THANK YOU, THANK YOU! Finally a mathematician that explains theoretical probability near the end of this video and doesn't try to seduce people with mathematical slight of hand by distorting reality. Leonardo Barichello for the win!
Interestingly you can easily make this game fair by adding one simple rule. 3) If both dice come up a five or six then a re roll is required. This means that Player 1 still has a chance of winning of 16/36 with one roll. Player 2 however now also only has a 16/36 chance of winning. There is a 4/36 chance you will need a reroll that leads to a situation with the exact same chance on either side of winning. So about half the time player 1 will win, and half the time player 2 will win. There is an very minute chance that you will require rerolls until you are both sick of the game.
+DJ Syntic Instead of rerolling, you could make the rule that Player 1 wins if both dice come up the same (i.e. two 5s or two 6s). Easier still, though, you could just roll one die and if it comes up odd, Player 1 wins, and if it comes up even, Player 2 wins.
+Mark Ten Well, that's just not complicated enough. Frankly, I'm insulted by the idea of having any less than 50 dice of varying polyhedral shape with you on a deserted island.
+DJ Syntic Umm that doesn't even work? There's only a 2/36 chance of rolling two fives or two sixes; there is a 2/36 chance that a reroll would occur. If player 1 would win with these two outcomes then both players would have 18/36 = 50%. With your suggestion, player2 has 50% of winning plus a chance at the reroll. Besides, mathematically it would be possible that the game would go on until the end of time with infinite rerolls.
Here's an improvement in the method which can give you 78% probability to win. Be the second player so there are 56% chances you will win. If you win, just take the banana, but if you lose (44% ) start trying to beat the other player physically, assuming both are equally strong, there are another 22% (50% of 44%) chances you will win, so in total there are 78% chances you will win.
only if you assume the other player is more trustworthy than you have turned out to be. if they fight you when you win like you plan to do to them the original game actually becomes irrelevant and it reduces to a 50% chance for both.
Right. Who's to say the other guy won't fight you when you win and will be fast enough to eat the banana before you even start a fight when you lose? (assuming you have an equal chance to be fast enough to just eat the banana when you win before the other guy starts a fight with you.)
Math Utility helped me so much! That proportions with variables video went through the whole process in less than a minute! They tromp the competition, and I will always rely on these videos! Give them a check-out, because honestly you wont regret it
Only in a math problem can a cat and a kid decide to play a game this complex instead of splitting the banana or making a much more simple game. . . Great video! :)
Using a computer simulation in python, I tried this out: Results: Player A: 44.4457644446% (expected: 44.444444444%) Player B: 55.5542355554% (expected: 55.555555555%) from random import randint pA = 0 pB = 0 trials = 9999999 def experiment(): global pA, pB nA = randint(1,6) nB = randint(1,6) if nA == 5 or nA == 6 or nB == 5 or nB == 6: pB += 1 else: pA += 1 return None for i in range(trials): experiment() print 'Results:' print 'Player A: ', str(100*(float(pA)/trials)) + '%' print 'Player B: ', str(100*(float(pB)/trials)) + '%'
+lee melymick python did a good job? This has nothing to do with python lol. If anything, the formula that generated the pseudo-random numbers in the function randint did a good job.
It took me a while, (I'm still a beginner in Python) but I got it! But is the return statement really necessary though? Agh, for loops are so confusing for some reason lol. :P
Greed? You dont wonder why you would have only one banana on an island but wonder why they are greedy? Besides people would probably kill each other if they are starving and just one banana is out there... Wonder if some psychologist would call a starving one killing another starving one so he can get the banana "a greedy one"...
Musti C. In reality, one banana isn't going to sustain you for very long and you're most likely going to die. So there wouldn't be any point in making the last thing you did with your life killing someone for a single banana.
Doug McQuade yep. That is the best way. But not many ppl out there will do it. I mean... If u would ask me.. now I would say that I would never ever kill any one. But my dear... when rly something like that happens. I wonder if I will still think the same way I do right now. Ate pasta and a cookie and enjoying my tea right now.. no starving I sight but not excluded either. Starving is the hardest death someone can have, imo. There are ppl who ate their hair cuz they were starving. And a banana is definitely better than hair ^^'.
