How Games Twist Probability - Extra Credits Gaming
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- Опубликовано: 30 авг 2022
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Probability! A cornerstone for the design of any game. It can build excitement and suspense but, it can also confuse and upset players. So how do you keep probability and players' choices in games meaningful without bogging them down with the specifics? Let's enter the psychology of gaming and how you can twist and obscure probability to build anticipation and emotion for your next game design.
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The Monty Hall Problem is a famous and often missunderstood problem. In one way, the intuition, that switching makes no sens because it's now simply a 50/50 chance between your original pick and the remaining door is corect. But there is one thing that is often overlooked, that changes the ods. Wenn one of the doors is revealed, it is always a goat(or turd in this video). This means that the host never reveals the main prize. This makes sens since if he did, switching would be meaningless. But it also means, that since the revealed door is not random and we are ignoring the cases where the main price would have been revealed, the ods change. Now it's like instead of a 50/50, you should see it as the original 1/3 but you and the host get a pick, meaning it becommes said 2/3 or 66%.
Thanks to YOU!!!
Fire Emblem actually has done interesting things with their RNG. Due to the permadeath mechanic the games have historically had, the RNG can feel particularly punishing. As a result, starting about halfway through the series, the developers started using a system the community dubbed "true hit," which is essentially just averaging 2 random numbers on every roll. Crucially, the games that use this system still display the chance as if only 1 number was being rolled. This results in displayed hit values over 50% being more likely to hit than their displayed chance, with the inverse for values under 50. Ironically, this actually lines up better with human perceptions of probability, making the game feel less punishing
Now I'm curious if tabletop RPGs like D&D has you roll two d20 on tests and take the average of the two. How more or less balanced would their combat be?
@@KnucklesEki considering how much it would nerf crits I am not sure it would work as well. Critical hits in FE are completely seperate from the hit roll, having their own role (which iirc is not affected by the true hit systems in some of the games). In ttrpgs, nat 20s would be exceptionally rare. FE doesnt have this issue
@@KnucklesEki That's equivalent (up to slight variation due to rounding) to the "roll 2d10 and sum", which is used by some tables.
Though this is not the most popular variant like that, the most popular variant like that is "roll 3d6 and sum". It gives a widly different gameplay, and while I don't like its effect on combats and other "chaotic" situation, but it works reasonably well for skill checks or any situation where you'd expect an expert to almost never fail.
I thought True Hit was the name for the original formula, with the new one being '2RN' or '2RNG'
Oh, that explains that option you can set for matches in Fire Emblem Tactics Online.
"Clarity when you want to make them think; obfuscation when you want to make them feel" explains so much of the world right now, well beyond video games.
A very interesting wrinkle in the Monte Hall problem that I've never seen explored is that if Monte were truly a random actor making decisions blindly then switching and staying would indeed have equal probabilities of winning. If Monte were to reveal doors at random, then there would be three equally likely possibilities: 1/3rd chance that the prize is behind the door you picked, 1/3rd chance that the prize is behind the door Monte offers as a switch, and 1/3rd chance that Monte accidentally reveals the door with the prize before offering the switch. This means that both switching and staying win 1/3rd of the time, and 1/3rd of the time you lose with no chance to switch. However, in the actual problem Monte is not acting randomly and has perfect information, and he will never allow that last outcome to occur. As a result you get 1/3rd chance that staying wins and a 2/3rds chance that switching wins, because Monte will always pick the situation where switching wins over the one where you just lose without getting to switch at all.
I think that's a fascinating probability lesson, as without knowing Monte's behavior the two situations are indistinguishable. If Monte revealed a dud by random chance then it's indeed a 50/50, but if Monte has perfect knowledge and used it to ensure he revealed a dud before offering the switch then the odds are 33/66. There are actually many tellings of the Monte Hall problem that aren't explicit about this, so the 50/50 interpretation is completely valid in those cases. It shows a lot of the nuance in statistics, that two situations that look identical can in fact be very different!
I haven't done the math since you said you haven't, but I'm pretty sure if you chose random actors like that it turns out to be a 25% to 75% outcome, since the 33% chance of picking the prize makes it an 100% chance for the randomly opened door to become a dud.
This is a common issue in statistics IIRC, because people 'stop testing when a positive result is reached'
So overall while it might look like a 1/3 chance for the randomly opened door to be the prize, the first 33% chance completely cancels out the probability of the next rolls, since both doors become a 0%% chance and not a 50/50
If door A has the prize and you picked A at random, there is no other 33% or 50/50 between B and C; the area of operation becomes invalid.
Lets say you were shooting a gun at a target and you had a specific number of shots to get a bullseye. Because people who 'succeed' don't take the rest of their shots; the statistics look more like a staircase than a bell curve if we see if are counting if a person shot each bullet or not on a graph. If it is truly random while testing for a positive result the distribution ends up changed.
