[Highlight] Shaq and Kenny argue over how much it costs to fill up half a tank of gas
One of the most infuriating segments to watch lol
it’s not as bad as Shaq stating they were closer to moon than LA
https://youtu.be/GCKFO6a1FBU?si=TqxpPlYhuGsIGSW5
As a fellow former LSU student… I can see why he thinks this
College men from LSU, went in dumb, come out dumb, too.
Hustlin’ round Atlanta in their alligator shoes
Gettin’ drunk every weekend at the barbecues
And– well, I’ll just stop right there, you go ahead and look up the rest of the lyrics for yourself.
Lmaooo
They let Shaq go when he said he was off by 26 hours lmao
Kenny has a car. Assuming gas prices are fixed, it costs Kenny $80 to fill up his car from zero to a full tank of gas. It takes about 5 days of driving, Monday to Friday to empty Kenny’s gas tank to zero. If Kenny fills up his car halfway through the week on Wednesday, how does it affect LeBron’s legacy?
LeBron couldn’t gas Kenny’s car without a star pump assistant on his team
Depends on if you think Kenny is a superstar or not
A soon to be SAT question
Good callback
Is Kenny the driver?
One of the rare hilarious segments where Chuck doesn’t really say anything
Prob couldn’t figure out who was right
Monkey banging cymbals was going through his head again
My reaction to what Shaq is saying is exactly how I feel when anyone explains the Monte Hall problem to me.
There’s 1,000,000 doors. Let’s say I have you pick one randomly and tell you there’s a car inside. Now after you pick, I eliminate 999,998 doors. One door left. What are the odds that you correctly picked the door with the car in it? .0001% are you THAT confident in your luck? If I offer you the opportunity to switch you should take it.
Holy shit I finally understand this now.. thank you. I remember learning this proof in Probability Theory in undergrad but conceptually could not wrap my head around it. It has finally clicked.
I’ve had arguments with math educated engineers about this. I tried to tell them that probability changes because you have more information in the way of doors being removed and they still didn’t understand and argued with me I’m wrong because probability shouldn’t change or some shit like that
The probability doesn’t change with more information. You had a 1 in 3 chance to choose the correct door, once you’ve made that choice the two doors you don’t pick have a 2 in 3 chance of being correct. Whether or not they open one of the two doors you don’t choose, that group of two doors will always have a 2 in 3 chance of having a winner. Since Monty always opens the non-winning door out of the two, that effectively shoves the two doors’ 2 in 3 chance of winning onto the remaining door you didn’t choose, but the choice is always between a 1 in 3 chance of winning with your original choice vs the 2 in 3 chance of winning you get by effectively switching to two doors’ worth of chances.
Yes the probability doesn’t change if you are a pedantic robot but the Monty Hall Problem wouldn’t be a Problem if humans hand an innate understanding of the issue. The easiest way I had of connecting with people on this is to say “You probably started by picking the wrong door. Now Monty took away the other wrong door. With the information you now have do you think it’s a good idea to switch your choice?” That connects way more often with them than yelling about probability
Stat classes drill into you the concept of independent events but never really teach you how to tell the difference where it’s not. Or maybe I just wasn’t paying enough attention…
They definitely do lol. Dependent variables/event are a big thing.
I would argue that the probably actually doesn’t change for the chosen door, which is why you should switch doors. The chosen door always stays at 1⁄3 chance, but the other door will always have a 1⁄2 chance. Hence, you should switch.
Probably of the original choice doesn’t change but your probability of making a correct decision changes once you have more information/eliminate certain possibilities. That’s the part people get so caught up on and focusing too hard on the original decision seems to be why this Problem is an actual Problem
You just gotta keep fighting the good fight and show them why they are wrong
It’s still 50⁄50 that my door or the door you left remaining are the correct door.
I DONT UNDERSTAND WHY IT WOULDN’T BE 50⁄50.
It would be 50⁄50 if I randomized the location of the car again after eliminating the 999,998 other doors. If you don’t switch you’re betting on a one in a million chance you picked the right door.
I respect you as human and I thank you for explaining this to me.
I honestly feel that one of us is of superior intellect and sees a truth. I’m not saying it’s me or you, but I am saying it can’t be me AND you.
The goddamn Monte Hall problem.
It’s easy to build a simulation that proves that switching is the right choice. I had to code it in my stochastic models class.
Another way to think of this - there are three doors, one has a prize behind it, the other two don’t. I let you choose a door. Then rather than opening a losing door and letting you switch to the other closed door, I tell you that you can swap your door for the other two doors. Would you still keep your original choice or would you take the two doors? Any rational person would take the two doors, since they have no information and now they get twice as many chances at having the winning door.
