How long does a risk die last (on average)?
I wanted to know for my ongoing blog post series (http://dieheart.net/lr-mmz-01/).
TBH steps down the die type on a roll of 1-2. MM on a 1-3.
https://codepen.io/sophiabrandt/full/OgzqzL/
Credit goes to Stefan Matthias Aust for the original TBH version: https://codepen.io/eibaan/full/zZxQzY/
Mine is just a simple fork where I changed one line of code and the text.
https://codepen.io/sophiabrandt/full/OgzqzL/
Ooh handy!
ReplyDeleteWhidou made the calculations once and I have the table somewhere, but it's nice to have a place to try out non platonic dice.
Eric Nieudan Be wary when using this calculator with dice other than {4, 6, 8, 10, 12, 20} because the set of dice used impacts the result. For example, the number of uses of a dR20 is different with the sets {4, 6, 8, 10, 12, 20} and {4, 6, 8, 10, 12, 14, 16, 18, 20}, for obvious reasons.
ReplyDeleteThis calculator uses the set of all dice with less side than the first dR counted down two by two to a minimum of 4, except for dR20 which will skip 18, 16 and 14. Thus, dR11 will use the set {5, 7, 9, 11}, dR16 the set {4, 6, 8, 10, 12, 14, 16} and dR20 the set {4, 6, 8, 10, 12, 20}.
This calculator is simulating a large number of rolls in order to get an a result. This is in my opinion a worse practise than getting to the solution through analytical reasoning. But well, it's good enough for this kind of usage is suppose.
For anyone wondering, any dRX lasts on average X/3 times before being stepped down. Thus a dR4 will last on average 1.33 times. A dR6 will last 2 times before being stepped down to dR4, therefore you can get 3.33 uses out of it (Udr6 + Udr4). A dR8 will step down to dR6 after 2.67 uses, therefore it's equivalent to a total of 6 uses (UdR8 + Udr6 + UdR4) and so on.
Yep yep. It's not really the best method to get a reliable result but for eyeballing it should be enough.
ReplyDeleteGood to know! Thanks Whidou. The things are like magic to me.
ReplyDeleteSo we'll have to wait for dR5s and dR16s in MM :)
Just to be clear this is the average number of times the dice exists, regardless of it's level.
ReplyDeleteIf you want to calculate the average number of times a die remains at that level before it drops it is a simple power series. [ 50% = (sides-3/sides)^(N-1); solve for N ]
So each die will on average last the following rolls before it drops.
d4: 1.5 uses
d6: 2 uses
d8: 2.475 uses
d10: 2.943 uses
d12: 3.409 uses
d20: 5.265 uses
Summing these values produces the number of times a dice can be rolled before it disappears entirely:
d4: 1.5 uses (ie effectively 1 use because rolls are discrete)
d6: 3.5 uses (ie 3 uses)
d8: 5.97 uses (ie almost 6 uses)
d10: 8.91 uses (ie almost 9 uses)
d12: 12.31 uses (ie effectively 12 uses)
d20: 17.58 uses (ie effectively 18 uses) [assuming a d20 drops to d12]
[Note that the higher the number of sides the less precipitous the variation before and after the 50% mark. For example, for the d12 N=3 has a value 56% and N=4 has a value of 42%, which explains the variation with the Monte Carlo method used in the original post, especially as only 1000 die rolls were used.]
For Eric's d5 and d16 the average number of uses it gets before it drops is:
d5: 1.756 uses
d16: 4.338 uses
No idea where these would fit in a chain though. =8)
Ian Borchardt is a magician too! Thanks for pitching in, this is all very useful.
ReplyDeleteThis is super helpful. Thanks!
ReplyDeleteEric Nieudan Break out the DCC dice!
ReplyDeleteThank you! This is very helpful.
ReplyDelete