The Monty Hall Probability Problem from Let's Make a Deal
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- Опубліковано 6 вер 2024
- The Monty Hall Problem is a famous probability problem. It was featured in the 2008 film "21." In the 70s, the game show Let's Make a Deal was hosted by Monty Hall, and sometimes contestants would guess which door among three hid a large prize. After making their guess, Monty Hall would show the contestant one of the doors that did not contain the prize and offer the contestant a chance to change their choice of doors. Many people are unsure if switching at that point is wise. For those that do think a switch would be wise, they are usually wrong about the improvement in likelihood of winning the prize. This video explains the best strategy and the reason why it is the best.
This video is part of the content available for free at www.statsprofe...
I never understood this until I saw this video! Bravo! Excellent explanation!
The fact the host knows what’s behind all the doors, factors in. He can never expose the door with the prize.
I don't think so. However, one should SWITCH doors.
Easiest way to explain it in my opinion is the contestant should NOT want to guess correctly at the start and because 2 of the 3 doors are empty that means the contestant has a 67% chance of ending up with an empty door which is good. If you guessed an empty door correctly, the only door left after the next step is the door with the prize.
Your explanation of the probabilities is good but this was never a segment on Let's Make a Deal. It was invented by a math professor (Steve Selvin) and named after the game show host. It's incredible how many people insist the two doors must have equal probability and refuse to accept the correct answer 🙂
The equation changes as soon as you offer the option of switching, the option is either blue box or yellow box in reality, which is a 50/50 chance of being both right and wrong. very simple.
No. 1/3 of the time you pick the correct one the first round.
The only way to win when staying is if you were correct in the first round.
That means you only win 1/3 of the time when staying.
You pick the wrong one 2/3 of the time in the first round.
You always win if you switch when your original pick was wrong.
Which means you win 2/3 of the times when switching.
You could think of it as choosing between the door you picked or both of the other doors. He just opens one of them before you've answered.
cjpatz's point is relevant. If the host doesn't know, and if he (not knowing) picks a door without the prize, I think the odds on the remaining 2 doors really then is 50/50 as between sticking and switching. Is that right?
Yes.
Nothing like starting the new year with some probabilities. New Year's Resolution:
Learning how to figure out which choices have high probability of making money.
Question: When Monty Hall helps us out by eliminating one choice, does it become a conditional probability problem given that one of the empty rooms has now been revealed? Is Bayes Theorem in play?
Also, is there a separate probability problem associated with Monty Hall's choice that could be incorporated into our final decision?
The intuition that the problem involves conditional probability is not a bad one; however, there isn’t really randomness to base probability on given the possible conditions. For example, p(win|change) doesn’t work as a condition since your choice to change is not random. There is no p(change) because you do it with either 0 or 100% probability. Likewise, p(win|chose a goat initially) = 1 and the other scenario is zero p(win|chose correct door)=0
@@dmcguckian wait, there needs to be randomness as an initial condition for the conditional probability method to work? I always thought the given is specific. I need to learn more about the effects randomness has on probabilities. Thanks.
@@dodgingdurangos924 the given that condition is the thing you know to be true or to have happened, but it has to be the result of an experiment with a random outcome. Remember the formula for conditional probability is P(A|B)=p(AandB)/p(B) the p(B) is the probability that the condition would be met in the original experiment. The idea of deciding to change or not change isn’t random (unless we chose by tossing a coin or something).
@@dmcguckian A non-statically-minded contestant may be more vulnerable to anxiety and pressure when a surprise choice like that pops up, tho. So, more often than not, those contestants might actually be making random & impulsive, "oh-what-the-heck" choices then, right?
Steph, you're still a winner in my book :)
Why did you not include the 4 scenarios
Money is behind A
You should B , he shows C is a loss, you switch to A and win
50% of the time you win and 50% of the time you lose. All 4 are possible scenarios too. Not considering all options does skew the facts
You choose C, he shows B is a loss, you switch to A and win
you choose A, he shows B is a loss, you switch to C and lose
You choose A, he shows C is a loss, you switch to B and lose
I think you have a misunderstanding about how the game works. There are only three scenarios, not four. There are some typos in your comment, so I am not certain I understand what you were trying to say. However, if you were saying that you could pick the door with the prize and win without the host showing you another door, that's not correct. If the money is behind door A and you choose A, the host will show you another door (like B that has no prize) and the host will ask if you'd like to switch. The same thing happens for the other 2 scenarios (you picked B first or door c first). My video shows every scenario and explains the corresponding outcome. I haven't left anything out. If you still aren't convinced, there are likely dozens of videos on this problem. It's very famous.
@@dmcguckian I see four scenarios.
Assume the prize is behind door 1 for every scenario.
Scenario 1:
Contestant picks door 2. Host reveals door 3 does not have the prize. Contestant switches from door 2 to door 1 and wins the prize.
Scenario 2:
Contestant picks door 3. Host reveals that doles 2 does not have the prize. Contestant switches from door 3 to door 1 and wins the prize.
Scenario 3.
Contestant chooses door 1. Host reveals down 2 does not have the prize. Contestant switches from door 1 to door 3 and loses.
Contestant chooses door 1. Host reveals down 3 does not contain the prize. Contestant switches from door 1 to door 2 and loses.
When the contestant initially picks the winning box. The host can reveals either of the two remaining doors as not containing the prize. In either of the two possible doors that the host exposes to not contain the prize, when the contestant switches they lose.
@@jasonmatheson8824 ahh I understand now what you were pointing to. The tricky thing about this problem is that all of the randomness that matters to us happens on the first selection of doors. The switching part is only seemingly random to an external observer but not to the contestant. The contestants choose to switch or not to switch. Their choice is deterministic on the outcome. There’s no randomness left for the contestants after their first choice. They either have or have not selected the door with a prize. After that, it’s just a matter of them flipping the winning chances in their favor or choosing not to. Their choice is made by looking back on the first selection and realizing that they win by switching unless they had the right door upfront which was a 1/3 chance at the time they selected. Whatever the host does cannot alter that. Therefore, we cannot view the three scenarios as four.
For people who want to test it themselves, here's a Simulator => zbbright.github.io/montyhall/
Nice work!
Thays among! It's interesting because I got mostly goats but my stats still said 30ish stay to 60ish switch each time...idk. ill just choose switch them. I'm on the show next week via video. Idk If I'll get chosen as a contestant tho :(