21 – Blackjack Movie
Ξ May 6th, 2009 | → 0 Comments | ∇ Books, Geeky, Maths, Oxford |
Watched 21 last night – not a bad film, in fact, regarding entertainment vs. cost value [it ran us just £3 from Matalan!] it was rather good.
The film is based upon the MIT Blackjack team, and as I’ve read/and-seen quite a lot about them before, I was quite happy to have the film thicken the plot [maybe that should be 'have one'?] – and there’s a nice twist or two at the end. As to the film’s inspiration, you can’t do much better than watch the BBC Horizon documentary on this:
I wish ‘Oxford types’ would get up to stuff like this [of course, they might do (would they tell their lecturers?)]; it’d be so much more fun!
Probability
I was quite pleased to see some probability stuff in the film being partially explained, i.e., their running through the Monty Hall problem [although the implied cleverness of the student here is a bit hard to swallow really].
Anyway, here’s the problem:
Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the other two, goats. You pick a door, say number 1, and the host, who knows what’s behind all of the doors, always opens another door – to reveal a goat – let’s say that’s door number 3. He then says to you, “Do you want to swap?”, i.e., swap your initial choice of door number 1, and change to door number 2? The crux being – is it to your advantage to swap?
In the film, the problem is presented in this clip. By the way, the answer is in this too, so if you want to think about it, get ready to hit the pause button at the 44s mark!
The explanation I find that works quite well here [and I've had my Oxford stats students scratching their heads over this problem initially (as do most people I believe)] is:
Given this scenario: when you pick a door, you’re more likely to pick a goat-door than sole car-door, i.e., you’ve a probability of .66 [or 66% chance if you prefer] of picking one of the two goat-doors vs. the only car-door. Hopefully, that’s obvious.
So, if chance [substitute luck of the draw/fate/the odds/divine-intervention ...] did the right thing here, and you picked a goat-door, you know with a decent probability that the car is behind one of the two remaining doors – but which one? Now, when the host reveals another goat behind one of the two remaining doors, the car’s obviously [again, if the odds etc worked for your initial pick] behind the other door!
Basically, it comes down to this: if you picked a goat-door initially [which you will 66% of the time], by swapping later, you’ll always win the car. Conversely, and given once again that you initially picked a goat-door, if you don’t swap, you’ll lose 66% of the time. Or, one other way … by swapping, you’ll only lose if you picked the sole car-door as your first pick [which you're likely not to have done].
Update: Just found a nice little simulator of this [Internet Explorer only] at http://www.grand-illusions.com/simulator/montysim.htm