what’s crazy is that i think when he played with beef boss, he beat standard in 1 or 2 tries. i think dudy dude either has a personal vendetta against poof or against the game.
Technically he is right initially though. The only factor being challenged is the ability to get over the bridge, you’d need to roll a 1,3 or 5 to do it, meaning you have a 50/50 shot. If someone else rolled and missed, and you go up, you also have a 50/50 shot. Your chances don’t get better, it’s only a 50/50 try, but it’s 6/12, instead of 3/6. There are more possibilities for you to get across on the second try, than specifically on the first try. However, like I stated before, your chances don’t increase. Say Jimmy has a 29% chance at making it to the green light on time. The only way for his chances to increase is if the car in-front of him keeps going straight and doesn’t turn at any of the intersections before the light. His chances increase the closer he gets to the light. It’s not the same with the game, since each player gets a turn at trying to turn the door, and each player has the same odds.
@@falcon1378 there's no way your this stupid. the door stays down after it's been opened. The first guy has a 50% shot of opening the door. If it gets to poof's turn then yes he also has a 50% chance of opening the door; however, we are looking at the whole picture here. If poof goes second there is a 75% chance he'll make it across the bridge 50% chance the first guy gets it and poof gets to cross and an additional 25% (0.5 * 0.5) chance to cross it by rolling it himself.
@@ciarasantee7138I am not religious and I can’t stand when people comment “repent to Jesus” and all that stuff cuz it’s so just annoying, but Jesus was actually a real person
You're ignoring that if the first place opens the door you don't have to open in again@@falcon1378 It's not 6/12 it's 3/6 + 3/6 probabilities going from a 50% chance to a 75% chance
@@saoirsecdoherty the title of the video is "Wii party U beginner difficulty doesn't want me to win". Tried to send the link of the video but it doesn't work for some reason
Wasn't in the first wii party u video, but he did get even more screwed in that one. He got UFO'ed from the end to last place like 140 spaces back and then the CPU rolled a 50.
Biostatistics major here. Though you would be just as likely to roll an odd number going in any position, the bridge would more likely be accessible after your turn if you went second vs. first (75% vs. 50%)
Out of curiosity where does 75% come from? I was thinking it would be 66% vs 50% sense going second would have 2 of the 3 possible outcomes result in poof getting through
@@SporkNotFork You have 50% chance to roll an odd number for both, thus making 4 situations since there's 2 rolls: Odd - Odd Odd - Even Even - Odd Even - Even Each one has 0.5*0.5=0.25 (so 25%) chance to happen, and all you have to do now is add up the probabilities, so 25 + 25 + 25 = 75% chance to any of the 2 players to roll an odd number, thus opening the bridge and pass if you go second You thought that were would be only 3 outcomes because of the Odd - Odd one is logically impossible since if you opened the way, there's no need to open it again. But for stats you gotta sometimes take count of impossible or illogical outcomes to find the correct answer, such as one you counted where if he rolled an odd number, there's no need to check if there's an even number right after since he can already pass
@@SporkNotFork Say O is odd and E is even. They need 1 O to pass, and both have .5 chance. With you going second after P1, this either occurs by P1 rolling O (.5) or by P1 rolling E (.5) AND you rolling O (.5); since rolls are independent, the probability of the latter sequence occurring is .5 x .5 = .25. Total prob of ending in O = prob of O from P1 + prob of O from you = .5 + .25 = .75 = 75%
9:44 Let’s talk about what the optimal play would have been for the door here. There is a 50% chance that a person who rolls for the bride will get through and unlock for everyone behind them. Because of this, assuming all 4 players are on the space just before the door, first place has a 50% chance of getting through, second place 75%, third has 82.5%, and last has 88.75% odds of getting past the door. Let’s also assume that everyone will get an average roll on their dice. If we do this, then the sides of the dice don’t matter, and we can represent them just as the amount that they are. 1st has 10 dice, 2nd 7, 3rd gets 4, and last gets 2. Combing these, we can see that 50% of the time, 1st place gets to move forward with an amount of 10 (not nessasary 10 spaces, just 10 times as much as one dice). 50% of 10 is 5, so first place, on average, would move 5 spaces. Second place gets their dice, 7, multiplied by 0.75 to get 5.25, meaning it is better to get second than first in this situation Third has 4 dice multiplied by .825, making a measly 3.3 on average. Last place gets 2 multiplied by .8875, granting them 1.775. According to these values, it would be best for you to get second place in the situation that all 4 players are stuck before the door.
so in the scenario that first rolls open the door, then you have 100% chance to get through, rolling all 7 dice giving you a score of 7 for your scoring system. in the chance this doesn’t happen, you get a 50% chance to open the bridge as they are separate, concurrent events, scoring 3.5. so in theory, first is always best in terms of getting the best bank for your buck, scoring you 5>3.5 however, going second increases your OVERALL odds of making it through the door by 1.5x, 3rd by 1.75x over first and so on. (as you said btw!) the event of the dice however, is obviously as you know a bit more complex, so i’m going to do a simple method. let’s take 7 dice and calculate the score of the dice being equal or less than half its total maximum roll value (7x6)/2 giving 21. this is rolled typically 25% of the time with 7 dice. (25% of the dice roll number combinations will equal or be less than or equal to 21) eg. 1,1,1,1,1,1,1 4,3,2,1,1,6,3 with 10 dice to get less than or equal to 21 you would need to be unlucky as you roll that 0.5% of the time. (1in200) let’s do the same but with more than/equal to 30, which is median to 10 dice. the odds are for 7, 13% roughly. the odds for 10, 84% roughly. so fact is, you have 1.4x the amount of dice in first than you do in second. if we simulated thousands of games, we would see first ahead of 2nd more times than 2nd ahead of first. due to the simple fact that 50% of the overall chance of the gate to open for 2nds relies on 1st to make it through before hand, and allowing them to get a significantly higher dice roll more often than not.
must clarify: the chances of the bridge being open for a specific player is what changes. not the odds of them opening it themselves. the bridge being open for 3rd is when it gets to there turn would be 75%. for the bridge to be open after their turn would be 87.5%. in other words, odds are 1-(0.5^n)that the bridge is open on a specific turn, but each person to open the bridge is a split 50-50.