The monte hall problem is specifically geared on the change of condition vs innate bias. The fact the third party reveals a dud is part of the required problem, changing that removes the problem all together and becomes what is basically a coin flip. The attachment people have to their initial choice even if they are aware of the solution; as well as how people operate and think is a major part of it. Most of these statistical philosophy things rely on the participants having 'a choice' rather than 'random chance'
I don't think in that scenario it was actually Monty picking the door. There's probably someone in the directors booth, who knows where the prize is, telling Monty which door to reveal. Monty likely didn't know where the prize was either, as that could allow someone to gather some information about the location.
I have abandoned many games with character development trees because I had no idea what effect the various choices would have, and I was worried about choosing the "wrong" one. Stuff like: "Do you want Necro or Gia?" The categories were completely obscure.
This is exactly why I've always gone to PokemonDB before deciding which moves to let my Pokémon keep, or to a wiki that lets me know what the upgrade paths are for weapons in Monster Hunter. I have no way of knowing, on my own, which path is better. It makes me tense to think that I could be moving away from the build I actually want, without even knowing it. I like being able to make informed decisions instead.
@@LamanKnight It's not even about knowing which path is 'better'. In any well-designed game (single-player at least, so you don't have to worry about the 'meta') both paths should be more or less equally 'good', equally powerful and feasible to play. But you don't know which one _you_ would like better and you can't know without consulting a wiki.
Terry Cavanagh, the designer of Dicey Dungeons, has mentioned that he decided to only use 6-sided dice in Dicey Dungeons because people tend to be able to better estimate probabilities when using them compared to using other dice. At the very least, it’s definitely easier to intuitively understand how those dice rolls effect gameplay then something like a weapon in an RPG having a +1.5% crit chance or something like that
Great to see some Dicey Dungeons love, it's a great game!
I swear that game is rigged, I've put a fair number of hours into it and the sheer number of times probability screws me is just wild
an ode to the D10, whos statistics are so plain they couldn't be confused. alternatively my favorite, the d12: a d6 with more options.
If you ever feel baffled by the Monty Hall problem, remember that it's a relatively small effect with just 3 doors. "You should always SWITCH" does NOT mean "switching will always lead to a WIN", but I think we sometimes hear it that way. With 3 doors, you'll still lose a lot by switching and win sometimes by not switching, it's just that a completely rational probability computer would recognize that switching always has an edge.
I find most intersting the paradox with the 2 door version.
Though I disagree with the montey hall problem being statistics. It's psycology too, as you have to assume an all knowing host that wouldn't reveal the key as door 3. They aint just blindly revealing door 3, they're revealing a GOAT, and your taking information from the position of that goat.
I just don't understand what logical leap they are trying to make here. Whether or not the door is revealed is meaningless for two reasons. First, you have no way of knowing which of the 3 doors is correct, therefore neither has better odds of being the correct answer (god knows that the definition of "probability" is, but one thing is for sure, neither answer is more correct than the other when it comes to choosing the prize). Second, revealing that one of the doors you didn't pick was a dud only increase the odds of the door you picked being correct (so why would a person feel more nervous? They have no reason to feel more nervous. The only reason they feel more nervous is because the people organizing the 'Game Show' are trying to bluff you).
I never really liked probability. I'm good with math, NOT probability. Probability never made sense. It probably works, but I'm incompatible with it. This 'Game Show' problem is smoke and mirrors to me.
Can someone explain how it improves your odds? Cause when you take a step back and examine all the parts, it was ALWAYS a 50/50, just disguised as a 1/3. You pick an option and then a wrong option is removed, so really you were always picking between a wrong and a right option; with a "fake" third option that never actually existed, but you were lead to think it does.
Fun fact: choosing whether or not to switch at random will give you a 50/50 chance of winning. This seems like it should be identical to choosing a door at random of the two remaining.
@@Leivve Imagine your strategy is to stay, not switch. The only scenario where you win by staying, is if your first choice was correct. In a game with three doors, the chance that you initially picked the door with the prize behind it is 1/3. The host's actions after your pick do not change this probability at all. And since the sum of all probabilities is always 1, the door left closed by the host must have a 2/3 chance of containing the prize.
When the two doors remain, there is indeed one correct, and one incorrect choice, but just because there are two choices does NOT mean the odds are 50/50.
Critical hits is something you need to be careful with as a designer. Diablo 3 for example lets you stack both bonuses to critical damage and critical hit change to the point that doing anything else is substantially suboptimal.
