Here is what I have learned after applying a little math to the free game that I referenced the other day; that and some knowledge of the game.
I decided to narrow my choices down to a select group of players each day and my selection has nothing to do with batting averages, at least initially. All you need is for your player to get a hit on the day you choose him, or them if you are choosing two. Multiple hits in a game, the longest homerun, or a dramatic hit to win the game have no bearing whatsoever. You need at-bats, chances, and the more the better. One of the reasons Pete Rose is the all-time hits leader, in addition to being a good hitter over his entire career, is that the Reds cemented him in the lead-off batting position forever because of two things, his ability to get a hit and the fact that he was a switch-hitter. He is also the all-time leader for at-bats because of that lead-off position. The man who currently resides in a tie for 2nd place behind Joe DiMaggio’s 56 consecutive games with a hit is none other than Pete Rose at 44 games in 1978. The other guy did it in 1897.
My first criteria is to look for lead-off hitters, but I prefer almost anyone in the top of that day’s line-up, meaning the top three of the batting order. My second criteria is to have a player(s) on a road team. My reasoning behind that choice is that if you choose someone from the home team, and the home team is winning after the top of the 9th inning, uh, game over. All things being equal over time, meaning road teams and home teams win about the same number of games in baseball (researched a long time ago; trust me on that one), the road team hitters at the top of the line-up will see more at-bats than the home teams. This problem has already affected me. I had a player on a home team who was set to lead off the bottom of the 9th but his team was leading the game going into the 9th. I needed the away team to score three runs just to give my guy that extra at-bat as he was 0-4 on the night. Didn’t happen. My streak ended.
Here is what I learned after checking on how those batters do over time. Granted, my sample size is very small and I am still watching this on a daily basis, but the amount of time spent checking these numbers every day going forward is a lot less of a hassle than going backward. After the first two weeks I notice that on average approximately 70% of the three guys who are on the road and at the top of the batting order get a hit on any given day. Last night it was 84%. There were 15 games, which means there were 45 players to choose from and 38 of them got a hit. I have seen the percentage as low as 63% during my first two weeks watching, so there is a big variance. But for now I will use the 70% number to make my point.
In order to tie DiMaggio’s streak you need 56, which means choosing 56 single players in a row, 28 nights of choosing 2 players, or some combination of the two. In the long run, meaning 56 winners, the odds are the same at the starting point regardless of your combination and here is the math:
[1 ÷ (0.7 ^ 56)] – 1 = 472,617,456:1 odds against winning this game. No wonder MLB.com has not been able to find a winner after 15 years. The odds are worse than the PowerBall Lottery. The big difference of course is that this game is free to play. The only cost to you is a sign-up (free), register into the game (free), and spend a little time each day (they advertise that you could spend 15 seconds per day so call that a freebie also), and if you are lucky enough to win then $5.6 million could be yours.
Now recall that I am only using players in the top three positions in the road-team line-up. I am certain that the odds would be a lot worse if you take the total population of all hitters, especially when considering the power hitters who strike out more than the lead-off guys. Getting back to the math, the odds in the long-run are the same regardless of your choices, 1 or 2 players on any given night. But in the short-run you have a problem if you consistently go for 2. This is where conditional probabilities come into play.
If I choose 1 player on any given night then he has ~70% chance of getting a hit based on the numbers that I have observed in which he resides in my chosen population: road team, top of the batting order. I have so far ignored his personal batting average, whether or not he is on a streak of his own, pitching matchup, weather conditions, did he get enough sleep, did I get enough sleep, you name it. Top of the order. Road team.
I choose my first guy who has a 70% chance of getting a hit, 1 of the 31 players who most likely will get a hit on a night when Major League Baseball has a full slate of 15 games, or 45 players in my chosen sweet spot. Now I decide to choose another player. In order for both of them to have a hit, they have to both reside in the population of 31 players to get a hit that night. I have to discount my first guy, which means the 2nd guy has to be 1 of 30 players in the remaining population of 44 players, or 68%.
In order for my streak to stay alive, they both have to get a hit. That drops the probability down as follows:
1 player = 0.7 times number of total players = 70% chance the streak stays alive.
2 players = 0.7 * 0.68 = 0.476 or 48% chance the streak stays alive.
In the long run we’re all dead. No matter how you slice it the odds are worse than PowerBall but it’s free to play.
In the short run you make it that much more difficult by choosing 2 players instead of 1. Sometimes though, like an actual manager who gets to choose players based on situations, you just have to go with your gut and common sense.
Last night the Dodgers were visiting Colorado, a park in which all batters look a lot healthier and pitchers’ ERAs go way up; that and the Dodgers were playing well recently. I had to pick 2.
I was listening to the game on the radio and my players had continued my streak within the first five pitches of the game. It rarely gets better than that.