9:00 Poof you're right, if someone goes before you, it is more likely that you'll get through the open door than if you went first. One way to think about it is to extend it to a thousand people. If a thousand people went before you, then it's super likely that at least one of them will open the door for you to go through.
funniest part of each section! pregame; poof drops his gamepad: 0:00 minigame 1; Juliette ties dudydude: 1:16 round 1; poof is last after first: 2:38 minigame 2; rigged?: 3:25 round 2: worse NAAWW: 3:39 1v3-round 2 #1; you can't skip stupidity: 4:22 after 1v1 #1; Putin's brother rolls doubles: 4:49 1v3- round 2 #2; rigging failed: 5:26 minigame 3; they all fail: 6:45 round 3; poofesures highest roll cut short: 7:50 minigame 4; dudydude breaks his face: 8:18 round 4; very advanced data analytics: 9:04 1v3-round 4 #1; stop talking: 10:24 minigame 5; dab: 10:57 round 5; 5 dice = NAAAWWW: 11:33 minigame 6; just caught a log: 11:48 round 6; though the door, to leave the door: 12:56 minigame 7; dead fish: 14:15 round 7; so climactic: 14:38 minigame 8; they shoot the same: 15:16 round 8; history repeats itself: 16:10 minigame 9; you shouldn't o' done that: 17:19 round 9; putins brothers 11s: 19:01 minigame 10; buff this: 20:36 round 10; by a hair: 21:15 post game; NOT EVEN CLOSE-technoblade: 21:41 took almost a hour to write
9:05 here are some odds: SPINNING BRIDGE Odds of getting through the spinning platform if you get to do it 1st: 50% If you get there second: 75% because there’s a 50% chance the door is already unlocked Odds you both advance: 50% Odds only the second one advances: 25% Odds none of you advance: 25% GREEN LIGHT Odds of getting at least one green light rolling 5 dices: 60% FINAL CHALLENGE Odds of rolling 40+ 20.5% Odds of rolling 35+ 53.6% If you can decide wether to try it first or second, go second, you have a 42.6% chance to win (not 53% because you won’t get to roll the dices if the person before you gets 40) whereas 1st has a 20.5% chance
9:00 I took college level intro to probability - you're absolutely right. Assuming you attempt the door before someone else, you have a 50% chance of going through (3 out of 6 possible dice rolls will let you through). Now let's assume someone else attempts the door first. You have a 50% chance of going through the door if it is closed and a 100% chance of going through assuming it is open. In other words, P(go through | door closed) = 0.5, and P(go through | door open) = 1. Since the opponent has the same 50% chance that you would normally, P(door open) is 0.5 as is P(door closed). To find the total probability, we can look at all outcomes: P(go through | door closed) * P(door closed) + P(go through | door open) * P(door open) = 0.5 * 0.5 + 1 * 0.5 = 0.75. You would have a 75% chance of going through the door if someone else were to attempt it first. Your odds improve by 25%.
Lets break down the probability problem you were talking about at the bridge. If someone goes before you, the probability of you getting over it is at 75 %. 50% for first runner to leave it open and in case the bridge does not keep open there ks another chance of 50/50. in a probability tree you could now add 50% plus 25% and get 75%
@@___ooobodybagooo___The 25% comes from the 50% chance from the first guys roll. When the first guy rolls, there’s a 50/50 chance of getting it or not. If he doesn’t get it, that’s when poof goes and there’s another 50/50 chance. The 25% comes from multiplying 50%, which is from the first guy, and multiplying it again by 50%, which is from poof’s roll, which gets us 25%. And then we can add the 25% to the 50% chance the first guy does get the bridge, giving us 75%.
Of course 0.5 prob to go through after coming first second: P(first gets) + P(poof gets| first doesn’t get) = 0.5 + 0.5(1-0.5) by independence of rolls = 0.75 But then u have to look at expected value of steps to decide whether coming first or second is better, since of course first gives u more dice and so your roll will likely be higher
I don't need sleep, I need to see if he actually won. Also I have a new job to fund college and a dolphinbar to emulate, and I happen to be short a wii remote. That new controller themed after Dudydude looks pretty damn tempting.
Congratulations Poofesure, you finally did it, I love these Wii Party series that you do man, it's so enjoyable and funny to watch you! Hope to see you play Spin Off next in Wii Party, or Mii Fashion Plaza in Wii Party U!
@@njgskgkensidukukibnalt7372”Wii Party U beginner difficulty doesn’t want me to win.” It’s his 2nd ever Wii Party U video. The thumbnail is a super close up of Beef Boss’s surprised face.
In case someone wants to know the statistics for the final doors, here it is: The odds of rolling a 40 or higher with ten dice is a 1 in 4.9 chance However, the odds of rolling a 35 or higher with ten dice is a 1 in 1.9 chance Approximately speaking, the first person who goes gets a ~20% chance to win, when the second person who goes gets a ~50% chance to win.
Statistics guy here, regarding the comment around 9:20 - The events are independent, meaning just because the person in front of you didn’t roll an odd number doesn’t make you more likely to do it yourself, you still have a 50% chance. However, being in the back of the line would still give you the best shot, because more attempts would be made ahead of you. The odds of rolling one even is 50%, the odds of two evens in a row is 25%, the odds of the first 3 players all rolling even is 12.5% and the odds of all 4 of you rolling even is 6.25%.
You are right of the dice having the same possibility to roll an even or odd number for every player, but the result of the rolls before the player affect the possibilities of having the door open or not: If you are going first, there is a 50/50 chance to open the door. If you are second, there is 3 possibilities: First player rolls an odd number (door is open): 50% First player rolls and even number (door is closed): - You roll an odd number: 25% - You roll and even number: 25% (that adds up to 50%) So the chance you have to have the door open is: the door is already open (+50%) or the door is closed but you open it (+25%) = 75% (the possibility of at least happening one thing of a set of events is the sum of its possibilities) Or what its the same: 100% minus the chance of the door not being open and you not opening the door after (25%) = 75% So if you are the third player, the possibility of not opening the door could be: First player rolls an even and second player rolls and even and you roll an even The possibility of three things happening at the same time is calculated by multiplying the three chances: 50% x 50% x 50% = 12'5% And the chance to open the door: 100% - 12'5% = 87'5% And for the fourth player, with the same method: even - even - even - even 50% x 50% x 50% x 50% = 6'25% So you have a 100% - 6'25% = 93'75% chance to open the door
I really enjoy your videos and i look forward to them every day after work man. Glad you chose to start making videos, but i do often wish i knew more about you
chat gpt to the rescue: User so say im playing at the easiest level, but i want to get second place on purpose because then someone would half to roll before me, and the chances of rolling an odd is fifty fifty. would my chances of being able to go through the door be higher? ChatGPT If you're playing Wii Party U on the easiest level and your goal is to strategically position yourself for the chance to cross the bridge by intentionally aiming for second place, here's how your strategy might work: Intentional Second Place: By aiming for second place, you ensure that another player rolls before you. This can be a strategic move to increase the chances of having an odd number rolled before your turn. Rolling Odds: If the player who goes before you rolls an odd number, then you would have a 50% chance of being able to pass the bridge successfully, assuming that rolling an odd number is the requirement for crossing the bridge. Adjusting Strategy: Your strategy depends on the odds of the other player rolling an odd number. If the odds are in your favor (50% chance for an odd roll), intentionally positioning yourself in second place could work well. Mini-Games: Keep in mind that the game may also include mini-games or other elements that could influence the outcome. Some mini-games may have an element of chance, affecting the overall strategy. It's important to note that the effectiveness of this strategy depends on the specific rules and mechanics of the game, as well as any random elements introduced by mini-games or events on the board. Experimenting with different strategies and observing the outcomes in different game sessions can help you refine your approach. User but just by logic, lemme