Xcom commits a big sin when it comes to RNG: it rounds numbers up without telling the player. I was playing the final level of the game with my veteran squad, when a Cyberdisc appeared out of the fog. It flies up to one of my injured soldiers and will instantly kill them as soon as it becomes the aliens' turn again, so I have to kill it in a single turn. I spend a lot of time thinking how to optimize my damage so I can deal with the other aliens also attacking me. According to the game, my sniper has a 100% chance to hit the Cyberdisc so I plan around that. I take my 100% chance shot... and it misses. A guaranteed 100% chance shot missed. Years later, I find out that the game rounds numbers up, so in actuality I had like a 99,99% chance to hit. Still sucks to miss a shot like that, but I wouldn't have been nearly as upset if it didn't present it as a guaranteed hit. They should've just rounded it down to 99% instead.
EU4 does the opposite with sieges, they always make the shown chance of success smaller than the actual chance, when its 50/50 it displays it as 49. That way it doesn't feel as bad when you fail twice in a row.
Heh, a similar effect happened in the original Red/Green/Blue Pokemon games, where most attacks have at least a 1/256 chance of missing without any modifiers (other than Swift which always hits). Always frustrating to have that critical last-chance attack that shouldn't miss suddenly whiff.
@@barryfraser831 It's worth pointing out that this is because eu4 sieges work in incriments of 7. If you pass a 21% chance without a critical success it will increase to a 28%, and so on. That's why 50/50 is displayed as 49%, eu4 never shows 50% on a siege. Also interesstingly with eu4 is that the reason why you almost always win a siege before it has a 60% chance or higher is cause by the time you get to 56% you have had at least 4 chances at succeeding. Pretty neat.
It's reasonable to round out numbers given there are so few 99.XX chances in the game that it will almost never come up, and the odds that this scenario is also one significant to the outcome of the round is vanishingly small. Annoying an inevitable few players out of millions is less of a problem than introducing decimal places to the probability calculation; not because of issues with engine or display but simply because that doubles down on the far more common problem of players mentally rounding out numbers.
@@nathanaelsallhageriksson1719 That said, when luck isn't with you, the sieges can tun stupidly long - and those are the ones the player is going to remember, not the ones where they managed to land 7%. This is probably why CK3 uses a siege system where there's a base level of progress gained automatically with RNG mostly serving to shorten the timer. The final result is still driven by RNG, but the worst case scenario is capped, so you don't get the aggravation of those really bad runs, and they just feel better.
Thanks for talking about probability in games. Next, can you explain how I missed a 100% shot in XCOM?
@@extrahistory what an amazing and cursed emote
It was rounded up and wasn't actually 100%
I think there's an important insight missing in your description of the Monty Hall problem. If I didn't already understand it, I don't know that I would have been convinced that your argument was valid.
Your example with 1000 doors came close, but then I felt like it didn't follow through. It's not that you'd have the choice between 999 and the initial door (which doesn't feel as obviously analogous to the three door situation), it's that you would open 998 dud doors. Which should make the informational asymmetry between your initial pick and the other unopened door much clearer. The odds of you happening to pick the right door are much lower than the odds that it's just another dud door, and now all the other duds are eliminated and the remaining door is the prize door.
The Monty Hall problem is an interesting thing. I'm glad you mentioned in the beginning the need for it to be a constant that one of the other doors opens after a choice, that ends up being one of the stickier parts, and the nature of the whole thing can change. Supposedly at one point the man himself was told about the problem and asked about it, and his answer to whether someone should swap or not was that it actually came down to the host and contestant trying to read each other. He might choose to open the other door or not depending on what the chosen door actually contained, or maybe just to mess with the contestant.
There are some games that can get great value from obfuscation. However, those tend to do so knowingly, so it's still very important to know what is and isn't important to distance from understanding, and how to do so.
That example with 1,000 doors really helped me understand the math. I’ve been told about the probabilities behind the Monty Hall problem, but it was never intuitive until that example. Thanks!
Thanks! It's hard to get your mind around it without the pictures and context. We're really glad to have helped!
Yeah, the 1000 door example is what I use when teaching the Monty Hall problem to children, as well. The issue with the 3-door example is that it deals with such a small sample size that it can seem completely arbitrary to someone hearing about it for the first time, since it directly contradicts what feels like common sense. This leaves them feeling like they don't really get 'why' it works like that.
1000 doors makes it a lot easier for a layman to appreciate the effect, even though the mathmatical principle is the same regardless.
We had a big debate about the Monty Hall with a friend who insisted the probabilities are 50/50. We finally coded the problem and simulated 10 million rounds. But even before running the simulation his confidence (that had not budged an inch during the day) was shaken just by seeing the code.
Basically if you code it you have to:
1) randomly pick an item from the list [1,2,3] and store it as prize_door.
2) randomly pick an item from the list [1,2,3] and store it as player_choice.
3) remove prize_door and player_choice from the list [1,2,3] and then pick a random item from the remaining list and store it as opened_door.