BONUS QUESTION: Three people roll one die each. if the smallest number rolled is one - player 1 wins. If the smallest number rolled is two or three - player 2 wins. If the smallest number rolled is four, five or six - player 3 wins. Who has the best probability of winning the game?
+scarletovergods Player 2 has the highest probability; his probability of winning is 4/9. The probability for player 1 is 31/72, which is very close to that of player 2, but for player 3 it is only 1/8.
+Mark Ten DETAILED SOLUTION For player 1 to win, there must be *at least* a 1. The probability of getting at least a 1 is complementary to the probability of not getting a one. Probability of not getting a one = probability of getting all 2-6s = (5/6)^3 Hence, probability of getting at least a one = 1 - (5/6)^3 = 91/216 For player 3 to win, there can only be 4, 5 and 6. Thus the probability is simply (3/6)^3 = 1/8. Since the probabilities of each player winning all add up to 1, the probability of player 2 winning is simply = 1 - 91/216 - 1/8 = 98/216.
@@matttejada7381 Actually no, Cats are obligate carnivores If they eat grass, it is most likely because they are trying to induce vomiting to clear out something irritating it's stomach (fur, parasites, etc.) That said, even obligate carnivores can consume plant based matter in small quantities, most cat foods are a blend of animal protein and starch (the plant matter in cat food is processed to be more readily digested, if that same plant matter were encountered in the wild, the cat's stomach would have a hard time breaking it down)
Okay. If number 1-4 comes up on the first dice, there's a 2/6 chance that it will be 5 or 6 on the second one, but if number 5-6 comes out on the first dice, player two have already won. So the total probability P(player one winning) = (1/6 * 4/6) + (1/6 * 4/6) + (1/6 * 4/6) + (1/6 * 4/6) + (1/6 * 0) + (1/6 * 0) = 4(1/6 * 4/6) + 2(1/6 * 0) = 4/6 * 4/6 + 0 = 16/36 = 4/9 ≈ 0.44. And P(player two winning) = 1 - P(player one winning) = 1 - 4/9 = 5/9 ≈ 0.56. We can also count the numbers of successful outcomes for player one and divide that by the total amount of outcomes, like so: P(player one winning) = (4 + 4 + 4 + 4)/36 = 16/36 = 4/9.
Another bonus answer: If the biggest number is 1, there is 1 possibility to win for player one; if it's 2, there are two; if three, three, etc. So, player one's possibilitys are ((1+2+3+4)*2)-4=10*2-4=20-4=16 and player two's are ((5+6)*2)-2=11*2-2=22-2=20. The *2 part is for the cases where whats explained before happens with the other dice. The -4 is for the ones that are repeated (1 and 1, 2 and 2, 3 and 3, 4 and 4), and -2 is for 5 and 5, 6 and 6
Good Morning! watched this last night and slept on it, this morning it hit me on how one might even the odds... add that all doubles go to the first player. The only change will be the 2 matches which would have been going to the second player so 16-20 become 18-18.
And this is why I have rolled quite a few critical misses in the past few weeks of D&D... Nothing like having three critical misses in a row with a 5% chance that I am going to roll a 1 on a D20 each time I roll. It's almost to the point that I have a huge amount of anxiety come up whenever I am rolling my attacks.
If you make it so player 1 wins if the highest number is 1, 2, 3, or 5 and player to wins if it's 4 or 6, they'll each have a 50% chance of winning. Also, if you use dice with sides that are divisible by 4, you can give the lower 4th and the upper 4th to one player and the middle half to the other and they'll both have a 50% chance of winning.
+Alvin Lee I'm glad I didn't have too much college, I've been learning from this guy: dfpolis R-9 Have Reason Will Travel. That's his 13th response video to naturalists who learned that the brain produces the mind (they learned it in college), but that doesn't make any sense, we are aware before our brain could process awareness!
You can please ALL of the people SOME of the time. You can please SOME of the people ALL of the time. But there are SOME people that you just CAN'T please!