The reveal of a losing door doesn’t change the scenario because if you’re getting two doors you already know one of them is a losing door, it just gives information that clouds the player’s judgement because they can see the outcome from one of their doors.
It’s technically 50⁄50 on the 2nd guess but the guy got rid of 999,998 doors for you.
I always use a deck of cards to show people. Here’s a deck of cards. Without looking draw the Ace of spade. You select one card. Now I get rid of 50 other cards that are NOT the ace of spade. Do you want to keep your card, or take the one I have left to pick from. Originally you had a 1⁄52 chance of getting it right. That’s horrible odds. I just got rid of every card in the deck except one, I have the ace of spades, not you.
People tell me it’s been scientifically proven but I can’t wrap my pea brain around it.
I mean just go grab a deck of cards and go do what I said. It’s the exact same concept but with 3 doors the numbers are smaller
Your example just doesn’t hit me. At the on-set I had a 1⁄52 chance of selecting the ace of spades.
That never at any point changes during your scenario.
Exactly! And with the MHP, you had a 1⁄3 chance of selecting the right door, and that never changes.
So if you stick with your original pick, you have a 1⁄3 chance of being right. If you change to the other door, you have a 2⁄3 chance of being right.
Let’s say that you never switch, you are physically incapable even if you wanted to. Alternate universe pizzapromise (lets call them evil pizzapromise) always switches, they cannot keep their original choice even should they want to.
You will have a 1⁄52 chance at picking the ace of spades, and this will not change since you don’t switch cards.
Evil pizzapromise always switches at round 2, they have a 51⁄52 chance at picking the ace of spades, because the only way for them to lose is if they picked the ace of spades initially - a 1⁄52 chance.
I’ve seen a lot of good explanations of the Monty Hall Problem, and this is one of the better ones.
In round 2 they have a 1⁄2 chance of picking the ace of spades, but the variables have completely changed since there are now only 2 options. Round 1 never changed - that always remained a 1⁄52 chance, even after the changing variable in round 2.
Now I’m just an evil version of myself and I still had the same chances of picking the ace of spades in round 1!!!!
Yeah, I’m trying to illustrate a situation where you/evil you never had a decision to make in round 2, because you are not allowed to. It’s not really the point of Monty Hall, but I’m trying to reduce the question for the sake of explanation down to a single decision: the one you make in round 1. Sidenote, this thread got derailed real quick. Glad to see everyone is being civil, though.
Yes it does, because I got rid of 50 wrong answers. You changed your odds from 1⁄52 to 1⁄2. You had horrible odds, I got rid of all the wrong answers for you, I’m holding the ace. Unless you got insanely lucky. But you don’t need luck now because I’m showing you which card it is. But with the doors it’s not as drastic. You’re taking a 1⁄3 chance and changing it to 1⁄2
The best way to think about it is to think that once you select a card (to continue this example), you split the deck into two groups. A) your chosen card, and B) all the other cards. As you said, A has probably 1⁄52 of containing the ace of spades. B has probability 51⁄52. The key insight here is that when we discard cards from group B, we are specifically discarding cards that we know aren’t the act of spades, ie it’s not random. So group B still has a 51⁄52 probability of having the act of spades, since we know for sure that we didn’t get ride of it.
I also like this blog post by a statistician that shows an explanation of it. Helps if you’ve encountered basic game theory before, but the hand drawn diagram at the end of it summarizes it pretty well.
https://statmodeling.stat.columbia.edu/2025/05/17/an-alternative-monty-hall-problem-as-with-the-usual-monty-hall-problem-just-set-it-up-as-a-probability-tree-and-it-all-works-out/
Thank you i will read the link you sent.
Just do it with cards pick a random card look at it, is it the ace spades? Probably not, so you should switch to the other card every time. Now do this with three cards pick a card is it the ace of spades? Probably not, so pick the other card offered. Basically just do this multiple times and you should understand it without getting into all the math.
Look at it this way.
The card you selected has a 1⁄52 chance of being the Ace of Spades.
That means the Ace of Spades has a 51⁄52 chance of still being in the deck. Now, when the dealer eliminates 50 other cards from the deck, they’re doing so with the knowledge of where the Ace is. And thus the probability doesn’t change.
Your card still has a 1⁄52 chance of being the Ace of Spades, while the dealer’s card still has a 51⁄52 chance.
Hey, you’re actually close to getting it.