4) if prize_door == player_choice increment a counter
5) run the whole thing a large number of time and at the end divide the counter by the number of runs.
The result will converge to 1/3.
Now notice that you can remove step 3 as it has no bearing on what we are counting in step 4. And the result will be the same.
**The host's action is irrelevant to the chances of your initial choice being the winning choice. Nothing he does changes the chances of that door when you picked it out of 3 doors being the winning door. It cannot magically jump up to 50%.**
@@TheSirmousavi I really like that explanation - very straightforward.
@@Stirdix 🙏
Ironically my introduction to the monty hall problem was zero time dilemma with 10 doors instead of three
I'm sorry. Did you say that I can win goats in this game?
Lord I hate the monty hall problem
I would much prefer a goat .
One time, Monty paid a total of $1000 for three different costumes, and then--later--promised a contestant a prize "worth a thousand dollars." Which, of course, was those costumes.
@@bobross547
Math is complex but a goat... now that's a statement!
Jokes on you. I'd take the revealed door because I want the goat.
There's two other on-average situations with crit in games:
1) Crit as an effect on damage formula
In some games, dmg is calculated in lines with total dmg - defensive modifier, so crits are meant to penetrate high defense (relative to character dmg) better. Another way is like in pokemon, where crits ignore defensive buffs. This is generally meant as a design that limits the effectiveness of "walling" with defensive stats.
2) Crit as a stacking effect
Many games have both crit chance and crit damage, which means stacking both effects multiply on themselves. For a simple example, imagine no base chance/effect at all - on average, +50% crit dmg and +50% crit chance is +25% dmg, while +100% crit dmg and +100% crit chance is +100% dmg! With careful control, this can favor crit for a pure-damage set of modifiers (favoring pure-dmg builds too) while hybrid builds would prefer a straight attack/dmg boost instead.
Second can clearly be seen in Monster Hunter where the meta for the past few games has revolved around getting 100% crit chance (or Affinity, as it's known in game) and then maxing out crit damage as much as possible. No matter how many changes Capcom makes to the system, the fact that you can build for 100% crit means that crit is always going to be meta-something both the in-game tools to test builds for damage and player-made tools continue to verify.
My kind of game:
A game where the complexity does not come from the single elements it is made of, but from their interactions with eachother.
This channel has gotten me through so so many hard times please please keep making videos
I remember the mtyhbusters episode on the "monty hall paradox". Worth a watch if you find it.
0:50 My years of watching VSauce have paid off. I'm gonna switch, Matt!
(Here's hoping I get a goat. I think they're adorable and good company)
We fully support your goat decision!
I have watched this Monty Hall problem in Brain Games years ago but never bother to think about it deeper cause random is random, you either get or don't. However, the last part about human perception is interesting cause this is what's special about this problem.
If you want a game that's all about probability, the Mounty Hall problem and stuff like that, I highly recommend checking out Zero Escape: Zero Time Dilemma! (And the previous two games, to understand the context)
I think the most intuitive way to view this for me is to make it a 50/50 choice between all the doors that got opened and your first choice, and the other choice. The new choice gets the whole 50% assigned to that half, but your first choice only gets the 50%/the number of doors on that side.
This doesn't give accurate probabilities (you'd have to do some math to the two percentages to get those, since the open doors don't actually take up probability space), but it's pretty intuitive for me.
I.e. if you have 6 doors, and you pick one, and 4 duds are revealed, you have a 50 chance for the new one (50 chance for a group of one door), and a 50/5 (50 chance for a group of 5 doors) = 10 chance of the first pick. Doing the math to make it actual percentages gives (50/60 or) an 84% chance that switching is the correct option, and a (10/60 or) a 16% chance that staying is correct.
There's one thing that could potentially be happening under the hood in a video game which puts the maths of this into question: is that "switch or not" question asked always, or sometimes? And what are the conditions for that? If you add an extra step there, maybe you are offered to switch most often when you DID pick the right door? But if you balance this subtly enough, there's barely any way of tellling other than looking into the game's code.
So... if I understand well, there are three possibilities, by staying you can only pick possibility 1, but by switching you are simultaneously picking possibilities two and three, since one of them was already revealed... If I have ever said that I don't know how something works is now. I understand that you have 1/3 that your initial pick would be correct, then your probability rises to 50%, but by staying with your original pick, wouldn't it also rise to 50%?, I mean the whole difference between one possibility, another and even certainty is to know the outcomes. I think that the only way to know for sure is to run this experiment a lot of times to see if it makes any difference on the final outcome.
Thank you for the video.
Dear God this explanation was so much better than when I learned it in college.
