Please feel free to remove any offensive comments I have made, and I will refrain from adding comments to this blog.

Pulling it back up again – and especially with language like But how can you possibly think that your simulation can prove that “One is a 4? means “at least one is a 4??? Could it be that you couldn’t figure out how to program anything else? seems a little bit like digging for a dead horse in order to be able to whip it.

I appreciate that you sat down and produced a more accurate simulation; but it seems it might have thrived at least as well in a blog of your own with a link here, not to mention with less randomly thrown insults.

Besides, it’s a 2 year old post – you expect me to still even remember the argument I was making back then?

n4=0;
n7=0;

for i=1:1000000
x=ceil(6*rand(1));
y=ceil(6*rand(1));
z=ceil(2*rand(1));
if (z==2 & y==4)
n4=n4+1;
if x+y==7
n7=n7+1;
end
end
if (z==1 & x==4)
n4=n4+1;
if x+y==7
n7=n7+1;
end
end
end

[(n7/n4)*11 (n7/n4)*6]

In this simulatin z randomly chooses which of the die the “one is a 4″ refers to. But since you randomly assigned the x and y to the dice in the first place, a simpler program would be just as appropriate (and will give the same answer)

n4=0;
n7=0;

for i=1:1000000
x=ceil(6*rand(1));
y=ceil(6*rand(1));
if x==4
n4=n4+1;
if x+y==7
n7=n7+1;
end
end
end

[(n7/n4)*11 (n7/n4)*6]

If the original problem had stated “at least one is a 4″ instead of “one is a 4″, your simulations would be accurate. But how can you possibly think that your simulation can prove that “One is a 4″ means “at least one is a 4″?? Could it be that you couldn’t figure out how to program anything else?

n4=0;
n7=0;

for i=1:1000000
x=ceil(6*rand(1));
y=ceil(6*rand(1));
if (x==4 & y~=4)
n4=n4+1;
if x+y==7
n7=n7+1;
end
end
if (y==4 & x~=4)
n4=n4+1;
if x+y==7
n7=n7+1;
end
end
if (x==4 & y==4)
n4=n4+1;
end
end

[(n7/n4)*11 (n7/n4)*6]

The only trick is to not to count [4 4] more than once. If the reason for this is not obvious, think of the rolls as ordered pairs, the dice as red and green, with 1st dice red, 2nd green.

Ah well, it was nothing amazing, just a simple list comprehension to generate the die roll space, followed by a filter to reduce that to the conditional space, and another to reduce that to the events space.

Have a good one!

Interesting post though, I should have a look into that probability library. Sounds like it could be usefull/interesting.

For fun, here’s another (slightly less than optimal) proof in the form ghci commands followed by results.

– The conditional probability space is:

let space = filter (\(x,y) -> x==4 || y==4) [(x,y)|x x+y == 7) space
-- events = [(3,4),(4,3)]

– The probability is therefore (all events in the space being equally likely):

let prob = (length events)/(length space)
– prob = 2/11

And the reason I do it this way instead of by actual probability analysis is that the core of the matter was a disagreement in the analysis of the problem, where the original author, Heath, claimed that 2/11 was invalid.

But yeah, the purely mathematical reasoning IS cleaner and better.

You could generate all outcomes pretty quick with list comprehension, and then just fold the sucker into a counting function. Heck, if you use a strict foldl, it won’t even consume any more space than that required to hold one outcome while doing the calculation.

Of course you could just as easily just do the math:

X = Outcome of first dice roll
Y = Outcome of second dice roll

P(X=4&&Y=3 || X=3&&Y=4 given X=4||Y=4) =
P(X=4&&Y=3 || X=3&&Y=4)/P(X=4||Y=4)

P(X=4||Y=4) = P(X=4) + P(Y=4) – P(X=4&&Y=4) = 2/6-1/36 = 11/36
(these events are not mutually exclusive, so you have to subtract one copy of the combined event as otherwise you have counted it twice because it is included in each of the independent events)

P(X=4&&Y=3 || X=3&&Y=4) = P(X=4&&Y=3) + P(X=3&&Y=4) = 2/36
(these two events are mutually exclusive, so nothing has been counted twice — equally, the term you would subtract is P(X=4&&Y=3 && X=3&&Y=4)=0 [being obviously impossible])

Therefore, the answer is 2/36 * 36/11 = 2/11.

Heath: I’m putting an answer up at Alon’s.