Question 382862
a die is tossed 4 times. what is the odds in favor of the die showing a 6 all four times. 

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P(6 four times) = (1/6)^4  
P(not getting 6 four times) = 1 - (1/6)^4
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Odds in favor of "6 four times" = [1/6^4]/[1 - (1/6^4)] 
= [1/6^4]/[(6^4-1)/6^4]
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= 1:1295
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Cheers,
Stan H.