explain. ***i want to emphasize this a lot, the number of spaces i go is not what im talking about, i only care about the chances of me getting passed the gate*** (also i forgot to mention, all 4 players have gotten to the bridge but all failed the roll, meaning all 4 players are on the spot right before the bridge, so after we finish the mini game, the first person would have to roll the dice again, so if he failed, the next person would roll and would continue until someone rolls and odd) what are my changes of only getting passed the gate if i instead of trying to get first, i let someone get infront of so they would have to roll for the bridge before me. ChatGPT I understand now. You're focused on the specific mechanic where players need to roll an odd number to pass the gate, and you're considering intentionally positioning yourself second to increase the chances of someone rolling an odd number before you. In this scenario, here's how you can analyze the probabilities: Initial Situation: All four players are on the spot right before the bridge, and none of them have successfully rolled an odd number to pass the gate. Intentional Second Place: If you intentionally aim for second place, the player in front of you will roll first for the chance to pass the gate. Odd Number Requirement: Assuming the requirement is to roll an odd number to pass the gate, there is a 50% chance of any player rolling an odd number on a fair six-sided die (1, 3, 5). Success Scenario: If the player in front of you rolls an odd number, the gate will open, and you won't need to roll for the gate. This occurs with a 50% probability. Failure Scenario: If the player in front of you rolls an even number, the gate will not open, and the next player will roll. This continues until someone rolls an odd number, allowing players to pass the gate. In this specific case, intentionally positioning yourself in second place gives you a 50% chance that the player in front of you will roll an odd number, opening the gate without you having to roll. It doesn't guarantee success, but it increases your chances compared to being in the first position. Keep in mind that this analysis assumes a fair six-sided die and that the odd number requirement is consistent across attempts. If there are additional factors or variations in the game mechanics, those would need to be considered as well. User yes obviously id have higher chances, but would you be able to give me an acurate percentage ChatGPT Certainly, let's break down the probabilities step by step: First Roll (Player in front of you): The chance of rolling an odd number is 50%. Second Roll (You): If the first player rolls an odd number, the gate opens, and you don't need to roll. This occurs with a probability of 50%. Overall Probability: The overall probability that both events happen (the player in front of you rolls an odd number and you don't need to roll) is the product of the individual probabilities. 0.5 × 0.5 = 0.25 0.5×0.5=0.25 So, in this specific scenario, there's a 25% chance that you won't need to roll for the gate if you intentionally position yourself in second place and the player in front of you rolls an odd number. Keep in mind that this is a simplified analysis based on the information provided, and the actual gameplay might involve additional elements or variations.
8:24 Here is my "technical advanced data driven analysis": Poof in fact does has a better chance of getting through the door if someone else goes first, more specifically a 75% chance (50% chance the computer passes + 25% chance poof passes it after the computer player fails) rather than a 50% chance himself. Or it's rigged I mean who knows really
Doing the domino mini game I liked I was like I'm such a nerd. Then proceed to call the dominoes dice. Anyway I love watching him play his party games. I wish he would go back to Mario Party 8.
poof i just noticed on the ”run to the sun” mini game: where there’s bigger particle stars in the background, that’s where the rocks come! meaning that’s what to avoid. for that one round, that theory worked. i hope that’s how the game is actually played and i hope you read this and never lose that mini game again. my life would be complete. (if ppl notice that or even prove i’m wrong plz comment!)
I understand the theory that Poofesure is suggesting. The idea is that the person rolling second is more likely to get through the challenge than the player moving first because the player moving first only has their roll to depend on before the next minigame and turn while the second player to move has their own roll and the roll of the person before them to rely on.
The probability of rolling the odd number after someone else rolls an even is, I think, 25% in the real world but probably different in Wii Party U. You have to roll odds, a 3 in 6 chance. You technically take two tries rolling from a 3 in 6 chance, but all 6 numbers are still there. You can roll a two and then another two, because the two can still be rolled unless you use a different die or different rules. You could theoretically roll evens forever in perfect randomness, but that happening is almost impossible. It’s definitely not possible in Wii Party where the odds seem to have some skew. (The tie roll-offs are never the same number, likely for this reason.) In regular odds, each number has a 16.67% chance of being rolled, but I think they programmed different odds after an even number is rolled to make odds more likely so the game moves on. The odds in-game are likely skewed to open the gate sooner, but in the real world, it’s a 50/50 shot on the first try, with increasing odds each try after. More depth below. I took a stats class in college, and majored in business analytics. I would need more info to do a better analysis, but this is what my gut tells me. The way I’m looking at it is like this: you roll for evens (Es) and odds (Os). It’s essentially a coin flip. Rolling an O on the first try has a 50% chance, just like an E. If you roll an E, you would have to reroll for an O, another 50% chance. These chances multiply together in statistics, so .5 * .5 = .25. So, the probabilty of O is 50%. The probability of EO is 25%. This means the probability of EEO is 12.5%. This function represents these odds: 1-(1/2)^r, where r is the number of rolls. My in-my-head math states you have an 87.5% chance of making it on 3 rolls. This number approaches 100% as you increase rolls. So E*10+O (passing the gate on the 11th roll) is already so close to 100% that it kind of doesn’t matter (99.95%, or 1999 in 2000). Likely some time before that, Nintendo changes the roll probability. In a perfect world, you could go to 1 million rolls and not pass the gate, hence why they would change the probability. For the roll-offs I mentioned you can roll the same number in the real world, not Wii Party U. My reasoning for the gate is also that it doesn’t matter if I roll a 2 then a 5. That would be (1/6)*(1/6) chance, but for the gate, the number doesn’t matter, just that it’s even or odd is what matters. The roll-offs are where the other roll matters. That function is ((1/6)*)(1/6))^r, where r is roll-offs. This function finds the probability of rolling a tie for r roll-offs in the real world, where the previous rolls are irrelevant. In the real world, each die can roll any number 1 to 6, including tying. But, Nintendo has the first die roll 1-6, then removes that result for the second die, thus never having a tie. If the first roll is a 5, the second die can roll 1, 2, 3, 4, or 6. Never 5.
Poof 3 episodes ago: "Rolling first at the start is always bad, it's better if you roll 2nd." Poof every time since then: "I'm going for 1st no matter what [insert complaints about that being bad and 2nd place doing better]."
9:05 If you're going first you have a 50/50 chance to get it, if you're second you hav a 75% chance, first the bot before you has a 50/50 chance, so you have two cases: Case 1: - Bot hits (you have a 100% chance to get through, so we start with the 50% chance of the bot for you when you didn't throw yet) Case 2: - Bot doesnt hit (you have also a 50/50 chance, so we add the probability of that times the probability that you even have to throw, so 50%^2 and get 25%) Now just add these together and you'll have the probability of getting though, which is 75% If you're going last the chance of not getting through is 6,25% btw
I played World Of Goo on the Wii, and I always found it incredibly hard (until I got it on Switch but whatever), so I honestly want to see Poof play it, regardless of if he does on Wii or Switch
That sneeze at 11:11 was the most satisfying moment in TH-cam history, not just because he finally got to cure his sneeze blueballs, but also because it _perfectly_ lines up with the magic number, 11:11.