I've seen an example of the opening problem applied to bathrooms at a festival. Once you've tried a stall, should you go with it or keep opening stalls until you've found a better one? The answer is yes. They had to lay out all the various possibilities for me to be convinced (like in this vid) :p , but it makes so much sense once you get it.
The festival bathroom problem is an example of the 37% rule. With that, you evaluate and reject the first 37% of possibilities. After that you pick the first thing you get that's better than anything you have previously seen. It's actually been found to work pretty reliably with tons of different things when you have to accept/reject a series of 'things' one at a time.
I'm not sure that I understand. In the example with three doors, door number three is opened regardless of whether the player chooses to switch right? If the player chooses to stay, he has effectively opened doors one and three. If he chooses to switch, it's doors two and three. Two out of the three doors are opened in both circumstances. I understand that having more doors opened gives the player a better chance of winning, but I don't understand how switching accomplishes this.
It doesn't. I'm half convinced this video is itself an example of how humans are bad at math. The psychological dilemma is nothing more than a farce because no matter what you choose the show can ALWAYS show you a bad door. While the probability becomes 50% once one door is opened at the beginning you always had a 33% chance that will always be reduced to a 50% chance by the show itself.
I honestly really dislike it when people go for the 100 door variant of the problem as I feel it doesn't really explain what the choice of switching or not switching affects.
Imagine that the doors were made out of glass and you could clearly see what is behind them. Of course now it's easy to just pick the prize. If you were to switch you would change to a goat instead and lose, so of course you wouldn't do that. But if you had to switch, what door would you pick? One with a goat behind it, since the other goat will be opened switching always results in you getting the prize.
If you won't switch you need to select the winning door first to win. If you are planning on switching you need to pick a goat door to get the prize.
So when the doors aren't see-through you want to use the switching strategy as it's more likely that you picked a goat than the prize.
This was my second favorite goat-themes game show in the past year, right after the one in Psychonauts 2.
the thing about crits in video game is that often they are multipliers and stacking many multiplying effects gets more powerful fast.
Ie you have 10 dmg and a +50% bonus, adding a +10 damage two times gives you 45 dmg, but adding two x2 effects, 10 x2=20 x2=40x1.5= 60 Total damage. Even tho adding 10's looks like you are getting the same thing looking at base dmg each +10 is double your starting damage adding ends up being worse
That's right. You'll see that in almost any game where "crit builds" are possible, that they are more optimal. When you can increase both your crit chance and crit damage, they are multiplicative and strictly better than upgrading your attack. The example Extra Credits discuss looks like it is only possible to improve the damage of critical hits. Even then, in most games enemies don't die in 1 hit and the argument about excess damage doesn't work.
love all youre subjects
I am the ANTITHESIS of this example: My school class (I believe it was middle-school, but it has been a long time) presented this, and AFTER switching when one of the bad doors was revealed, I found that I had originally picked the correct door. I hated myself all day afterwards because of this.
I knew the right answer, having seen the problem in a couple other places but I'll be honest it never actually made sense to me until watching this video. Thanks!
The Monty Haul problem has been explained to me before, and it still makes no sense to me.
Tf2 crits are always a nice thing when they come in clutch
goddamned monty haul problem. I STILL can't wrap my head around it.
Ah crockets, my favorite
This video is great!
OG views of GameTheory would recognize this same concept being taught but I like this video better.
Thanks for the compliment!
Wonder what the game mentioned at the start was?
A gygaxian dungeon is like the world's most fucked up game show.
Behind door number one: INSTANT DEATH!
Behind door number 2: A magic crown!
Behind door number 3: ten pounds of sugar being guarded by six giant KILLER BEES!
This sounds like a Ted-Ed explanation of a riddle
The key to understanding the Monty Hall problem is that, at the point you get to change your choice, the two doors are not equivalent, the situation is not symmetric with regards to them. The one you first picked was, at the time, equivalent with the other two. After the elimination of one of the wrong doors though, the symmetry breaks down. The third door is a survivor of this elimination, unlike the first door. Which means that the third door is more likely to be the right one, since it always survives the elimination.
always love hearing about the monty hall problem and probability
I first discovered this channel by history know im even more interested in another sections great video as usual
Thank you!!!
I literally learned the opening segment last year in science weird to think you find these random things everywhere
When you choose to "stick" you are not sticking with the original probabilities - you're making a fresh choice between 1 of 2 doors, regardless of which you choose. It's not staying with the same decision made previously and retaining the previous chance of success, it is a new decision with new probabilities either way.
Not quite. I'll tell you why, but we have to change the game a little using a die.
I roll a d3 and ask you to guess what I rolled. You guess one. I tell you that I didn't roll a three. Stop here!
