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I saw this on a video slot machine, there are 13 pigs in a circle like so...
P P
P P
P P
P P
P P
P P
P
Two dice are rolled and whatever number comes up, that many pigs are counted off and the pig you land on turns into a bomb. So let's say you start at the bottom pig and roll a 4-3, so 7 pigs are counted off clockwise and you would land on the top pig on the right and it would turn into a bomb. If you now roll 3-3, you land on the pig at the bottom and it turns into a bomb. If you now roll a 4-3 again, you land on the first pig you were on which is now a bomb and the game is over. So the goal is to land on all 13 pigs without landing on any of them twice or the game is over.
So what is the probability of making it all the way through 13 pigs?
No homework problem, just for fun but I haven't a clue how to solve it.
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Hi Fruityloop;
I ran a simulation of 26 000 000 games and was only successful 104 times. This is a probability of ≈.000004. Suggesting that it is very rare to get through all the piggies.
Without an analytical solution and if the simulation ran correctly, I can bound the answer between
with a 99.7% confidence level., meaning their is 1 chance in 300 that the answer is not bounded by the above inequality.
Admittedly answers provided by simulation are not as pleasing as 3/17 for instance, but in the analysis of very complicated systems they are often the only way to go. I will try to settle the question with a mathematical solution.
Last edited by bobbym (2009-07-18 22:42:31)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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