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Bajillion Stats


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Bajillion Stats
So, what is the chance of making a roll?

So if the target number is 14, 4 players and using 2 dice.

If you attempted the roll of 14 by 1 player, you would succeed 9.4% of the time.
If all 4 attempted the roll with 2 dice, you would succeed 32.5% of the time.

If instead, all 4 attempted to reduce the roll by 1, you would succeed in reducing it by 1, 32.5% of the time.
This would improve your odds of success per roll from 9.4% to 15.6%. If all attempted, you would get to 49.5% (50/50).
But, losing the first 4 attempts is too much to overcome.

You would succeed in reducing the roll by 2, 17.9% of the time.
This would improve your odds of success per roll from 9.4% to 23.4%. (All attempting the number is now 66%.)

In short, with target number of 14, if everyone attempted the Success all the time roll with 2 dice, it would be 1/3, 1/3 = > 55%.

If you went to reduce the target number by 1 or 2 with a 14 Support number, then 4 attempts at success, the number would only be ~41% (the odds are wonky).

In this case, everyone should just roll for success.

If the Support Number is a 12 then then reduce 4 attempts, then roll Success is ~50%.

If the Support Number is a 10 then then reduce 4 attempts, then roll Success is ~59%.

So, the break even point is about an 11 for a 14 Success Number. Anything greater than that, everyone should just roll for success.

The statistics are a bit more complicated (so I put in the ~ and the binomial distribution in excel is wonky).


Yours,

   IronConrad



Message Replies:
Can we get Two decimal places? -- red (posted: 3/4/2018) 
The Problem of course is predicting N .... -- IronConrad (posted: 3/4/2018) 
 
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