You are playing a pick 4 numbers game. For a "straight" you much get all four numbers in the exact order . (ex: you bet 4119, and can only win if 4119 is selected) Odds of winning are 1 in 10,000. For a "Box" there are four ways of winning. (ex: your # is 3338, but you win if the number selected is 3338, 3383, 3833, or 8333) the odds of winning this way are 1 in 2, 500. What are the probabilities of winning each type of game?

Question

You are playing a pick 4 numbers game.  For a "straight" you much get all four numbers in the exact order . (ex: you bet 4119, and can only win if 4119 is selected) Odds of winning are 1 in 10,000.  For a "Box" there are four ways of winning.  (ex:  your # is 3338, but you win if the number selected is 3338, 3383, 3833, or 8333) the odds of winning this way are 1 in 2, 500.  What are the probabilities of winning each type of game?

anonymous · Answer

you are given the odds as 1:2,500 which means you win one time, lose 2,500 times, probability is $\frac{1}{2501}$

anonymous · Answer

@satellite73 that's how I did the problem too, but was told it was wrong.  it somehow has to do with either a permutation or a combination.

anonymous · Answer

I think I may have figured this out.  If anyone is so inclined to check my work, that would be great.  I forgot to mention in the original problem that for the box, 3 out of your four numbers will be repeated.  
So....
For the Straight, since there is only one way to win=1/10000= .01% probability of winning
4 way box= 10*1*1*9=90
90/2500=3.6% probability.