Two strong-man competitors, Atlas and Brutus, are smashing large rocks with great blows of a hammer. Their goal is to smash as many rocks as they can with four strikes. They each get one point for each crumbled rock.
An unstruck rock has a 20% chance of crumbling, a rock that has been struck once but didn't crumble has a 40% chance of crumbling on its second strike, a rock struck twice has a 80% chance of crumbling on its third strike, and a rock that survives 3 strikes is guaranteed to crumble on the fourth.
What is the probability that the final score is a tie?
Note: This problem is highly adapted from a regional math competition for students up to grade 9 and calculators were not allowed. The solution was requested to two decimal places.
I'll assume that each party will keep pounding at a given rock until it's crushed. The following program counts for four hammer swings, the probability that one, two, three or four rocks are crushed, starting with all fresh rocks:
% keep pounding till it's crushed strategy
pCrumble=[.2,.4,.8,1];
pN=zeros(1:5); % subscript 1 if for zero crushed
for n=0:4
ways=combinator(4,n,'c');
for i=1:size(ways,1)
pThisWay=1;
w=ways(i,:);
tp=1;prev=0;
for j=1:n
fail=w(j)-prev-1;
for k=1:fail
pThisWay=pThisWay*(1-pCrumble(k));
end
pThisWay=pThisWay*pCrumble(fail+1);
prev=w(j);
end
fail=4-prev;
for k=1:fail
pThisWay=pThisWay*(1-pCrumble(k));
end
pN(n+1)=pN(n+1)+pThisWay;
end
disp([n pN(n+1)])
end
For each number of successful crushes, it adds up the probabilities of all the ways of scoring that number of crushes.
for this many rocks crushed this is the probability
0 0
1 0.576
2 0.3776
3 0.0448
4 0.0016
Adding in the square of each probability and summing these gives
0 0 0
1 0.576 0.331776 tied at 1 each
2 0.3776 0.14258176 tied at 2 each
3 0.0448 0.00200704 tied at 3 each
4 0.0016 0.00000256 tied at 4 each
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0.47636736 any tie
It's almost 50% probable there's a tie.
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Posted by Charlie
on 2023-05-08 14:22:56 |
computer solution
