So there's a certain Texas-themed steakhouse chain that's in the habit of giving away prizes during the holidays. They let you spin a big wheel with 32 spots on it, and if you're lucky you win free steaks for a year and they make a big hurrah. Last year John Reemtsen (COSI's Web Manager) won the year-long steaks not once, but twice in two attempts. He got another three chances this year and, lo and behold, he will not be wanting for steaks in 2017.
John knew he got very lucky, but he asked me just how lucky he was to get three meaty wins out of five chances, when there was only a 1/32 (3.125%) chance to win at every spin.
Well then, game on.
To work out probabilities you count up all the ways that you get what you want and divide that by all the ways of getting any possible result. Assuming the wheely-spinny thing is fair and each spin is independent, we start by counting up the ways to get 3 wins in 5 attempts.
For example, John could win on his 1st, 2nd, and 3rd attempt. Or he could win on his 2nd, 4th, and 5th attempt. And so on. The situation is small enough that we can just type out all the possible combinations of getting three wins:
There are 10 different ways to win three times with five chances. Of course there are handy mathematical formulae for easily calculating larger problems, and for the curious I recommend looking up "combinatorics".
Now we need all combinations for all results. There are 32 spots on the wheel and 5 different spins, giving 32*32*32*32*32 possible outcomes in total.
So John's final probability of winning 3 times out of 5 is 10/(32*32*32*32*32), or .0000298923%, or about one in three million.
We can all agree that John was a lucky dude. And that he owes me a steak.