KrisBlueNZ Doesn't the statement "There are some people you can't please" directly contradict the statement "You can please all of the people some of the time"?
Why would I "at first glance" think that player one has an advantage? Player two could win on either die. Even if I didn't sit down and work out the odds like twenty different ways (really, did it take this long to explain?), it would seem like player two had the clear advantage.
Shawn Ravenfire The reason it might look that way on first glance is that Player One has more sides in favor; any of the first 4 numbers are more likely to turn up than any of the last two, appearing to put the odds in Player One's favor with 2/3 chance of winning. Of course, when one looks at it more carefully, the latter two numbers actually hold a stronger chance for victory when they do turn up; any of Player One's numbers can be overpowered by Player Two's, while Player Two can never be overpowered (Even if Player One gets a 6, it counts as a win for Player Two).
George Dolley If you need to explain why it immediately looks like player one has the advantage, then it doesn't immediately look like player one has the advantage.
You can simply do these calculations in your head if you calculate the chance of NOT rolling a winning number. Each die has a 2/3 chance of not rolling a winning number, multiply them and you have a 4/9 chance of not winning, meaning you have a 5/9 chance of winning. This works for any number of winning numbers and any number of dice.
Actually the coin toss is also unfair when you catch the coin after tossed. You can experiment by tossing 3 times and results are the side which is up when hold 2 times is the result. So to be fair it is wiser to let coin dropped on the ground in cricket games.
Just to point out the math for Player 2: *Player 2* has two winning numbers, but he only needs *one* of the dice to be a winning number to win. So his chances are: (5-6 on the first die) OR (1-4 on the first die AND 5-6 on the second die) Pr(Player 2 wins) = 2/6 + (2/6 * 4/6) = 5/9 : )
+Copydot ........just saw this after posting my comment up there lol. How come the win requirement for player 2 is not just % of first roll >4 OR % of second roll >4 = 2/6 + 2/6 = 2/3. Is this 2/6 * 4/6 part of the win requirement for player 2?
+dzpak Well, think of it like this: if you roll a 5 or 6 on either die, you can stop rolling right? So the odds of rolling the second die at all are dependent on getting a 4 or less on the first. So even though your odds of rolling a 5 or 6 on the second die are the same, there's only a 4/6 chance that you'll *have to* roll that second die. And if you could just add 2/6+2/6... for each die, then for 4 dice you'd end up with an 8/6 chance of rolling 5 or 6. That doesn't make much sense ;)
You can launch a dice in a centrifuge with rockets from the moon, make it land in a fast flowing river, picking up the centrifuge after it has fallen from a waterfall and see what the dice has "rolled", and it still wouldnt be random, as long as you calculate all the variables. Now compared to a flat surface and a simple hand throw to roll some dices wouldnt be hard to calculate.
Besides making a table is there an easy way to see right away that it was the case player 2 had the better probability? At first when asked I mistakenly thought player 1 but now it's obvious. How can you calculate the odds right away with no table or listing.?
+Chandler C You can calculate in your head the probability that the girl wins, which is 4/6 * 4/6 = 16/36. From this you can quickly see that it is less than half, so the cat has a better chance.
I have a question for very smart math nuts, what is the probability of getting any 7 random numbers from 0-50 vs the probability of getting a single number 7 times in a row??? Intuition says 7 of the same number is less probable however, aren't all numbers equally as improbable even if it's repeated? Ahhh it's killing me, which has a lower probability? Or are they equal?
+95TurboSol The reason getting "any 7 random numbers" seems more probable is that you aren't thinking of any specific numbers. For instance, if you were thinking of the numbers "1, 2, 3, 4, 5, 6, 7" vs "7, 7, 7, 7, 7, 7, 7" it would be at least a little more obvious that they have the same probability of showing up. We just tend to think of certain strings of numbers as "special," whereas other number combinations are boring and unremarkable, so the boring numbers SEEM more likely to show up even though they aren't, because that category of numbers is much larger. Does that help?
eljazeccle19 Exactly what I was thinking. I would love to see someone do the math on it though just to be sure, sometimes intuitions can be misleading.