9:25 As far as my statistics knowledge goes is that each dice roll's probability doesn't change but we would expect the likely hood of the first occurrence of some result to increase with each roll. So you're right in a practical sense but in a literal sense the actual probabilities don't change
Regarding the theory with everyone on the bridge space: 1st place: Avg. dice roll of 3.5 x 10 dice x 50% chance to advance = 17.5 expected spaces 2nd place: 3.5 x 7 dice x 75% chance to advance = 18.375 expected spaces So while first obviously has the highest ceiling you were right in thinking second has an advantage
9:09 Me, a math major: “While you’re not twice as likely, you are 1.5 times as likely.” 1st PLACE - 2 scenarios happen: a=you get the bridge and go through {p(a)=3/6=1/2} b=you don’t get the bridge {p(b)=1/2} X=Probability you get through the bridge {P(X)=1/2=0.5} 2nd PLACE - 3 scenarios happen: c=1st place gets the bridge therefore you go through {p(c)=1/2} d=1st place doesn’t get the bridge but you get the bridge and go through {p(d)=1/2*1/2=1/4} e=1st place doesn’t get the bridge and you don’t get the bridge {p(e)=1/2*1/2=1/4} Y=Probability you get through the bridge {P(Y)=1/2+1/4=3/4=0.75} …sorry I couldn’t resist
Being in second place you are more likely to get past the door HOWEVER, it’s only better to get second because the expected value is higher To calculate the expected value for a given place the calculation is: 3.5 x (num of dice) x p(pass door) In the case of player 1 it’s 3.5 x 10 x .5 = 17.5 For player 2 it’s 3.5 x 7 x (.5 + .5(.5)) = 18.375 For player 3 it’s the same process but because they only have 4 dice the expected value is not high enough regardless of how likely it is for the door. Same case applies for player 5. Player 2 is the best spot because the expected roll for that player is > than that for player 1
OK there's a 50% chance of getting through door if you are first which would give you 10 dice (vs 0 if you lose the door) giving an expected value ('average') of (using average value of a die throw of 3.5) 0.5*10*3.5+0.5*0*3.5=17.5 steps forward Now if you are second you have a 50% chance the door is already open and a 50% of opening the door if it is not already open. That gives a 0.5+(0.5*0.5)=0.75 probability of passing the door. But you get 7 dice as you are second giving an expected value of 0.75*7*3.5=18.375 steps... You would be better off going second assuming my maths is right. Big question is why do I do this with my time?
This is ONLY IF he KNOWS every other person in front of him rolled even. If everyone has not rolled yet, the odds of getting passed the bridge are NOT 50/50
The more people roll the dice to open the door, the higher the probability the door will open. You can use this: percent chance of door opening in n rolls is (1 - 1/2^n) * 100%. E.g. after 3 rolls, there will be a (1 - 1/2^3) * 100% = 7/8 * 100% = 87.5% chance of the door being open. This is because the ONLY way the door will remain closed is for those three rolls to all be failures, which have a 50% chance. So, in order for the door to stay closed, you have to hit 3 50% chances in a row, which is 1/2 * 1/2 * 1/2, which = 1/2^3 = 12.5%.
Man, it probably only took Beef Boss to beat Beginner and Standard difficulty 2 times. Dudy Dude has been struggling so many times. This game must love or hate Poofesure.
I like you poof, because you always put a comment at the top so people on mobile don’t get the video spoiled for them. You really are the best. Keep up the great work
![BOWKYLION - วิงวอน (ex-change) [Official MV]](http://i.ytimg.com/vi/cLdnZWDaO-w/mqdefault.jpg)
what happened
congrats
ido
O
bice
😮
Friendly reminder for poof: You have indeed rolled a 50 before, but it was on a ×10 backward dice roll.
actually insane
What video
@@Souc._ "Wii Party U beginner difficulty doesn't want me to win" at the 11 minute mark.
lol
@@OhMyGucciGames you’re a legend
Watching Poofesure has become like a nightly routine for me, some people listen to white noise, beach noises, etc. I however watch Poofesure videos.
Yeah. He is also a nightly routine for me. He uploads at 4 am in my timezone which fits my sleep schedule perfectly.
I also watch poofesure scream compilitations before sleeping it helps a lot
I once attempted to watch poof while sleeping and woke up scared to him rage screaming at Mkwii 😅
sometimes the screams are comforting
my family made fun of me because i did the same lol. poof screaming at 3am was not their idea of background noise lol
what’s crazy is that i think when he played with beef boss, he beat standard in 1 or 2 tries. i think dudy dude either has a personal vendetta against poof or against the game.
Dudy dude is still mad poof made him say "YOU F*CKING A$$HOLE" and "Oh f*ck yeah I'm hard as a rock" in family feud with the voice-line feature 😂
Poof debating himself on statistics and probability is the true content here
Technically he is right initially though.
The only factor being challenged is the ability to get over the bridge, you’d need to roll a 1,3 or 5 to do it, meaning you have a 50/50 shot. If someone else rolled and missed, and you go up, you also have a 50/50 shot.
Your chances don’t get better, it’s only a 50/50 try, but it’s 6/12, instead of 3/6. There are more possibilities for you to get across on the second try, than specifically on the first try.
However, like I stated before, your chances don’t increase. Say Jimmy has a 29% chance at making it to the green light on time. The only way for his chances to increase is if the car in-front of him keeps going straight and doesn’t turn at any of the intersections before the light. His chances increase the closer he gets to the light. It’s not the same with the game, since each player gets a turn at trying to turn the door, and each player has the same odds.
@RepentandbelieveinJesusChrist-jesus is not real
@@falcon1378 there's no way your this stupid.
the door stays down after it's been opened. The first guy has a 50% shot of opening the door. If it gets to poof's turn then yes he also has a 50% chance of opening the door; however, we are looking at the whole picture here.
If poof goes second there is a 75% chance he'll make it across the bridge 50% chance the first guy gets it and poof gets to cross and an additional 25% (0.5 * 0.5) chance to cross it by rolling it himself.
@@ciarasantee7138I am not religious and I can’t stand when people comment “repent to Jesus” and all that stuff cuz it’s so just annoying, but Jesus was actually a real person
You're ignoring that if the first place opens the door you don't have to open in again@@falcon1378
It's not 6/12 it's 3/6 + 3/6 probabilities going from a 50% chance to a 75% chance
Btw, the biggest role you ever rolled was a 50. It was a 10x dice backwards and I believe it was actually your first wii part u video lol
Apparently that roll was a bad omen to him.
would you happen to have a link I need to see this rage lmao
fr I need that link
@@saoirsecdoherty
@@saoirsecdoherty the title of the video is "Wii party U beginner difficulty doesn't want me to win". Tried to send the link of the video but it doesn't work for some reason
Wasn't in the first wii party u video, but he did get even more screwed in that one. He got UFO'ed from the end to last place like 140 spaces back and then the CPU rolled a 50.
If I remember correctly, Poofesure's largest roll was a 50.
Going backwards.
Video?
Well, -50 is *still* less than any positive, INCLUDING the rest.
@@PajanimationsWell, no. He rolled 50. It’s just that he went 50 backwards. The roll is still 50
You can tell this game was unexpectedly easy by how quiet Poof is.
