Given d dice each with n sides, find the probability that when they are rolled at once, there are no two consecutive numbers.
This task may be quite difficult. For a warm-up, try finding the numerators for fixed d such as {1,2,3} or for fixed n such as {2,3,4}.
Note 1: I don't have a formula so much as an algorithm.
Note 2: This problem arose as an attempt to solve http://perplexus.info/show.php?pid=12342 by Larry which uses non-independent cards instead of dice.
(In reply to
simulations for comparison by Charlie)
clc
global sides dice fail succ die
for sides=2:10
fprintf('%3d ',sides)
for dice=2:6
die=zeros(1,dice);
succ=0; fail=0;
addOn(1)
prob=succ/(succ+fail);
fprintf('%3d %8.6f ',dice,prob)
end
fprintf('\n')
end
function addOn(wh)
global sides dice die fail succ
for i=1:sides
die(wh)=i;
if wh<dice
addOn(wh+1)
else
s=sort(die);
good=true;
for j=1:dice-1
if abs(s(j)-s(j+1))==1
good=false;
break
end
end
if good
succ=succ+1;
else
fail=fail+1;
end
end
end
end
2 2 0.500000 3 0.250000 4 0.125000 5 0.062500 6 0.031250
3 2 0.555556 3 0.333333 4 0.209877 5 0.135802 6 0.089163
4 2 0.625000 3 0.343750 4 0.179688 5 0.091797 6 0.046387
5 2 0.680000 3 0.376000 4 0.200000 5 0.107200 6 0.058688
6 2 0.722222 3 0.416667 4 0.223765 5 0.116512 6 0.059714
7 2 0.755102 3 0.457726 4 0.250312 5 0.130719 6 0.067123
8 2 0.781250 3 0.496094 4 0.278809 5 0.147644 6 0.075951
9 2 0.802469 3 0.530864 4 0.308032 5 0.166286 6 0.086265
10 2 0.820000 3 0.562000 4 0.337000 5 0.186100 6 0.097882
and four 5-sided dice yields 125/625 = 1/5.
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Posted by Charlie
on 2021-01-26 11:54:00 |
by algorithm
