Thanks to surreptitious57, Lacewing, HexHammer, and vegetariantaxidermy, for your interest! I'll try to make your gracious attention to my post both interesting and rewarding.
Practical applications of probability -- games of chance such as card games and other casino games of chance, for example, are amenable to exact calculation of odds, so the gambling house will know its long-run profit. Luckily for the non-mathematical types, such applications are also amenable to simulation, to observe the long-run odds.
In this case (the three doors), if you don't want (or can't stand) to go through the math calculation, take three cards from a deck, one red suit and two black. The red suit represents the prize door. Have a friend keep track of which card is red. Mix the cards face down, spread them out, and point to one you guess is red. Have your friend turn up a black card. Two cards remain face down.
1. To simulate
not switching: Stay with your first choice and flip the cards up. After, say, 50 trials of this, you'll find that you guessed red only about a third of the time. Not so surprising.
2. To simulate
switching: Run another 50 trials, but this time, for each trial, switch to the other card that is still face down. What do you see happening?
I hope these simulations may be more enjoyable, at least for awhile, than, say, playing solitaire. If your friend isn't available, you can do the simulations by yourself and pretend you don't know which card is red.
surreptitious57 wrote: ↑Tue Jul 03, 2018 4:00 am
I keep to my original choice simply because I have no reason to change my mind
Of course it might actually be the wrong door but I have no way of knowing it is
surreptitious57, you have one chance in three of guessing the prize door (two chances in three of having the wrong door) at the outset. You're right, no way of knowing it, but please think in terms of odds, or chances. It's still one in three to win, if you don't switch. You'll find out that you do have a reason to switch! Not that you'll know for sure, in any case, until the game is over, but you'll actually improve your chances by switching.
Lacewing wrote: ↑Tue Jul 03, 2018 4:11 am
The more doors, the greater the probability that switching to Monty's remaining switch-door would be the right way to go, since odds are against you for picking the ONE correct door in the first place, while Monty can eliminate any number of doors to bring you to a choice between 2 doors, giving the illusion that you have a 50/50 chance. Right?
Good thought, Lacewing, increasing the number of doors -- but better (or worse) than 50/50, if you switch (don't switch). Say there are ten doors. If Monty opened eight non-prize doors, leaving two doors -- your first pick and one other door -- maybe you can see that switching is a good idea. There is a 9 in 10 chance that your first pick is wrong, because you only have a 1 in 10 chance of guessing correctly the first time. This means there is a 9 in 10 chance that if you switch, you'll get the prize! Switching is good even with the minimum three doors.
HexHammer wrote: ↑Tue Jul 03, 2018 5:54 am
I have yet to see actual proof, instead of the usual theoretical proof.
HexHammer, I hope you try the simulation described above. But my reply to Lacewing should give you a good clue to the proof in the three-door case.
vegetariantaxidermy wrote: ↑Tue Jul 03, 2018 7:08 am
Absolutely right. As no one ever seems to change their choice then there is no way of proving whether the theory translates into the practical. Logic would tell you that it makes no difference to change, since the prize is still behind the same door that it was in the first place.
vegetarian ..., there is a proof that changing (switching your choice) is the way to go, even with just three doors. Check out the ten-door situation above in my reply to HexHammer, which dramatizes the virtues of switching. The mathematical theory of probability provides such proof, and casinos depend on such proofs and the corresponding calculations, for their ill-gotten gains.
Now, I hope we'll all be able to calculate, or observe in a simulation, the odds of winning by switching in the three-door game!
Future fun(?) -- Ten doors, Monty opens less than 8 non-prize doors. Do you want to switch your choice to one of the other doors remaining unopened? How? You can use a random device. For example, if there are 7 doors left, yours and six others, throw a die to select one of the other six doors. Does this improve your odds of winning?