Fr but if it is a stretched thumbnail and it’s him loosing prepare ur ears
Biostatistics major here. Though you would be just as likely to roll an odd number going in any position, the bridge would more likely be accessible after your turn if you went second vs. first (75% vs. 50%)
Needed this comment
Out of curiosity where does 75% come from? I was thinking it would be 66% vs 50% sense going second would have 2 of the 3 possible outcomes result in poof getting through
@@SporkNotFork You have 50% chance to roll an odd number for both, thus making 4 situations since there's 2 rolls:
Odd - Odd
Odd - Even
Even - Odd
Even - Even
Each one has 0.5*0.5=0.25 (so 25%) chance to happen, and all you have to do now is add up the probabilities, so 25 + 25 + 25 = 75% chance to any of the 2 players to roll an odd number, thus opening the bridge and pass if you go second
You thought that were would be only 3 outcomes because of the Odd - Odd one is logically impossible since if you opened the way, there's no need to open it again. But for stats you gotta sometimes take count of impossible or illogical outcomes to find the correct answer, such as one you counted where if he rolled an odd number, there's no need to check if there's an even number right after since he can already pass
@@SporkNotFork
Say O is odd and E is even. They need 1 O to pass, and both have .5 chance. With you going second after P1, this either occurs by P1 rolling O (.5) or by P1 rolling E (.5) AND you rolling O (.5); since rolls are independent, the probability of the latter sequence occurring is .5 x .5 = .25. Total prob of ending in O = prob of O from P1 + prob of O from you = .5 + .25 = .75 = 75%
Btw for anyone wondering for more math, you have 87.5% chance to have the bridge open by going 3rd and 93.75% chance by going 4th
9:44
Let’s talk about what the optimal play would have been for the door here.
There is a 50% chance that a person who rolls for the bride will get through and unlock for everyone behind them. Because of this, assuming all 4 players are on the space just before the door, first place has a 50% chance of getting through, second place 75%, third has 82.5%, and last has 88.75% odds of getting past the door.
Let’s also assume that everyone will get an average roll on their dice. If we do this, then the sides of the dice don’t matter, and we can represent them just as the amount that they are. 1st has 10 dice, 2nd 7, 3rd gets 4, and last gets 2.
Combing these, we can see that 50% of the time, 1st place gets to move forward with an amount of 10 (not nessasary 10 spaces, just 10 times as much as one dice). 50% of 10 is 5, so first place, on average, would move 5 spaces.
Second place gets their dice, 7, multiplied by 0.75 to get 5.25, meaning it is better to get second than first in this situation
Third has 4 dice multiplied by .825, making a measly 3.3 on average.
Last place gets 2 multiplied by .8875, granting them 1.775.
According to these values, it would be best for you to get second place in the situation that all 4 players are stuck before the door.
This is excellent
This is sensational
wrong!
so in the scenario that first rolls open the door, then you have 100% chance to get through, rolling all 7 dice giving you a score of 7 for your scoring system. in the chance this doesn’t happen, you get a 50% chance to open the bridge as they are separate, concurrent events, scoring 3.5. so in theory, first is always best in terms of getting the best bank for your buck, scoring you 5>3.5 however, going second increases your OVERALL odds of making it through the door by 1.5x, 3rd by 1.75x over first and so on. (as you said btw!)
the event of the dice however, is obviously as you know a bit more complex, so i’m going to do a simple method. let’s take 7 dice and calculate the score of the dice being equal or less than half its total maximum roll value (7x6)/2 giving 21. this is rolled typically 25% of the time with 7 dice. (25% of the dice roll number combinations will equal or be less than or equal to 21)
eg. 1,1,1,1,1,1,1
4,3,2,1,1,6,3
with 10 dice to get less than or equal to 21 you would need to be unlucky as you roll that 0.5% of the time. (1in200)
let’s do the same but with more than/equal to 30, which is median to 10 dice.
the odds are for 7, 13% roughly.
the odds for 10, 84% roughly.
so fact is, you have 1.4x the amount of dice in first than you do in second.
if we simulated thousands of games, we would see first ahead of 2nd more times than 2nd ahead of first. due to the simple fact that 50% of the overall chance of the gate to open for 2nds relies on 1st to make it through before hand, and allowing them to get a significantly higher dice roll more often than not.
must clarify: the chances of the bridge being open for a specific player is what changes. not the odds of them opening it themselves. the bridge being open for 3rd is when it gets to there turn would be 75%. for the bridge to be open after their turn would be 87.5%. in other words, odds are 1-(0.5^n)that the bridge is open on a specific turn, but each person to open the bridge is a split 50-50.
he's finally getting better at 1 2 punch after all these years. im so proud of you poof
Fr
9:00 Poof you're right, if someone goes before you, it is more likely that you'll get through the open door than if you went first. One way to think about it is to extend it to a thousand people. If a thousand people went before you, then it's super likely that at least one of them will open the door for you to go through.
Poof’s Wii Party series has seriously given me something to look forward to every week 🐐
Agreed, this is my favorite series on his channel. I remember his first play through of Wii party u master difficulty, he’s come so far
@RepentandbelieveinJesusChrist- shut up pls
Proof's Wii Party series is actually amazing, I agree! But on an unrelated note. Cipher is Soo hot 🥵
Fr
Poof using the word "nah" as a verb, adjective, and a noun during this series is so hilarious to me 😂
I swear, the games either love him or hate him. There is no in-between.
Lol
funniest part of each section!
pregame; poof drops his gamepad: 0:00
minigame 1; Juliette ties dudydude: 1:16
round 1; poof is last after first: 2:38
minigame 2; rigged?: 3:25
round 2: worse NAAWW: 3:39
1v3-round 2 #1; you can't skip stupidity: 4:22
after 1v1 #1; Putin's brother rolls doubles: 4:49
1v3- round 2 #2; rigging failed: 5:26
minigame 3; they all fail: 6:45
round 3; poofesures highest roll cut short: 7:50
minigame 4; dudydude breaks his face: 8:18
round 4; very advanced data analytics: 9:04
1v3-round 4 #1; stop talking: 10:24
minigame 5; dab: 10:57
round 5; 5 dice = NAAAWWW: 11:33
minigame 6; just caught a log: 11:48
round 6; though the door, to leave the door: 12:56
minigame 7; dead fish: 14:15
round 7; so climactic: 14:38
minigame 8; they shoot the same: 15:16
round 8; history repeats itself: 16:10
minigame 9; you shouldn't o' done that: 17:19
round 9; putins brothers 11s: 19:01
minigame 10; buff this: 20:36
round 10; by a hair: 21:15
post game; NOT EVEN CLOSE-technoblade: 21:41
took almost a hour to write
Holy shit thats a fucking essay
i ain't reading allat
I love when poof gushes us with content until we can’t take it any more 😊😊
Pause
Shut up
Dude. Come on now.
Ayo
🤨🤨🤨🤨
I think this is the calmest Wii party u video I’ve ever seen from poof
I can’t believe I found your channel over 4 years ago and I have loved every single video
POOFESURE FINALLY DID IT.
