Kevin Spacey in "21" and Strat Cards

[ugrant]

In the movie "21" the professor played by Kevin Spacey presents a scenario to his class: a TV game show has three doors, behind the doors are 1 winning and 2 losing lots. The contestant picks one of the three doors, in this case door #1, hoping that he has randomly picked the winner. Before revealing the winner, the Host opens one of the other two doors, let's say door #2, and reveals it is a loser. The Host then gives the contestant a chance to switch his pick to door #3.

What pick gives the contestant the greater chance of winning, staying with door #1 or switching to door #3?

In the movie, Professor Spacey postulates that switching to door #3 is the correct answer, since during the original pick between the three doors the odds were the contestant had a 1/3 chance of being correct and 2/3 chance of being wrong. In the professor's world those odds remain valid although it is known door #2 is a loser - hence there is a 2/3 chance the contestant wins by switching from door #1 to door #3.

So what does that have to do with Strat? We play with six doors (columns) instead of three, so the odds are a bit different - but the statistical dilemma remains the same.

When making out a lineup, do you decrease your chances of getting back to back OBP (hits, walks, HBP) by placing nearly identical cards (cards having their hits/walks in the same columns) sequentially? Or is it better to alternate cards with hits/walks in different columns?

In Strat there is a 1/6 chance of rolling a particular column, followed by a 1/6 chance of rolling it again on the next roll. That means there is a 5/6 chance of not rolling that column again in sequence.

I'll use an example from the '69 set since that's the only game I'm playing at the moment (the () next to the player show the type card he has, where 1-3 means his hits/walks are in those columns). What lineup makes more sense:

Lineup A Lineup B

Morgan (1-2) Bonds (2-3)
Bonds (2-3) Morgan (1-2)
Reggie (2-3) McCovey (2-3)
McCovey (2-3) Howard (1-3)
Howard (1-3) Jackson (2-3)
Petrocelli (1-3) Petrocelli (1-3)

Lineup A represents what a player not looking at the cards would probably construct, based on overall performance. Lineup B represents the lineup the professor would probably use, based solely on card construction.

There was a huge discussion/debate about this subject back when TSN Strat first started with the 2001 season. One side agreed with Professor Spacey and the other argued vehemently that every roll is independent.

Comments?

[apolivka]

Ummmm, it won't make any difference. It doesn't really work that way, the "paradox" in the movie is very misleading. In order for it to be valuable to switch doors, the host must know what door the prize is in so he can reveal a losing door ON PURPOSE. HAL has know such knowledge, so it's not valuable for you to move things around for your next roll. Each roll occurs as a completely isolated event.

Never rely on Hollywood to provide useful information.

[franky35]

These two cases are not comparable. The Spacey example is trivial: the initial odds of picking a losing door are 2/3. So you pick a door and the odds are 2/3 it is a loser. The game show then reveals one of the losing doors - if the winner is door 1, then the game show can reveal door 2 or door 3; if the winner is door 3, then the game show can only reveal door 2. So the fact that door 3 is not revealed means that it is 2/3 likely to be the winner.

I think this is correct, but it may not be. I wish I could express this mathematically.

If the rules of the game show are: come hell or high water we are going to reveal door 2; then there would be a 50/50 chance of the winner being door 1 or door 3.

In SOM, the three dice are independent. So the chances of each column are the same.