This may sound like more of a General Question, but my last math problem provoked a healthy discussion in MPSIMS, so I’ll start this one here and see if it sticks.
The Cubs radio network/WGN radio have this promotion called the “Cub Foods Attendance Game.” Basically they ask for the seventh caller, who then must guess which of the two Cubs play-by-play announcers will be closer in guessing the attendance at tonight’s game, either Ron Santo or Pat Hughes. During the game, Ron & Pat will write their guesses. Whoever is closer wins, and if the lucky caller guessed the right guy, he/she wins two tickets to an upcoming Cubs game. I won a couple of weeks ago, by guessing Ron Santo.
The kicker is that all of the winners get entered into an end of the year drawing for $10,000. The worst-case scenario is that 162 people guess correctly (162 games), thus making the odds of winning the $10,000 1 in 162. Best-case scenario is that I’m the only winner and my odds of winning the $10,000 is 1 in 1. Of course, neither option is going to happen, so can we determine what the actual odds are likely to be?
Randomness alone would say that contestants are likely to pick the right guy 50% of the time (since they are choosing from two guys), making the odds 1 in 81. BUT, randomness would only rule here IF both men (Ron and Pat) were closest in guessing the attendance in equal numbers. As it stands, however, Pat Hughes tends to guess correctly about three times for every two times Ron does. IOW, Pat dominates by a 3-2 margain.
However, contestants in the game seem to pick Ron or Pat more or less 50/50; if anything they favor Ron slightly, perhaps out of pity.
So, how many people are likely to end up in the end of the year drawing, given a 50/50 likelihood that contestants will pick either guy, and given Pat Hughes’ 3-2 dominance in the game?