+95TurboSol The first one is 100% because saying 'any' implies there's no limitations on the numbers you can receive, so you can receive 7 0-50 numbers, unless you mean receiving up to 7, then that would be (8/51)^7 which equals about 0.00023%. The second one would just simply be (1/51)^6 (6 because all the numbers are based off the first number you receive, so you don't take the first number into account) which equals about 0.000000006%; you'd have a better chance at the lottery at 0.000007%.
Chicken Surprise Thanks, yeah I meant for instance if I did an auto pick of any 7 random numbers up to 50 and allowing repeat numbers. Now suppose I got 2 19 46 31 28 10 39 on one random pick, then doing it again suppose I got 5 5 5 5 5 5 5, which is less probable? Or are they the same?
player 2 also wins because each side is slightly heavier or lighter than another, like 6 is the lightest because it has 6 holes, unlike 1 has only 1 hole
Took this to VBA for a test of 10.000.000. Giving the following wins Ta = 5555139 Tb = 4444861 So the Rnd of VBA seems to work just as expected. Sub banana() Dim Ta, Tb, count, A, B As Long For count = 1 To 100 Ta = Int(6 * Rnd) + 1 Tb = Int(6 * Rnd) + 1 If Ta >= 5 Or Tb >= 5 Then A = A + 1 Else B = B + 1 Next count Debug.Print A, B End Sub
But I've always thought that the number 1 side is slightly heavier because they indent holes into the surface for the number of spots, if this is the case number 6 (the opposite side) would come up more frequently than its 'probability' would claim?
You can also say that if they play forever then eventually player 1 would win 99% of the time since if you have an infinite amount of time then you can just keep playing until player one gets insanely lucky, but by that time they may need to win 1000s of games in a row, yet it is theoretically possible given enough time.
Just eat the banana she's so hungry that she thinks that the cat is a person
Good observation
cats also need food but considering the island appears to be made out of spam, they're both delusional.
LOL
@@throatychunk bananas are bad for cats
better yet eat the cat
i'd just suggest we split the banana.
+ijfharvey Ray Comfort is the banana man.
TrustInJesus111 You mean Ray Cumfart?
TrustInJesus111 Ok, I'm blocking you now.
I don't remember that cats eat bananas or can stand on 2 legs and shake hands and throw a dice
Id just eat the banana lol
THANK YOU, THANK YOU! Finally a mathematician that explains theoretical probability near the end of this video and doesn't try to seduce people with mathematical slight of hand by distorting reality. Leonardo Barichello for the win!
Thank you :)
how do you find dice on a deserted island yet no food
And there is also the fact that you have vegetation all around you
...a cat also won't eat a banana.
at the walmart, but they may not be organic
They had dice with them.
It's OUR island, but we do die.
Interestingly you can easily make this game fair by adding one simple rule.
3) If both dice come up a five or six then a re roll is required. This means that Player 1 still has a chance of winning of 16/36 with one roll. Player 2 however now also only has a 16/36 chance of winning. There is a 4/36 chance you will need a reroll that leads to a situation with the exact same chance on either side of winning. So about half the time player 1 will win, and half the time player 2 will win. There is an very minute chance that you will require rerolls until you are both sick of the game.
+DJ Syntic Interesting idea!
+DJ Syntic Instead of rerolling, you could make the rule that Player 1 wins if both dice come up the same (i.e. two 5s or two 6s).
Easier still, though, you could just roll one die and if it comes up odd, Player 1 wins, and if it comes up even, Player 2 wins.
+DJ Syntic Or just throw one die since there is an even number of sides.
+Mark Ten Well, that's just not complicated enough. Frankly, I'm insulted by the idea of having any less than 50 dice of varying polyhedral shape with you on a deserted island.
+DJ Syntic Umm that doesn't even work? There's only a 2/36 chance of rolling two fives or two sixes; there is a 2/36 chance that a reroll would occur. If player 1 would win with these two outcomes then both players would have 18/36 = 50%. With your suggestion, player2 has 50% of winning plus a chance at the reroll. Besides, mathematically it would be possible that the game would go on until the end of time with infinite rerolls.
