# The Grey Labyrinth is a collection of puzzles, riddles, mind games, paradoxes and other intellectually challenging diversions. Related topics: puzzle games, logic puzzles, lateral thinking puzzles, philosophy, mind benders, brain teasers, word problems, conundrums, 3d puzzles, spatial reasoning, intelligence tests, mathematical diversions, paradoxes, physics problems, reasoning, math, science.

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 [quote="Ghost Post"]Well as I am new you all see that i wanted to reply to Luck of the Draw and not start a new topic. I am sorry for your inconvenience... Also since my post was not formated as i wanted it to be the P formula is a bit of the stuff over/under the P should be at the Sum function of course... Christian Schiess[/quote]
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Ghost Post
Posted: Thu Feb 24, 2000 1:38 pm    Post subject: 1

Well as I am new you all see that i wanted to reply to Luck of the Draw and not start a new topic. I am sorry for your inconvenience...

Also since my post was not formated as i wanted it to be the P formula is a bit of the stuff over/under the P should be at the Sum function of course...

Christian Schiess
Ghost Post
Posted: Thu Feb 24, 2000 1:34 pm    Post subject: 0

As previous posters have pointed out, if you know the suits of the pile there will always be a 5/9 chance that skinny wins. The first poster (sorry forgot the name) also pointed out, that after the first game skinny knows the suits of the piles and hence will always have a 5/9 chance of winning. But let assume the following :

The positions of the three piles are changed after every game.

Then skinny would have to randomly choose one of the other two piles you did not pick. Therefore he could pick the pile which has the 5/9 chance of winning but also the 4/9 chance pile.

so this equals the following probability :

1/2 * 5/9 + 1/2 * 4/9 = 5/18 + 4/18 = 9/18 = 1/2

So if you randomly change the piles after every game, its an absolutely random and thus fair game....

Buts lets assume skinnys opponent is not as good in statistics and doesnt change the piles.... then the probability would be the following : as i just pointed out the first game is all about luck because skinny doenst know the suits of the piles....
In all the other games skinny has a 5/9 chance of winning :

X is the number of played games

X-1
P(of Skinny winning) = 1/(2*X) + Sum((1/x)*5/9) = 1/(2*X) + ((X-1)/X)*5/9
1

That would result in the following scheme :

P1 = 1/2
P5 = 0,5444444
P10 = 0,55
P100 = 0,555
P1000 = 0,5555

Well if we are assuming they play infinite games Skinny will have a 5/9 chance if winning (doh) . Well heres the proof.... That is taking into account his opponent doesnt know statistics, though....

Christian Schiess