On Fri, 25 Feb 2000 kim@... wrote:

>
> If your group uses the /random 1 100 method of lottoing loot,
> and all N people in the group are in the lotto, what is the
> average value of the winning roll?

Whoops, I meant median value (the value you need to roll to
have a 50% chance of winning the lotto). The average roll is
skewed towards the low end due to there being more low numbers
than winning numbers for groups > 2.

Anyways, here's the answer:

The winning roll is the highest roll. That means all the
other rolls must be lower than it (duh!). If, say, the
winning roll were 87, that would mean all the other rolls
would have to be below 87. Well, the chance of a roll being
86 or less is 86% (surprise! :). So:

N = number of people in the group
W = odds of a non-winning roll (as a fraction, i.e. 86/100 = 0.86)
W+1 = the winning roll
P = chance of winning

Since W is just the chance of another roll being less than
yours, W^(N-1) is the chance of all other rolls in the group
being less than yours. Since the median is the 50% win mark,
the median winning roll will be a 50% chance of all other
rolls being less than yours, or P = 0.5 (50%)

P = 0.5 = W^(N-1)

W = 0.5^[1/(N-1)], round up since we're really after W+1

Here's how it works out based on the number of people in your
group:

N Winning roll
2 51
3 71
4 80
5 85
6 88
10 94
20 97

So if you're in a group of 6 and the group gets a FBSS, and
you roll an 88, you have a 50/50 chance of winning the lotto.
Likewise, for a group of 6, here are the odds of you winning
if you roll a certian number:

Roll Chance to win
2 0.0000001%
10 0.00059%
20 0.025%
30 0.21%
40 0.90%
50 2.82%
60 7.15%
70 15.64%
80 30.77%
85 41.82%
88 49.84%
90 55.84%
92 62.40%
95 73.39%
98 85.87%
100 95.10%

Seems unfair that you need to roll an 88+ just to have a 50/50
chance to win, eh? Well, take comfort in knowing that
everyone else in the group faces the same odds, and the chance
of one of you winning is 100%. :)

--
John H. Kim
kim@...