| Welcome to Toegoff. We hope you enjoy your visit. You're currently viewing our forum as a guest. This means you are limited to certain areas of the board and there are some features you can't use. If you join our community, you'll be able to access member-only sections, and use many member-only features such as customizing your profile, sending personal messages, and voting in polls. Registration is simple, fast, and completely free. Join our community! If you're already a member please log in to your account to access all of our features: |
| Monty Hall Problem; A Fun 'Paradox' | |
|---|---|
| Tweet Topic Started: Mar 14 2014, 02:13 PM (141 Views) | |
| Toegoff | Mar 14 2014, 02:13 PM Post #1 |
|
Administrator
![]()
|
I actually spent a pretty decent amount of time on this at first. It's a pretty cool trick that takes a bit to understand, and really only a paradox because it doesn't seems so counter intuitive. The problem states you are on a game show. You may choose 1 of 3 doors in front of you. One will contain a car, and the other two will contain goats. Upon selecting your door (let's say door number one) the host (who knows ahead of time what is behind all three doors) opens one of the incorrect doors (let us say, door two). He now gives you the choice to stay at Door Number 1, or move on to Door Number 3. What do you choose? If you said, "Stay where I am" ... you'd be very incorrect. Statistically, you should absolutely change to door number 3 (with a 66% chance of winning at door 3, and only a 33% chance at door one). How does this make any sense at all? Aren't you at 50/50? NOPE! That's where the fun of the math comes in. There are a couple ways to look at it - When you pick your first door, you have a 66% chance of being incorrect (there are, after all, two goats) and a 33% chance of being correct (with one car). The host will open up an incorrect door...obviously he won't open your door, that ruins the game...and he won't open the door with the car, you'd pick that in a heart beat. So he is forced to open up an incorrect door. Odds are, you are already standing in front of an incorrect door (there is a 66% chance of that because two doors are wrong) - and he MUST eliminate the other wrong door. Thus, by switching doors, you actually have a 66% chance of winning. You had a 66% chance of picking incorrectly the first time, one of the wrong answers is eliminated, so now, you can safely say that the other 'remaining door' is the likely option to switch to! ------------------- Let's blow up the scale a bit. I tell you I'm thinking of a number between 1 and 100. You tell me 23. I tell you "If you guessed correctly with that 23, I will throw it a random wrong number...however...if 23 was incorrect, I will give you the right answer. The other number is 87." Most everyone would think "Duh...go to 87. I only had a 1/100 chance on my first guess. It is far far more likely that the other number is the correct one." Shrink that down now to 1 in 3 and the same thing applies. Hope you enjoy the explanation. Let me know if it still remains confusing! |
![]() |
|
| savepoints | Mar 14 2014, 03:14 PM Post #2 |
![]()
Forum Hero
![]() ![]() ![]() ![]() ![]() ![]()
|
The way I was able to understand this the first time I saw it was more or less through plain math. Three doors: Door A, Door B, Door C At first glance, since we have no real information to go on, the odds for picking the correct door is obviously: Door A = 1/3 Door B = 1/3 Door C = 1/3 So, say you pick Door A. Another way to write the odds would be: Door A = 1/3 (Door B + Door C) = 2/3 The host will then open up one of the incorrect doors, as per the rules (let's say Door B). Since we actually have some information to go on now, we can do some simple math. Door B + Door C should equal 2/3. Since we've been told (and shown), that Door B is incorrect (0/3, that is), then (0/3 + C) = 2/3. Simply put, Door C must have a 2/3 chance of being correct, according to our math. ------------------- I think one of the biggest problems with people trying to understand this is they just assume whatever door was opened is now completely irrelevant, and out of the question, which is evidently not the case. There's also this website, where you can actually test this out. Try running it a couple of times, and you'll notice that changing has a much higher chance of being correct, and that percentage is very close to 2/3. |
![]() |
|
| ReptilePZ | Mar 14 2014, 04:04 PM Post #3 |
|
Big Sweaty Moose Bleepers
![]() ![]() ![]() ![]() ![]() ![]()
|
Probability 101 |
![]() |
|
| Unfie | Mar 14 2014, 07:53 PM Post #4 |
![]()
Big Sweaty Moose Bleepers
![]() ![]() ![]() ![]() ![]() ![]()
|
what is math |
![]() |
|
| « Previous Topic · General Forum · Next Topic » |











7:16 PM Jul 10