Now, again suppose that this rule set omitted to define the D20 as being marked with 0-19 or 1-20. How do you interpret the rules?
    One way generally to solve this (though it would take time) is to begin a convention of using d6, d10, d20 etc in which the numeral represents its highest score but not necessarily the number of sides it has. So a die markd with 0-19 might be called a d19, rather than a d20. Or maybe we incorporate low-high numbers in the expression: d1-6; d1-10; d0-9; d1-20; d0-19 etc.
    What do you think?
    Cheers,
    Ion

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