What, if anything, at this point, changes the odds that a d3 rolled a one? Nothing, that's what. The odds are still 1 in 3 to roll a one, which is your guess. All I did is say it isn't a three. The odds don't change to 1 in 2 that the die rolled a one, they remain 1 in 3, so it's a 2 in 3 chance that it's not the value you initially guessed.
That's where people mess up the Monty Hall paradox; they incorrectly assume the second choice is equal because it's the last choice, but the reveal of the dud has stacked the odds.
@@amwoodco3049 after you say you didn't roll a three, you ask me to stick with one or change to two. The odds that I rolled a 1 are the same. Yes. The odds that I choose the right option out of the remaining two is 1/2. Because I am choosing one of two options. One over two is the same as 50%. After the reveal, the second choice is made out of two options, not three.
The relevant percentage isn't the chance of rolling a 1. The relevant percentage is the chance of choosing correctly after the reveal.
Close, but not quite. In your scenario, the 1/3 chance only applies at the outset, because your guess was
*one of three, two of which* are wrong;
when a different option is removed, it's *no longer* the same problem or probability, because _now_ your _original guess_ is
*one of two, one of which* is wrong.
But if you're talking about a scenario in which a host or code can manipulate the results, then you might be right, but it's no longer a straight math problem, it's a guide to Las Vegas.
Another way to explain it: if all three options are equal from the start,
and if according to you, opening door 2 after door 3 is revealed is like opening both doors 2 & 3,
then re-choosing door 1 after door 3 is revealed is like opening both doors 1 & 3.
Same addition/multiplication of odds, same probability of outcomes.
Be logically consistent.
I think there is a movie where Kevin Spacey act as a professor of an university and a brilliant student answer correctly
21
I've used the Monty Hall problem as a hook for a presentation a couple of times, always a fun way to mess with people's intuitions
If critical hits don't entirely shake up the direction a battle is going (both ways!), you're doing it wrong. The Persona series got that.
Ngl I've always disliked crits in games, I really love games that have stagger mechanics where by playing well you make the enemy vulnerable to follow up damage. Or to a lesser extent simply targeting thier weakness, eg. holy damage against undead.
The CURRENT "let's make a deal" doesn't let you switch anymore unless it's cash for some reason. But, all of the prizes are pretty great now. (Unless it's gym equipment. It's nice gym equipment, but who wants gym equipment?)
Make more videos like this
I wonder if the probability isn't actually slightly better than 66.6%, because they don't randomly open a door, they specifically open one of the wrong doors.
Well, I guess that's not a change of probability though, huh? That's just a change in our ability to measure the probability. We have more information, and can adjust our assumptions... This was a really interesting video, made me think. Love it. lol
the remixed intro is good I'll say that.
"Nat 20 baby!" -Marcy Wu
Was waiting for this subject to be covered! Never know how to get it right. Thanks for this!
Thank you for watching!
In Soviet Russia Monty Hall problem is zero percent chance to win. Only the gulag behind both doors.
XCOM is a great example of this effect in action - even though the game has been unpacked completely and the exact RNG algorithm has been solved, even though the game offers strict RNG and the only exception is actually to BENEFIT the player on lower than the hard difficulties, people will die on the hill that the game is cheating. It's simply that most players mentally round out percentages - everyone has an innate point where they round out a percentage to 'yes', 'even', or 'no'. 80% feels close enough to 'yes' that even though people accept it missing ,a few misses in a row feels like a cheat even though it's statistically very likely over the course of a campaign. The role of perception is clear - because it's your small, elite squad versus a large, weaker group of aliens in an action-economy game, you the player have a small number of high percentage chances to hit or kill, versus the enemy's large number of low percentage chances to hit or kill. When the difference between perceived odds and actual odds comes up, it will usually be the player missing and the enemy hitting, and because these are very memorable, they take on outsize significance.
The joke's on Monty Hall... I actually *wanted* the goat!
Ha, this is literally the opening scene of the movie "21." I already knew the correct choice. B]
Because probability can feel exciting right until you lose.
Can you do some on American outlaws what Billy
0:50 switch, Mythbusters tested it and switching had a 70% higher chance to win
2:36 While it is correct, that you have 2/3 probability when switching, you cant increase your actual chance of winning (unless the open door is the prize, we can ignore that, since you would be either blind or an idiot if you dont choose the open prize). When there is leftover behind the open door, you have a 50/50 Chance to choose the correct door, since you cant say what is behind the other 2 doors for sure from the open one.
5:15 Paradox: HAHAHAHA.....Ha.... *NO*
That is what i was thinking too. If I am always shown the Bad Doors and never if I missed the Prize, how does this Matter if I want to win. I don't want to open 999 Doors, I want to open the winning door.
@@HexAF8B9 You don't have to open 999 doors. A better explanation would be...
You choose one out of 100, that's a 1% chance of winning.
The host then opens 998 doors, all of them duds. The prize doesn't move.