Omg this is HISTORIC
Funny enough. The only time Poof rolled a 50 was when he was rolling for how many spaces to go back
9:05 here are some odds:
SPINNING BRIDGE
Odds of getting through the spinning platform if you get to do it 1st: 50%
If you get there second: 75% because there’s a 50% chance the door is already unlocked
Odds you both advance: 50%
Odds only the second one advances: 25%
Odds none of you advance: 25%
GREEN LIGHT
Odds of getting at least one green light rolling 5 dices: 60%
FINAL CHALLENGE
Odds of rolling 40+ 20.5%
Odds of rolling 35+ 53.6%
If you can decide wether to try it first or second, go second, you have a 42.6% chance to win (not 53% because you won’t get to roll the dices if the person before you gets 40) whereas 1st has a 20.5% chance
12:17 “The key is just to watch the very top of the screen, and wait for it to come.” 🤣🤣🤣🤣🤣🤣🤣🤣
his giggle after it too 💀
9:00 I took college level intro to probability - you're absolutely right.
Assuming you attempt the door before someone else, you have a 50% chance of going through (3 out of 6 possible dice rolls will let you through).
Now let's assume someone else attempts the door first. You have a 50% chance of going through the door if it is closed and a 100% chance of going through assuming it is open. In other words, P(go through | door closed) = 0.5, and P(go through | door open) = 1. Since the opponent has the same 50% chance that you would normally, P(door open) is 0.5 as is P(door closed).
To find the total probability, we can look at all outcomes: P(go through | door closed) * P(door closed) + P(go through | door open) * P(door open) = 0.5 * 0.5 + 1 * 0.5 = 0.75.
You would have a 75% chance of going through the door if someone else were to attempt it first. Your odds improve by 25%.
It's 3:51 in the morning right now, at least where I live.
Sleepless nights with poof are quite a nice thing to have
Not sure how you didn’t turn that 3:51 into a timestamp tbh.
It's 3:51 in the morning right now, at least where I live.
Sleepless nights with poof are quite a nice thing to have
@@speedymatt1236 copy pasted the comment into a reply and it happened again
Strange!
11:11 I sneeze quite loudly, too! Bless you, Poofesure!
Lets break down the probability problem you were talking about at the bridge. If someone goes before you, the probability of you getting over it is at 75 %. 50% for first runner to leave it open and in case the bridge does not keep open there ks another chance of 50/50. in a probability tree you could now add 50% plus 25% and get 75%
50% then 25% where the 25% come from. Its 50% still. 3/6 die roll is 50% flip a coin 100 times. It's still 50% heads or tails.
@@___ooobodybagooo___The 25% comes from the 50% chance from the first guys roll. When the first guy rolls, there’s a 50/50 chance of getting it or not. If he doesn’t get it, that’s when poof goes and there’s another 50/50 chance. The 25% comes from multiplying 50%, which is from the first guy, and multiplying it again by 50%, which is from poof’s roll, which gets us 25%. And then we can add the 25% to the 50% chance the first guy does get the bridge, giving us 75%.
@@ohmpatel-qk1owcouldn‘t have explained it better thx!
Of course 0.5 prob to go through after coming first
second: P(first gets) + P(poof gets| first doesn’t get)
= 0.5 + 0.5(1-0.5) by independence of rolls
= 0.75
But then u have to look at expected value of steps to decide whether coming first or second is better, since of course first gives u more dice and so your roll will likely be higher
I don't need sleep, I need to see if he actually won.
Also I have a new job to fund college and a dolphinbar to emulate, and I happen to be short a wii remote. That new controller themed after Dudydude looks pretty damn tempting.
Congratulations Poofesure, you finally did it, I love these Wii Party series that you do man, it's so enjoyable and funny to watch you! Hope to see you play Spin Off next in Wii Party, or Mii Fashion Plaza in Wii Party U!
You got blessed by Wii Party U gods themselves for this one
7:46 you got a negative 50 once on the 10 dice going backwards
Lmao do you remember the video title
@@njgskgkensidukukibnalt7372 no but i am pretty sure it was in the beef boss days
@@njgskgkensidukukibnalt7372”Wii Party U beginner difficulty doesn’t want me to win.” It’s his 2nd ever Wii Party U video. The thumbnail is a super close up of Beef Boss’s surprised face.
In case someone wants to know the statistics for the final doors, here it is:
The odds of rolling a 40 or higher with ten dice is a 1 in 4.9 chance
However, the odds of rolling a 35 or higher with ten dice is a 1 in 1.9 chance
Approximately speaking, the first person who goes gets a ~20% chance to win, when the second person who goes gets a ~50% chance to win.
My favorite occurring series, thanks Poof.
Statistics guy here, regarding the comment around 9:20 -
The events are independent, meaning just because the person in front of you didn’t roll an odd number doesn’t make you more likely to do it yourself, you still have a 50% chance.
However, being in the back of the line would still give you the best shot, because more attempts would be made ahead of you. The odds of rolling one even is 50%, the odds of two evens in a row is 25%, the odds of the first 3 players all rolling even is 12.5% and the odds of all 4 of you rolling even is 6.25%.
You are right of the dice having the same possibility to roll an even or odd number for every player, but the result of the rolls before the player affect the possibilities of having the door open or not:
If you are going first, there is a 50/50 chance to open the door.
If you are second, there is 3 possibilities:
First player rolls an odd number (door is open): 50%
First player rolls and even number (door is closed):
- You roll an odd number: 25%
- You roll and even number: 25% (that adds up to 50%)
So the chance you have to have the door open is: the door is already open (+50%) or the door is closed but you open it (+25%) = 75% (the possibility of at least happening one thing of a set of events is the sum of its possibilities)
Or what its the same: 100% minus the chance of the door not being open and you not opening the door after (25%) = 75%
So if you are the third player, the possibility of not opening the door could be:
First player rolls an even and second player rolls and even and you roll an even
The possibility of three things happening at the same time is calculated by multiplying the three chances:
50% x 50% x 50% = 12'5%
And the chance to open the door: 100% - 12'5% = 87'5%
And for the fourth player, with the same method:
even - even - even - even
50% x 50% x 50% x 50% = 6'25%
So you have a 100% - 6'25% = 93'75% chance to open the door
You know we’re off to a good start when Poof starts the video by dropping his gamepad
Three years later and he finally looks at the top of the screen for One-Two Punch!
I really enjoy your videos and i look forward to them every day after work man. Glad you chose to start making videos, but i do often wish i knew more about you
chat gpt to the rescue:
User
so say im playing at the easiest level, but i want to get second place on purpose because then someone would half to roll before me, and the chances of rolling an odd is fifty fifty. would my chances of being able to go through the door be higher?
ChatGPT
If you're playing Wii Party U on the easiest level and your goal is to strategically position yourself for the chance to cross the bridge by intentionally aiming for second place, here's how your strategy might work:
Intentional Second Place: By aiming for second place, you ensure that another player rolls before you. This can be a strategic move to increase the chances of having an odd number rolled before your turn.
Rolling Odds: If the player who goes before you rolls an odd number, then you would have a 50% chance of being able to pass the bridge successfully, assuming that rolling an odd number is the requirement for crossing the bridge.
Adjusting Strategy: Your strategy depends on the odds of the other player rolling an odd number. If the odds are in your favor (50% chance for an odd roll), intentionally positioning yourself in second place could work well.
Mini-Games: Keep in mind that the game may also include mini-games or other elements that could influence the outcome. Some mini-games may have an element of chance, affecting the overall strategy.