Exactly as you said, it’s a 1⁄52 chance that you picked right and it doesn’t change at any point. So at the end, even if the other guy has tossed out 50 of the other cards, the card you picked still has a 1⁄52 chance to be the ace of spades. It’s not suddenly 1⁄2 just because the other guy is showing you 1 of the cards you didn’t pick.
If your card has a 1⁄52 chance to be the ace of spades, it also has a 51⁄52 chance to not be it. The question he asks you at the end isn’t really “Is the ace of spades your card or this specific one?”. It is “Is the ace of spades your card or is it not?”. Chances are it’s not. You should bet on the card in his hand, because it represents all other 51 cards that you didn’t pick. He just went through them, found the ace of spades and hid the others to mess with your brain.
I’m being kinda silly but that’s essentially what it amounts to. If you just think that the “opponent” knows what all the other cards are and is looking for the ace of spades just as you are, it becomes obvious why you should switch.
I get it, it’s really confusing for me too. I think the Monte Hall effect is tricky when it’s 3 options because it makes everything feel like a 50-50. The thing is the actual situation is that the host has to tell you which one the prize is NOT in until there’s only one door left. With 3 doors he’s only picking 1 so it seems like it’s a 50-50 left, but once you add more doors, the host has to remove all is them but 1.
So with 100 doors, when you pick 1, you have a 1⁄100 or a 1% chance of being right. That means that there’s a 99% chance it’s in the other 99 doors. Of those 99 doors, the host tells you it’s not in 98 is them, basically telling you “IF the prize were in one door it would be this one, and there’s a 99% chance of it”
There was a 0.0001% chance the door you picked was the right one which means the rest of the doors combined have a 99.9999% chance of car. After he opened the rest but one and there was no car, that probability stays in that one door You didn’t choose. So if you switch 999,999 out of 1000000 times you will get car
Think about it like this:
You pick a door, then the host gives you the option to open your door, or every other door.
Which situation do you choose? Your door? Or the 999,999 doors?
It’s because the person running the game knows where the big prize is, so they’ll never open that door.
If you picked the correct door right away, that doesn’t matter. But there’s a 2⁄3 chance you DIDN’T pick the correct door on the first chance, in which case the gameshow host opened the only door you could switch to that WASN’T the big prize.
So think of it this way: at the beginning of the game you have a 1⁄3 chance of picking the big prize at random, and a 2⁄3 chance of the gameshow host implicitly showing you which door has the big prize.
One good way to understand the Monty Hall problem is to simulate it.
This is how I break it down:
If you don’t switch, it is a 1⁄3 chance regardless. Showing the door means nothing.
If you do switch, you need to pick the door that doesn’t contain the prize, so the chance is 2⁄3.
To better understand this, let’s imagine a scenario where door 1 has the prize, but door 2 and 3 don’t have prizes.
If I pick door 1, Monty will show door 2 or 3, I switch, and then I lose
If I pick door 2, Monty will show door 3, I switch, and I win
If I pick door 3, Monty will show door 2, I switch, and I win.
If you stay, you have only chosen one door. If you switch, you have chosen every door except for one.
The odds of each door being correct is 1⁄3. When you pick one door, you’ve locked in a 1⁄3 probability of being correct, and the two doors you didn’t choose are locked in 2⁄3 probability of being correct. You should always trade your 1⁄3 probability of being correct for the 2⁄3 probability of being correct if given the opportunity.
I still feel dumb af… because you’re not given the option to look at both remaining doors. You’re given the option to take one of the remaining two… So, half of the 2/3rds probability… So 1⁄3 probability, which is the same as the probability of my original choice.
Like, I know mathematically you’re right, but I’ve never heard the convincing layman’s explanation.
You are missing that they always remove a door with a goat no matter what. So it’s basically like checking both doors.
If you pick A and C has the car, they remove B
If you pick A and B has the car, they remove C
You’re right, that makes sense remembering that detail.
Opening a door is a red herring, it doesn’t change the choice between keeping a 1⁄3 chance of winning vs trading it for a 2⁄3 chance of winning. The group of two doors keeps a 2⁄3 chance of winning whether or not one of them is opened.
For the Monte Hall problem, you choose 1 out of 3 doors. You have 1⁄3 chance of being correct, and 2⁄3 chance of being wrong. So if someone ask you, would you choose those 2 other doors or this one door, obviously you choose those 2 doors, right? And this is how you are choosing those 2 doors by switching:
After you initially chose the door, they will always reveal the wrong door, then let you choose again. So on the surface, it looks like 50⁄50. But in reality, you are choosing between 1 door or 2 doors because you’ve always known one of those 2 doors would be wrong. They merely reveal which one. But it doesn’t matter which door they reveal because it’s about picking that package of 2 doors over the original door that you choose.
