Question 394723
p(sandy wins/she goes first)=0.6
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reasoning
on her first try, she has 1/5 chance of getting the "3" tile and wins 
if she does not get the "3" tile, then clause has 3/4 of not getting the "3" tile since there are only 4 tles left and the game can continue.
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On Sandy's second try, there are only 3 tiles left and for Sandy to get a second chance to win, she must of failed the first time 4/5 chance and Claude must of failed 3/4 chance and for Sandy to win in the second try she has 1/3 chance so Sandy can win on the second try with 4/5*3/4*1/3=1/5
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If Sandy does not win of the second, she gets a third try if she failed on the first and Claude failed oh his first  and sandy failed on her second and cluade failed on his second and finally Sandy tries a 3rd time. This can happen by Sandy failing on first 4/5 chance, Claude failing first 3/4 chance, sandy failing on second 2/3 chance and claude failing on his second with 1/2 chance
or 4/5*3/4*2/3*1/2=1/5
after this there are no more tiles
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So sandy wins by going first if she wins on the first try, or second try or 3rd try = 1/5+1/5+1/5=3/5=0.6