It's important to note that the effectiveness of this strategy depends on the specific rules and mechanics of the game, as well as any random elements introduced by mini-games or events on the board. Experimenting with different strategies and observing the outcomes in different game sessions can help you refine your approach.
User
but just by logic, lemme explain. ***i want to emphasize this a lot, the number of spaces i go is not what im talking about, i only care about the chances of me getting passed the gate*** (also i forgot to mention, all 4 players have gotten to the bridge but all failed the roll, meaning all 4 players are on the spot right before the bridge, so after we finish the mini game, the first person would have to roll the dice again, so if he failed, the next person would roll and would continue until someone rolls and odd) what are my changes of only getting passed the gate if i instead of trying to get first, i let someone get infront of so they would have to roll for the bridge before me.
ChatGPT
I understand now. You're focused on the specific mechanic where players need to roll an odd number to pass the gate, and you're considering intentionally positioning yourself second to increase the chances of someone rolling an odd number before you. In this scenario, here's how you can analyze the probabilities:
Initial Situation: All four players are on the spot right before the bridge, and none of them have successfully rolled an odd number to pass the gate.
Intentional Second Place: If you intentionally aim for second place, the player in front of you will roll first for the chance to pass the gate.
Odd Number Requirement: Assuming the requirement is to roll an odd number to pass the gate, there is a 50% chance of any player rolling an odd number on a fair six-sided die (1, 3, 5).
Success Scenario: If the player in front of you rolls an odd number, the gate will open, and you won't need to roll for the gate. This occurs with a 50% probability.
Failure Scenario: If the player in front of you rolls an even number, the gate will not open, and the next player will roll. This continues until someone rolls an odd number, allowing players to pass the gate.
In this specific case, intentionally positioning yourself in second place gives you a 50% chance that the player in front of you will roll an odd number, opening the gate without you having to roll. It doesn't guarantee success, but it increases your chances compared to being in the first position.
Keep in mind that this analysis assumes a fair six-sided die and that the odd number requirement is consistent across attempts. If there are additional factors or variations in the game mechanics, those would need to be considered as well.
User
yes obviously id have higher chances, but would you be able to give me an acurate percentage
ChatGPT
Certainly, let's break down the probabilities step by step:
First Roll (Player in front of you): The chance of rolling an odd number is 50%.
Second Roll (You): If the first player rolls an odd number, the gate opens, and you don't need to roll. This occurs with a probability of 50%.
Overall Probability: The overall probability that both events happen (the player in front of you rolls an odd number and you don't need to roll) is the product of the individual probabilities.
0.5
×
0.5
=
0.25
0.5×0.5=0.25
So, in this specific scenario, there's a 25% chance that you won't need to roll for the gate if you intentionally position yourself in second place and the player in front of you rolls an odd number. Keep in mind that this is a simplified analysis based on the information provided, and the actual gameplay might involve additional elements or variations.
8:24 Here is my "technical advanced data driven analysis": Poof in fact does has a better chance of getting through the door if someone else goes first, more specifically a 75% chance (50% chance the computer passes + 25% chance poof passes it after the computer player fails) rather than a 50% chance himself.
Or it's rigged I mean who knows really
Doing the domino mini game I liked I was like I'm such a nerd. Then proceed to call the dominoes dice. Anyway I love watching him play his party games. I wish he would go back to Mario Party 8.
poof i just noticed on the ”run to the sun” mini game: where there’s bigger particle stars in the background, that’s where the rocks come! meaning that’s what to avoid.
for that one round, that theory worked. i hope that’s how the game is actually played and i hope you read this and never lose that mini game again. my life would be complete. (if ppl notice that or even prove i’m wrong plz comment!)
Board game island, highway rollers and globe trot is my favorite series to watch
I understand the theory that Poofesure is suggesting. The idea is that the person rolling second is more likely to get through the challenge than the player moving first because the player moving first only has their roll to depend on before the next minigame and turn while the second player to move has their own roll and the roll of the person before them to rely on.
The probability of rolling the odd number after someone else rolls an even is, I think, 25% in the real world but probably different in Wii Party U.
You have to roll odds, a 3 in 6 chance. You technically take two tries rolling from a 3 in 6 chance, but all 6 numbers are still there. You can roll a two and then another two, because the two can still be rolled unless you use a different die or different rules. You could theoretically roll evens forever in perfect randomness, but that happening is almost impossible.
It’s definitely not possible in Wii Party where the odds seem to have some skew. (The tie roll-offs are never the same number, likely for this reason.) In regular odds, each number has a 16.67% chance of being rolled, but I think they programmed different odds after an even number is rolled to make odds more likely so the game moves on. The odds in-game are likely skewed to open the gate sooner, but in the real world, it’s a 50/50 shot on the first try, with increasing odds each try after. More depth below.
I took a stats class in college, and majored in business analytics. I would need more info to do a better analysis, but this is what my gut tells me.
The way I’m looking at it is like this: you roll for evens (Es) and odds (Os). It’s essentially a coin flip. Rolling an O on the first try has a 50% chance, just like an E. If you roll an E, you would have to reroll for an O, another 50% chance. These chances multiply together in statistics, so .5 * .5 = .25. So, the probabilty of O is 50%. The probability of EO is 25%. This means the probability of EEO is 12.5%.
This function represents these odds: 1-(1/2)^r, where r is the number of rolls.
My in-my-head math states you have an 87.5% chance of making it on 3 rolls. This number approaches 100% as you increase rolls.
So E*10+O (passing the gate on the 11th roll) is already so close to 100% that it kind of doesn’t matter (99.95%, or 1999 in 2000). Likely some time before that, Nintendo changes the roll probability. In a perfect world, you could go to 1 million rolls and not pass the gate, hence why they would change the probability.
For the roll-offs I mentioned you can roll the same number in the real world, not Wii Party U.
My reasoning for the gate is also that it doesn’t matter if I roll a 2 then a 5. That would be (1/6)*(1/6) chance, but for the gate, the number doesn’t matter, just that it’s even or odd is what matters. The roll-offs are where the other roll matters. That function is ((1/6)*)(1/6))^r, where r is roll-offs. This function finds the probability of rolling a tie for r roll-offs in the real world, where the previous rolls are irrelevant. In the real world, each die can roll any number 1 to 6, including tying. But, Nintendo has the first die roll 1-6, then removes that result for the second die, thus never having a tie. If the first roll is a 5, the second die can roll 1, 2, 3, 4, or 6. Never 5.
Really love these videos, they always make my day, seeing poofesure post new videos gets my day better, lover your videos
Poof 3 episodes ago: "Rolling first at the start is always bad, it's better if you roll 2nd."
Poof every time since then: "I'm going for 1st no matter what [insert complaints about that being bad and 2nd place doing better]."
Oh, you’ve rolled over 50 before… in the wrong direction.
Don’t you hate it when that happens?
