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1,145,451 Views • Feb 19, 2023 • Click to toggle off description
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RYD date created : 2024-11-23T19:34:04.45517Z
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Top Comments of this video!! :3
In the end, it doesn't matter if you switch or not. You always had a 50% chance. The 3 door equation is different.
4.4K |
The variable X is actually referring to two different values in the same expression. Variables, of course, are allowed to have multiple possible values, but the value they take on must be consistent throughout a calculation in order to be valid.
In the expression (1/2)(2x) + (1/2)(x/2), the first x is the smaller possible value, and the second x is the larger possible value. This is not a valid calculation because the variable x is misused.
An easy way to correct for this is to let x be the smaller value, so that the larger value is always 2x. We don't know whether our envelope has x or 2x in it, but we can calculate the expected gain/loss of switching.
If we have x, then switching gets us 2x, and 2x - x = x is the amount of change. If we have 2x, then switching gets us x, and x - 2x = -x is the amount of change. Both possibilities are equally likely, so the expected change is
E = (1/2)(x) + (1/2)(-x) = 0
Thus, there's no expected increase or decrease by switching.
1.5K |
Its always a 50/50 chance. Doesnt matter if you swap or not. Its NOT the Monty Hall patadox. The MH paradox involves 3 doirs and the possibility of getting no money.
470 |
Haven't watched part 2 but here's my guess: This calculation isn't accounting for the contents of the other envelope. When you switch you are not just gaining a new envelope, you are also giving up the one you picked. And these envelopes aren't independent; they are of course linked. We've defined the amount of money in the envelope you're holding as x but x is not actually constant here because it can have two different values. You're either holding the smaller amount, call it y, or the larger amount, 2y. So the true possibilities are: you're holding y, so switching gives you double (2y) and sticking gives you y; you're holding 2y, so switching gives you half (y) and sticking gives you 2y. Add these up and switching has a 50% chance of giving 2y and a 50% chance of giving y, while sticking gives a 50% chance of giving y and a 50% chance of giving 2y. Hence they are equal outcomes.
1.7K |
There's still a 50/50 chance it's either half or double. You can't multiply arbitrary values with probability and expect that to mean something in the probability world
82 |
I feel this is for people that know the Monty Hall problem and now "know" that switching is always better...
The monetary amount is just a pointless distraction. There is a right, and a wrong envelope, or if you like, a winning and a losing envelope.
In the first choice, you have a 0.5 chance of picking the winning envelope. Te opportunity to switch gives you the exact same choice, with a 0.5 chance to pick up the winning envelope.
So no, it does not matter if you switch.
336 |
The real question isn't "should I switch?". The answer to that question is obvious. The real question is "Why is the simple math telling me I should switch wrong?" And that is much, much harder to answer and involves a deeper dive into relatively complex math for what seems to be a simple problem.
5 |
If you followed the calculation you’d have to apply it to the initial envelope you picked, which would also equal 5x/4. So the envelopes are equal to each other and there is no gain in switching
27 |
The sum of envelopes (S) is constant and not independent of choice. But in calculation for EV in first case S = 3x, in second S = (3/2)x which is changing interpretation for x. In the same manner we can ask how many is 5 apples plus 3 pears!
Ok, let's calculate EV using S which is constant.
Rewrite it in the context: EV = (1/2) 2x [where S = 3x] + (1/2)(x/2) [where S=3/2 x]
EV = (1/2) 2x [where x = 1/3 S] + (1/2)(x/2) [where x=2/3 S]
EV = x [where x = 1/3 S] + (x/4) [where x=2/3 S]
Let use S instead of x:
EV = 1/3 S + 2/3 (S/4) = 1/3 S + 1/6 S = 1/2 S
EV = 1/2 S - half of the sum of the envelopes. Simple.
4 |
the two Xs you're adding together are different. if the two values are say $100 vs $200. then you would get x/2 from switching if x = 200, and 2x from switching if x = 100.
228 |
Well if your choice ends up being an odd number…you know which envelope has the double amount now. That’s the only way I can think of.
1 |
Sure would be nice if Youtube Shorts had some kind of associated text field, a sort of "video description" if you will, where the uploader could, for example, provide a convenient link to the next part of a multi-part series. I know, sounds crazy
108 |
This isn’t the Monty hall problem, it’s 50/50
101 |
Let y be the amount in the other envelope. Either x=2y or y=2x. Therefore, either EV=0.75x or EV=1.5x respectively. But never does EV=1.25x in the finite cases. As some have deduced, the error seems to sneak in the cases of infinite, cardinal numbers and ignoring the finite cases.
9 |
But making the switch won't result in receiving the average.
1 |
The numbers in the equation are right but the original logic is wrong.
The variable (X) is not the same in both choices. If one envelope had $1 and the other had $2, then X would be either 1 or 2 depending on the first choice. So the equation would be
0.5*2($1) + 0.5*0.5($2) = $1.5
The average of the 2 choices, so switching or not doesn't matter [probability wise].
Psychologically, getting it wrong feels bad, but having it right then switching feels even worse.
38 |
Nope, BOTH envelopes have the EXACT SAME possibilities. There is ZERO proof that the second has any different probability if you consider that the person making the offer also has ZERO idea which envelope has more money. It ONLY changes when there are THREE or more choices
1 |
Your expectation value is 0.5 * x + 0.5 * 2x = 1.5x. It doesn't change even if you change the envelope, because it is same for both. If there were 3 envelopes like in Classical Monty Hall problem, it would be different.
18 |
To be honest I'll just choose the envelope which feels thicker and heavier.
1 |
@trucid2
1 year ago
Take two envelopes. Put $1 in one and $2 in the other. Pick one. Switch. Switch again. Keep switching. Each time you have 25% more money, on average. Gain infinite money. Retire.
8.3K |