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Help me Understand the Monty hall problem | ||||
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Seen this in two movies already Situation: There are three doors Door A, Door B and Door C Behind only one door is a car You pick Door A Host who knows where the car is opens Door C, its empty Offers you to change if you want What do you do? So first you have 33.33% chance to guess right After Door C is opened it means you have 50% chance But in movies they say now you have 66% if you change! But how? Why? Why is it not 50% The genius in movie says I switch cause its 66% now Could someone explain why is it better to switch to B I am thinking I had 33% first and now 50% Switching gives me 66%? Explain please | ||||
Because you had 3 choices. 33.3 per leaves you with a 66.6 choice with either one you pick. | ||||
@jarl, not quite. This is called the Monty hall problem. Best explained when using larger numbers. Let’s say I had 100 boxes. One had 100 bucks and the rest had nothing. If you picked the correct box you get to keep the 100 bucks. So you pick box #32. I open boxes 1-31, 33-67, and 69-100. They’re all empty. You can now pick to keep #32 or switch #68. Would you switch to 68? I would. You picked #32 and it was only a 1% chance. You can now pick 68 for a 99% chance instead(the sum of the others that we’re limited, plus it’s odds) | ||||
Phat: You're less lazy than myself lol I knew someone would explain it in detail. | ||||
This is fundamentally why I hate probabilities. A given chance relative to the start Vs chance relative to the current conditions. In both Jasmina’s and Moff’s cases, the end result is a 50/50 shot looking at purely the choices presented before you. But by removing the past cases, you’ve cleared 33% (in Jas’s case) and 98% (in Moff’s case) of the options, leaving you with the remainder. Just like any good problem, it’s all about framing :) in application, it’s a 50/50 in the current situation, though relative to the options being cleared, it’s a higher chance. But for the point of final decision, it’s 50/50. I hate probabilities with a passion, cause they can be severely misleading. | ||||
BUMP LOL | ||||
Thanks for speaking to that @percy, I believe the actual proofs behind it would argue each selection is isolated from the others. So when it’s 2 choices, it’s 50% | ||||
You guys are all so stupid, are you telling me that there's nobody guarding the doors? Can't you judge the way the HOST reacts (guy guarding the doors) to know which door it's behind? There is no way this one is 50/50 unless you're as dumb and dishonest as a scientist. You can be autistic and its still at least 51/49 if you have any sense at all about how the host reacts when you solicit a door Consider it like 'Poker' | ||||
Help me understand part II You play lottery(1-60). Random numbers; you chose 1,11,22,33,44,55 Is the chance of winning the same if you took any other random numbers? If you take 1,2,3,4,5,6 would it have the same chance of winning? Could someone explain? | ||||
Yes, you have the same 1/x chance of winning regardless of which numbers you pick. think of this on a smaller scale you pick a number 1 or 2. Whichever number you pick you have a 50% chance of winning. Same applies here just on a larger scale, statistically you have the same chances to win. | ||||
If you do 1 million rolls, and see the outcome. You're saying that once you add up all the million outcomes that came out, you will have approx the same amount of times that 1,2,3,4,5,6 will apear as numbers 11,22,33,44,55,66, right? | ||||
Well considering most lottery's are 1/100million or so, i'd say that you'd either see each sequence 0 or 1 time with 1million rolls. But theoretically - given enough random rolls - you'd see a pretty similar distribution. However that doesn't mean after your number not being rolled you're more likely to have it come up - it's still the same chance to hit every time. Theoretically - the same set of numbers can roll repeatedly forever. | ||||
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