Will you keep that door, or switch?
@@saxor96 so the Doors the Host will open could also contain the Prize? Only then I think the Percents adds up. Since I have won nothing if he 100% will not reveal the Prize. This would not change my Odds.
@@HexAF8B9 no, the host always opens a door without prize. So the choice always ends up in this:
Option 1: You got it right the first time. In a 3 door contest, that would be only a 33% chance. In this scenario, if you switch, you lose. (33% of winning by staying)
Option 2: You chose any of the wrong doors. In a 3 door problem, this is 66% likely to happen. If you switch in this scenario, you're guaranteed to win (so, 66% of winning by switching).
The thing you're letting to chance is guessing in which one of the two options you're in. You have a 66% chance of getting it wrong the first time, so the most plausible scenario suggests that you switch so you rectify your wrong, original selection.
@@saxor96 I do not get how my Odds to win get better if the Additional Doors I get for Free are bad anyway.
I get the Percent Stuff of 3 Doors each 33% and 2 Represent 66%. But if is is guaranteed that the right door will never be the one I get for free before choosing to switch, at which Points do my Odds increase that the Door I have chosen is the winning Door.
Sure, I had 33% of picking the right one, but how does having a bad one eliminated raise my odds of winning? At the End of the Day the Door that I could switch too did not have 66% of being the Winning Door at the Time of choosing. I do not get 2 Doors by switching if the one I additionally get as Information is not the Winner for certain.
In either Scenario I will see 66% of the Contents behind the Doors. Switching in a 50/50 Scenario which will be created after I have chosen does not benefit me, or does it?
I get hung up on the 100% a Bad Door gets eliminated. How does that impact my Odds of being in the right Scenario if I switch?
PANR has tuned in.
Thanks so much PANR!
So dexterity/agility builds are always the smart choice
- EC: You should always switch
- Me: Why?
- EC: Idk man
The example with 1000 doors gets actually even more intuitive when you use the original rules. You pick one of the doors, 998 get revealed. Do you want to change?
6:22 sounds like how gambling makes people feel
The 3 door problem is a worthless question. Both probability equations are identical if you redo the first equation with the new information (1 door being eliminated). Meaning switch or not your odds are identical. If you go based on common belief then all you did was make your original equation wrong since you didn't correctly account for the entire equation (lacking an accurate variable)
And just to stave off some of the useless counterpoints. I'm assuming that you are given another free choice (which you are) and can freely choose the same door again (instead of only being able to choose the other, which you can). This is just to keep the conceptual statistics to a minimum.
It should be noted that the door example only works if you know that the door that is opened is intended to always be a dud.
If the door opened is random, then there is 2/3 chance that your are in a world where you picked the correct option, and a 1/3 chance where you exist in a world where you didn't pick the correct option.
This makes the probability work back to be 50:50
In the parameters of the Monty Hall problem, it is always a dud that is opened, yes. Wouldn't be much fun as a game show if they opened the door and revealed the prize you could no longer have.
If the correct door can be randomly opened, you won't have a chance/reason to switch, so this is entirely a moot point.
"you should always switch" but what if I really like goats?
*hits that switch*
I never got what was so bad about winning a goat.
Can you do a video about the greek indepedence war?
I have great respect for Marilyn Vos Savant, the genius (highest IQ ever recorded) who answered the Monty Hall problem. But this logic, it's just wrong, and it's not even that difficult. The argument hinges on the notion that your probability of success (1/3) does not change once one door is opened. But it obviously does. That's new information, which requires a new equation.
@Jason
Allow me to point out your error. Think of it as me rolling a d3 in secret and guessing what I rolled. You guess one. I tell you it's not three. Stop here!
How does that information change the odds I rolled a one on a d3? The answer is it doesn't. That is the catch to the Monty Hall paradox, the new information doesn't change the odds of the initial choice, but stacks the odds in the second choice. Remember, the initial conditions set for all this is includes equal propability distribution and you aren't being tricked.
Mythbusters did an episode on this and brute forced the problem, showing that switching results in twice as many wins. They, however, also went into the psychology of the problem: most people don't switch.
Players are not stupid some of them will simply crunch the numbers and use their brains to crush the others forcing the rest to copy them no matter what your goal initially was.
That depends on the game, it's not always an option. Sometimes some dude with a Pachirisu comes along and just wrecks your ultimate tier plan anyways.
Hey I learned this in Math class!
But math class wasn't as cool as us right?..... Right!?
the critical argument made partway through the video does not account for any incoming damage, or any kind of health regeneration mechanics, and specifically the boss scenario(most players will base their characters around these) in which large amounts of damage are required
a direct dps mean average does not correlate with the experience of anyone who plays games
In the initial example you gave, you are essentially choosing either doors 1 and 3 or doors 2 and 3 so it's still a 50 50 chance of success. Especially since you as the player still don't have any new information about doors 1 and 2. In other words, the scenario at 3:08 is different from the original one and it's misleading to treat them as identical.