Fun fact Poof has rolled a 50 before, it was just going backwards…
It always good to watch poofesure playing Wii Party/Or Wii Party U best series of the Mii party’s of all time :)
9:05 If you're going first you have a 50/50 chance to get it, if you're second you hav a 75% chance, first the bot before you has a 50/50 chance, so you have two cases:
Case 1:
- Bot hits (you have a 100% chance to get through, so we start with the 50% chance of the bot for you when you didn't throw yet)
Case 2:
- Bot doesnt hit (you have also a 50/50 chance, so we add the probability of that times the probability that you even have to throw, so 50%^2 and get 25%)
Now just add these together and you'll have the probability of getting though, which is 75%
If you're going last the chance of not getting through is 6,25% btw
"I use advanced data analytics to get through doors" Best thing I've ever heard
I played World Of Goo on the Wii, and I always found it incredibly hard (until I got it on Switch but whatever), so I honestly want to see Poof play it, regardless of if he does on Wii or Switch
dudy dude is honestly my favourite poofesure mii
Love watching poofesure before bed, probably the best TH-camr out there in terms of entertaining hameplay
That sneeze at 11:11 was the most satisfying moment in TH-cam history, not just because he finally got to cure his sneeze blueballs, but also because it _perfectly_ lines up with the magic number, 11:11.
9:25 As far as my statistics knowledge goes is that each dice roll's probability doesn't change but we would expect the likely hood of the first occurrence of some result to increase with each roll. So you're right in a practical sense but in a literal sense the actual probabilities don't change
Had he not won this time, he would have lost not two, not three, but five times. 😂 But he got it !
Regarding the theory with everyone on the bridge space:
1st place: Avg. dice roll of 3.5 x 10 dice x 50% chance to advance = 17.5 expected spaces
2nd place: 3.5 x 7 dice x 75% chance to advance = 18.375 expected spaces
So while first obviously has the highest ceiling you were right in thinking second has an advantage
9:09
Me, a math major:
“While you’re not twice as likely, you are 1.5 times as likely.”
1st PLACE - 2 scenarios happen:
a=you get the bridge and go through {p(a)=3/6=1/2}
b=you don’t get the bridge {p(b)=1/2}
X=Probability you get through the bridge {P(X)=1/2=0.5}
2nd PLACE - 3 scenarios happen:
c=1st place gets the bridge therefore you go through {p(c)=1/2}
d=1st place doesn’t get the bridge but you get the bridge and go through {p(d)=1/2*1/2=1/4}
e=1st place doesn’t get the bridge and you don’t get the bridge {p(e)=1/2*1/2=1/4}
Y=Probability you get through the bridge {P(Y)=1/2+1/4=3/4=0.75}
…sorry I couldn’t resist
12:18 “the key is to just watch the very top of the screen for it to come…. Hehe🤭” that made me laugh so hard bc I’d laugh too
Poofesure never fails to bring entertainment to us.
7:06 “I’m gonna be a bottom now.”
ehehehehehehehe
Being in second place you are more likely to get past the door HOWEVER, it’s only better to get second because the expected value is higher
To calculate the expected value for a given place the calculation is:
3.5 x (num of dice) x p(pass door)
In the case of player 1 it’s
3.5 x 10 x .5 = 17.5
For player 2 it’s
3.5 x 7 x (.5 + .5(.5)) = 18.375
For player 3 it’s the same process but because they only have 4 dice the expected value is not high enough regardless of how likely it is for the door. Same case applies for player 5.
Player 2 is the best spot because the expected roll for that player is > than that for player 1
OK there's a 50% chance of getting through door if you are first which would give you 10 dice (vs 0 if you lose the door) giving an expected value ('average') of (using average value of a die throw of 3.5) 0.5*10*3.5+0.5*0*3.5=17.5 steps forward
Now if you are second you have a 50% chance the door is already open and a 50% of opening the door if it is not already open. That gives a 0.5+(0.5*0.5)=0.75 probability of passing the door. But you get 7 dice as you are second giving an expected value of 0.75*7*3.5=18.375 steps...
You would be better off going second assuming my maths is right. Big question is why do I do this with my time?
12:20 the secret is to watch the very top... yeah no shit where have you been looking?
Poof explaining the watermelon catching game 😭😭😭 who would’ve guessed you had to watch for whether or not it’s a log or a watermelon
12:20 his little giggle after saying “come” 💀
21:21 hair salesmen be like
This has to be the first time that he won every single minigame in Highway Rollers (1 vs 3's don't count)
Poofesure , your videos have helped me through dark times in the past . Keep up the good work 🫡
6:47-7:10
Dating as a Gay Vers be like
Best comment ever
i think poof is right. i think the game pre-determines who's gonna win, which is why he wins when he wins, but will get outright screwed elsewise!
You caught the melons wow! You are improving!
Next video we need to see your improved swimming skills 😂
@Poofesure_Giveawayss OMG is this the real Poofesure_Giveawayss??
Waiting for this to drop at 2:50am in the uk
Found a video of all your 50+ rolls, so its safe to say 48 isnt your highest. Found it by trying to find out if someone rolled a 60 before.
Finally, glad you finally beat STANDARD difficulty. CONGRATS
Poof it’s always gunna be a 50% no matter what position you roll at
This is ONLY IF he KNOWS every other person in front of him rolled even. If everyone has not rolled yet, the odds of getting passed the bridge are NOT 50/50
First off. Congrats. Second off, bless you.
Poof needs to be my stats teacher i swear
Greetings from Germany, i Love Wii Sports! Kann nicht drauf warten bis die Deutschen die Kommentarsektion einnehmen
yes poof, if you go second you gave a higher chance going through the door
The meteors go right, left, up, down, right so you have to do the opposite of that lol
At the first Door in this specific going second has a expected value of 18.375, while going first only 17,5 so going second is actually better
Imma start gambling because using the door logic: I either win or I don't and 50% is a passing grade, so I can basically win 50% of the time
You have a higher chance of getting past the bridge but the same chance of rolling the door if you got second.
The more people roll the dice to open the door, the higher the probability the door will open.
You can use this: percent chance of door opening in n rolls is (1 - 1/2^n) * 100%. E.g. after 3 rolls, there will be a (1 - 1/2^3) * 100% = 7/8 * 100% = 87.5% chance of the door being open.
This is because the ONLY way the door will remain closed is for those three rolls to all be failures, which have a 50% chance. So, in order for the door to stay closed, you have to hit 3 50% chances in a row, which is 1/2 * 1/2 * 1/2, which = 1/2^3 = 12.5%.
I like how he complains about the difficulty of the tube minigame and is the only one of the group to get immediately clapped
53 is biggest roll
Standard Arc Ending
The Wii gods finally rewarded his perseverance and mental sacrifice. That or the game felt bad and forced him to win.
It would be 75% chance as its a 50/50 chance that its open , and if its closed another 50/50 chance , so its ½ + ½ × ½ , so 0.5 + 0.25, so 0.75
Watching poofesure at night is my medicine
This makes my day better
Man, it probably only took Beef Boss to beat Beginner and Standard difficulty 2 times. Dudy Dude has been struggling so many times. This game must love or hate Poofesure.
2:21 alfonso staring at us in the background feels strangely disturbing
14:05 rare Wii Party U minigame! I dont think we've seen this one yet or if we have its been ages!
I like you poof, because you always put a comment at the top so people on mobile don’t get the video spoiled for them. You really are the best. Keep up the great work
7:45 i love the fact that the one time he got a 50 it was negative
Poof won so fast he had to show the entire final opponent roll to make the twenty.