I am currently working on a problem set discussing Pick 5 lottery problems. In summary, the lottery churns out a number between 00000 and 99999. In a straight bet, you win a greater prize by guessing the exact number, with the digits in the same order. This probability is easier to calculate.
For the box bet however, you win if you guess the same five digits as the winning number but the order does not matter. I am struggling to figure out how to calculate the probability of this.
The straight bet is clearly just (1/10)^5, where I think I am getting confused is that while order is important does it actually have an effect on the probability? After all you still have to guess each digit which is still a 1/10 chance repeated 5 times. Am I over thinking this?
Your probability of winning a box bet depends on the pattern of repeats in the digits you choose. If all $\ 5\ $ of the digits you choose are the same, for instance, then you can only win if you've guessed the exact number, which has probability $\ 10^{-5}\ $. If they're all different, on the other hand, you will win if any of the $\ 5!\ $ permutations of your chosen sequence of numbers comes up. Your probability of winning is therefore $\ 5!\times10^{-5}\ $ in this case. This is the best you can do. Your probability of winning the box bet for all possible patterns of repeated digits is given in the following table \begin{array}{|c|c|c|c|c|} \text{pattern}&\text{probability}\times10^5\\ \hline abcde&5!=120\\ \hline abcdd&3!\times{5\choose3}=60\\ \hline abbcc&{5\choose2}\times{3\choose1}=30\\ \hline abccc&2!\times{5\choose3}=20\\ \hline aabbb&{5\choose3}=10\\ \hline abbbb&5\\ \hline aaaaa&1\\ \hline \end{array}