You have the option to choose 2⁄3 odds every time you switch, which is better than 1⁄3.
Your first selection has a 1 in 3 chance of being correct.
Probability always has to sum to one.
So after the host opens an unselected door, the 2/3rd probability goes entirely to the unopened dood that you didn’t pick.
Would you rather spend 20 dollars every 3-4 days ? or spend 50 every 6-7 days?
/s
If you work out every other day, how many days per week would you end up working out?
Just don’t workout & you won’t have to answer this hard question.
are you stupid?
/s again LOL
Greatest YouTube video of all time and it’s not close
6-7 😂 😂 😂 🤣
LMFAOOOOOOOOOOOOOOOOOO.
I’m pushing 30 but I still get the reference.
I JUST BEND ON THE HIGHWAYYYY
LMFAOOOOOO, god man. It’s gonna suck when in like 5-10 years im gonna be so disconnected from modern trends / slangs that I’ll say something and be oblivious to it.
Shaq Math was years ahead of Girl Math
But not nearly as effective or accurate as Steiner Math
shaq dont do girl math. shaq a boy doctor.
Can someone explain to me what shaq was trying to say?
If I explain what he’s saying, then I’d be wrong.
Let’s say it takes 60 dollars to full tank your car. Shaq thinks that if you’re at half tank, you can fill it right back up again for 20. By the time you fill up twice, you’d only spend 40.
Shaq is wrong because it wouldn’t take 20 to fill it all the way up from half.
The crux of the fallacy is that he assumes he can assess accurately how much gas he has based on the gas indicator on his car. Which is a very general and nonlinear reading. The indicator might look like he has half, but his tank might actually have 60%. Likewise the indicator on F might only mean he has 85%. And I don’t know if newer cars solved this, but if you park on a slope the reading is affected too.
Even if he assumed right, it’s still a bad argument.
The thing is, there are people that truly believe this. I’ve heard that the first half of your tank burns slower than the second half because less fumes. So you save money by filling up when it’s half empty. Pretty sure this is the same myth someone has told Shaq and he just believes it.
so you think the fumes that are burning are the fumes inside the gas tank? ignition doesnt happen in the gas tank.
What Shaq did a poor job of explaining is that you don’t have to spend a full \(60 at once if you spread out your \)20 payments and drive less. At some point Shaq tells Kenny you don’t drive that much anyways lol.
Same mindset my cheap self uses lol. Dont have to pay a larger sum at once if you just pay smaller chunks frequently and if you don’t drive frequently then you’ll feel better lol.
Bad readings, Shaq thinks you spend less money when you fill at half, but he’d probably fill it around 60% so he assumes that specific half is cheaper lol
The absolutely most generous possible explanation is that you can avoid sticker shock of having to shell out a large amount all at once by splitting it into smaller amounts. However, he undermines that explanation by saying that you wouldn’t be stopping more often, so there really is no explanation other than Shaq’s a dumbass.
I need terrence Howard to explain shaq’s math here
If you want to be even more baffled, consider that Shaq has a PHD
low trade day? what the hell is goin on
Love these guys. Especially Shaq and Charles!
Shaq is worth hundreds of millions. With multiple businesses. WTF.
Shaq didn’t work out as much as he should’ve during his career cuz if he did, he’d have more rings than he’d be able to count
Shut up fool, shaq was a baller
i kinda get shaq on this one; kenny was complaining about spending \(80 in one go and his point was that spending \)20 four times would feel less expensive. it’s also very funny that nobody listens to him
Damn u dumb
He’s actually right though. Yes the total cost of gas is the same either way, but then why would you complain about having to spend $80 to fill a larger tank? That just means you have to go to the gas station less often.
The complaint doesn’t make sense, and Shaq is explaining how you can get around it.
I literally put together a spreadsheet on this and the argument comes down to the time value of money (holding onto as much cash now is better than spending it all in one go) vs. the opportunity cost of the time of stopping for gas more frequently.
The conclusion of that exercise was it doesn’t matter and it all comes out in the wash. Kenny’s right on this one in the sense of the gas tank can only get so full, so whether you stop more often and spend less per trip or vice versa, it shouldn’t matter.
oh yeah, it’s not a solution to anything at all other than the specific complaint of not wanting to spend $80 in one go. the problem is that they mistake that response for shaq thinking kenny’d be spending less overall