This is a good way of putting it and I think would resonate with some people!
I've not been able to get my dad to understand why you should switch with 1,000,000 boxes because he refuses to believe the 3 box and 1,000,000 box case is the same. I can't even convince him a 3 and 5 box problem are the same.
I actually think I could convince him switching is correct using this strategy because I suspect if he thinks he's tricking or cheating at the game he'll be more open to the explanation.
I think the only way to demonstrate it to lay people is to rephrase it as the odds of picking the wrong one right off the bat. If it's 2/3 odds that you chose wrong, and I remove one card, did that change the odds that you already chose wrong?
The trick is focusing on the 1/3 odds of winning, and not mentioning there is an alternate (and correct) way of looking at the question - the initial odds of losing.
My problem with Monty Hall is that it’s not a repeated game. Average probabilities don’t help me.
This is the point Taleb makes in a lot of his books. You are not the average probability. You only get one play through [life] and the only thing that matters is how the dice fall for you. Even a 90% average chance of getting $prize, is still a 10% chance of walking away with nothing and you better be ready to get that one because you very well might.
The strategy to increase your average probabilities works in poker. Because you play multiple hands and even multiple games. In Monty Hall, you’re just making a guess and no matter how clever you are about it, it’s still just a guess. There is no strategy because the game doesn’t last long enough for strategy to matter.
Now if you had an iterated monty hall, then yes, bring out the maths.
> The possibility that you get nothing is not an argument to ignore the context.
Ah but I’m not ignoring the context. I’m expanding the context beyond “fun math puzzle” to “wow this is a game not worth playing because there’s no edge”.
Again, only applies to the non-iterated version. If you find an analog of monty hall that gives you many at-bats, play away.
Well, even if I try to embrace your argument — life itself is an iterated version of a more general decision game. It doesn't have to be Monty Hall specifically.
Over the course of your life, if whenever you have a choice to make you choose the lower probability strategy (here with the Monty Hall example it's not even an expected-value vs. variance tradeoff, switching is just all-around more optimal) you will most definitely end up worse off in the end.
That's not correct, because if you agree that one kind of result tends to happen more than the other, then it is easier that the single game that you are currently playing belongs to the larger group. When you play once, it is like if you were randomly choosing a game of all the possible ones that you could play, in which case you are more likely to choose one of those that are more numerous.
It's the same reasoning that you would apply in other aspects of life. Imagine you have to travel to some town and there are only two possible roads: A and B. The stats tell that on road A there tend to occur 100 fatal accidents per year, while on road B only one accident has been registered in the last 10 years.
The causes of that disparity may be multiple, like, for example, that road A has many curves with precipices while B does not. But you don't even need to know those causes; the starts themselves tell you that as result it is easier to get a falal accident on road A than on road B. But, of course, it always exists the possibility that you survive in any of them, and also that you have an accident in any of them.
So, if you were to travel through one of them just once and had to choose, I don't think you would say that those stats don't matter only because you will travel once. You would prefer to take the safer road B.
In Monty Hall game, deciding to stay would be like taking road A, and deciding to switch would be like taking road B.
Moreover, if the game consisted in that instead of revealing a goat, the host gave you the opportunity to reject your first choice and instead check inside the other two doors and take which you prefer from them, it is obvious that it would be better to switch to the other two (unless you think 2 is not greater than 1), and that's true regardless of if you are playing just once or multiple times. Anyway, by checking those two doors you would necessarily find at least an incorrect one, as there is only one prize in total.
If you notice, as the host knows the locations of the contents and always removes a losing door from those that you did not pick, then it is like if he was doing that work for you of checking inside those other two doors, eliminating from them the incorrect one that you would have found anyway, and leaving closed exactly which you would have picked if you were who had checked inside those two doors.
So, if you agree that it would be better to switch to the other two doors, you must also agree that it is better to switch to the other single one that the host is offering, as both ways win in exactly the same cases.
If I tell you that I'm rolling a die, and ask you whether I rolled a six or not a six, offering you 100$ if you guess right, you very obviously should say "not a six", even if we only play the game once.
No, reading Taleb I realized that if it's not repeated it’s just a guess and winning or losing is a 50/50 thing. There is no strategy because the game doesn’t last long enough for strategy to matter. [/s]
It sounds like you embrace the frequentist interpretation of probability, but the bayesian aka subjectivist interpretation of probability says you can make predictions on the likelihood of single events based on your knowledge.
This does not work as an argument against the Monty Hall problem.
It does work against simplistic arguments about which of two or more probabilistic games one should prefer, because it's basically an argument that linearity if expectations doesn't necessarily apply in non-repeated games.
> You don't see any value in doubling your chances of a big win?
You’re improving your average chances. But you’re not improving this game’s chances. If you play 10 games, your average chances matter. If you’re playing 1 game, reality was already set before you started the game. The car doesn’t move because you improved your average probability.
Yes, the whole point is that the car doesn't move. When you initially chose, you had a 1-in-3 chance of correctly choosing the door with the car. The host is now offering you to switch over to, essentially, "all the doors you didn't choose", which means 1-(1/3). As you state, you've only got one opportunity to play, and the car itself hasn't moved, but you can switch your choice. Stick with the initial 1-in-3 chance, or go to the other side, i.e. 2-in-3 chance?
Unless you are arguing that you don't really have a choice because your actions are predetermined you are absolutely increasing your chances of winning this game.
I've not been able to get my dad to understand why you should switch with 1,000,000 boxes because he refuses to believe the 3 box and 1,000,000 box case is the same. I can't even convince him a 3 and 5 box problem are the same.
I actually think I could convince him switching is correct using this strategy because I suspect if he thinks he's tricking or cheating at the game he'll be more open to the explanation.