5 ms·
You have to be careful with the truth table method. It's easy to get it wrong, as your events might not have equal probability. For example, we'll label the th
by daveFNbuck 7y ago
You have to be careful with the truth table method. It's easy to get it wrong, as your events might not have equal probability.
For example, we'll label the three doors P, L, R, for the prize door, the leftmost goat door and the rightmost goat door. This gives possibilities
1. You choose P, Monty chooses L - switching loses
2. You choose P, Monty chooses R - switching loses
3. You choose L, Monty chooses R - switching wins
4. You choose R, Monty chooses L - switching wins
This seems to indicate that switching gives you a 50/50 likelihood of winning. In fact, this is the correct truth table to use if Monty doesn't know which door has the prize. In that case, these are all equally likely.
Instead, if he does know where the prize is and won't open that door, this truth table is misleading because the cases where switching loses are less likely. If you choose L, he'll definitely choose R, but if you choose P, he has two options to split the probability.
Similarly, if you model Monty as not knowing what's behind the doors, then the truth table you're thinking of doesn't work. Either way, the truth table on its own isn't convincing.
The thing I find more clear is using Bayes' rule, as it shows exactly how the model of Monty's behavior affects the probability that switching is beneficial.
Let H be the event that your door has a prize and G be the event that Monty shows you a goat door. Then P(H|G) = P(G|H) * P(H) / P(G). Obviously, P(H) = 1/3 and (assuming Monty must pick a door other than yours) P(G|H) = 1, so it just comes down to what P(G) is.
If he always shows a goat, then P(G) 1 and you should switch. If he doesn't know what's behind the doors, it's 2/3 and it doesn't matter whether you switch. If he knows but prefers to spoil the prize, it's 1/3 and you definitely have the prize door.
- Scarblac 7y agoBut you have two cases where you choose P, and only one case for each other choice. No wonder you do better than expected.
- daveFNbuck 7y agoThe problem with this logic is that it applies regardless of how we model Monty, and not every model of Monty has you doing better than expected. This reasoning is just as wrong as expecting that your chances are 50/50 in the standard form of the problem.
- hinkley 7y agoI think my statistics prof would have said that they are dependent variables. That was the part that always tripped me up, and I see it trip up plenty of others so I'm in good company, but it's definitely part of my mathematical heartbreak. I loved math in high school and a series of unfortunate events and terrible curricula ruined calculus for me. All I have to show for it is that I have more sympathy for people who 'hate' math (even though I think it's fear sublimated into frustration), and a deep and abiding hatred for Stephen fucking Wolfram. Those italics on Mathematica became my proxy nemesis and I figured out how to break the autocorrect by editing the word just so.
- Sean1708 7y agoI've always preferred to draw it out as a tree of what your your possible choices are at each stage, I find easier to get right and I find that people understand it more intuitively. ┌─────────────┼─────────────┐ G G C ┌┴┐ ┌┴┐ ┌┴┐ Stay│ │Switch Stay│ │Switch Stay│ │Switch G C G C C G
- daveFNbuck 7y agoThat has similar drawbacks of hiding the subtlety behind how Monty is being modeled. The tree you've drawn will tend to lead people to the wrong answer if you ask whether switching improves your odds when Monty doesn't know which door has the car.