I don't think you understand the Monty Hall problem then. I am assuming by initial example you mean I pick door one, door three gets opened and it is a dud and then I get to switch or not. If I switch it is roughly 66.6% that I win. That is why they call it the Monty Hall Problem, it appears 50/50 to most people, but is actually 66.6%. The example at 3:08 is also 66.6%.
Fundamentally, I disagree with the solution most people think is correct for the Monty Hall Problem. I understand the math involved and have some background in Probability, however, the Monty Hall Problem is not one problem, but two. In the first, you basically are given a 3-sided die with words "heads," "tails," and "body." You are then told to guess what the roll will be. After that, you find out the call didn't matter. The 3 sided die is then taken away from you and you are given a coin to call heads or tails. This creates an entirely new and independent event from the first with even odds heads vs tails (let's just ignore the fact that most coins aren't EXACTLY 50/50 due to imbalanced weighting of each side and physics).
This does not mean you always switch or always don't switch. What it means is you instead assess the situation and realize "I am now making a new decision between two things." Not recognizing that fact makes you run amok of the "Gambler's Fallacy" and you are making your decisions based on the wrong reasons. That is how I have always seen the problem.
tl:dr - I think the Monty Hall Problem is a dumb problem on a fundamental level.
It's a useful problem though. It has a very simple premise and execution, people's gut tells them to do one thing (staying) and the outcomes show a clear advantage in doing the opposite. This discrepancy is interesting.
Unfortunately, you are interpreting the problem incorrectly. One of the prerequisites of the Monty Hall Problem is that before any door is guessed, all three doors have an equal chance of being correct. What your example suggests is that the correct answer was only ever going to be between two doors, which simply isn't the problem.
FINALLY found somebody in here who also thinks the Monty Hall problem is ridiculous
@@garyermann But if they always reveal one of the duds from the doors you didn't pick then ask if you'd like to switch or stay, hasn't it now become a choice between two doors?
I still don’t get the problem since while there is 3 possible scenarios one of them is removed thus resulting in a 50/50?
Think of it this way. You have a 33% chance of picking correctly on your first guess (1-in-3). Why would your odds magically increase to 50% because a wrong door was opened? Since you can only pick one, there will always be a wrong door to open regardless of your selection and your odds of picking correctly on your first guess haven't changed, they are still 33%. Meanwhile, your initial odds of being wrong (66%) also haven't changed, but instead of those odds being split across two doors, there is now only one door representing that possibility. So switching doors will always give better odds (66% vs 33%).
It would only be 50/50 odds if a door was opened before you had a chance to pick, or you walked in to see two doors without prior information regarding which door was picked initially.
I have heard of this before.
Is the estate of Monte Hall that litigious?
Please do napoleonic wars
This explanation doesn’t make sense to me. Opening one of the “dud” doors doesn’t increase the probability you would win by switching; it just reduces the possibilities. In the first example with 3 doors you just go from a 1 in 3 chance of being right to 1 in 2. If instead the game always opened door #3 regardless of whether it was a winning or dud door, then if it did reveal the prize you would instantly know you had a 0 in 2 chance (assuming you couldn’t switch to the now-open door) and if it didn’t you’re still left with a 1 in 2 chance.
They don't always open the third door. They always open a door with a dud behind it.
Never mind, I'm teaching Father the math. Whatever, Rosa.
4:25 Tf2's random crits are broken. The more you kill, the more likley you are to get a random crit. It should be in the other way to balance it out. So if you are new, you have am evel lesser chance to get a kill. One of the many things that are killing tf2.
Nothing is killing TF2, it's literally doing better than ever before. Back in the day the average was about 40-50 000 players per month, nowadays it's between 80-100 000 average players.
For how long it stayed there, idk how much is "killing it".
But that's probably cuz people got used to it over the years. If something like that would have been added to a modern game, it most likely wouldn't last long
Games often don't even use true RNG, they will increase probabilities after repeated failures since long failure streaks will give the player the impression of a lower than indicated chance of success.
Ever been hit with a crit rocket in tf2
Just gotta say, the goats are a *great* prize
Does Brilliant teach music theory too?
Goat show
HELL YES MONTY HALL
I wonder if that is an electric goat... :/
Does anyone know it's price in the catalog?
Do the fourth crusade
How does the game change if there are two good doors and one bad door?
Maybe your reply is sarcasm. However, if you ares serious the game always eliminates a "bad" door, so you'd be left with two good doors. If you are turning this into they always eliminate one good door then staying should net you 66.6%, and switching 33.3%