Real solution: Don't buy odd number of bananas
distributing equal number of bananas to each individual doesn't make it fair.
Just split the banana is good enough
Valen Cheung
How is that not fair?
What if there is an odd number of people
You think they bought the banana at the Deserted Island Whole Foods?
Or don't buy bananas at all.
Here is a thought experiment: This is a coin toss game. It's called "Heads I win; Tails you lose." Enjoy.
+KnowJesusKnowPeace ... oh you.
LOL
My dad trolled me with this when I was around 8 years old. I didn't get why I was losing all the time.
If it falls on the side, he wins.
Roll the dice, don't think twice.
Here's an improvement in the method which can give you 78% probability to win. Be the second player so there are 56% chances you will win. If you win, just take the banana, but if you lose (44% ) start trying to beat the other player physically, assuming both are equally strong, there are another 22% (50% of 44%) chances you will win, so in total there are 78% chances you will win.
only if you assume the other player is more trustworthy than you have turned out to be. if they fight you when you win like you plan to do to them the original game actually becomes irrelevant and it reduces to a 50% chance for both.
Right. Who's to say the other guy won't fight you when you win and will be fast enough to eat the banana before you even start a fight when you lose? (assuming you have an equal chance to be fast enough to just eat the banana when you win before the other guy starts a fight with you.)
You know, I think a huge cat will beat a person of equal size (because claws and teeth).
Or you could roll the die first, and then just eat the banana while the other player rolls their die. 100% probability
Anthony Clohesy lol 69 likes
Math Utility helped me so much! That proportions with variables video went through the whole process in less than a minute! They tromp the competition, and I will always rely on these videos! Give them a check-out, because honestly you wont regret it
Only in a math problem can a cat and a kid decide to play a game this complex instead of splitting the banana or making a much more simple game. . .
Great video! :)
I just wanted to be the cat
furry
Aww
Hecking furries cheating us out on our food
But the (Schrodinger's) cat is in the box
Using a computer simulation in python, I tried this out:
Results:
Player A: 44.4457644446% (expected: 44.444444444%)
Player B: 55.5542355554% (expected: 55.555555555%)
from random import randint
pA = 0
pB = 0
trials = 9999999
def experiment():
global pA, pB
nA = randint(1,6)
nB = randint(1,6)
if nA == 5 or nA == 6 or nB == 5 or nB == 6:
pB += 1
else:
pA += 1
return None
for i in range(trials):
experiment()
print 'Results:'
print 'Player A: ', str(100*(float(pA)/trials)) + '%'
print 'Player B: ', str(100*(float(pB)/trials)) + '%'
+lee melymick python did a good job? This has nothing to do with python lol. If anything, the formula that generated the pseudo-random numbers in the function randint did a good job.
It took me a while, (I'm still a beginner in Python) but I got it! But is the return statement really necessary though?
Agh, for loops are so confusing for some reason lol. :P
Or player 1 and player 2 could not be greedy assholes and just split the banana in half...
"A thought experiment"
Сабыркулов Эмир He made a thought experiment aswell...
Greed? You dont wonder why you would have only one banana on an island but wonder why they are greedy? Besides people would probably kill each other if they are starving and just one banana is out there... Wonder if some psychologist would call a starving one killing another starving one so he can get the banana "a greedy one"...
Musti C. In reality, one banana isn't going to sustain you for very long and you're most likely going to die. So there wouldn't be any point in making the last thing you did with your life killing someone for a single banana.
Doug McQuade yep. That is the best way. But not many ppl out there will do it. I mean... If u would ask me.. now I would say that I would never ever kill any one. But my dear... when rly something like that happens. I wonder if I will still think the same way I do right now. Ate pasta and a cookie and enjoying my tea right now.. no starving I sight but not excluded either. Starving is the hardest death someone can have, imo. There are ppl who ate their hair cuz they were starving. And a banana is definitely better than hair ^^'.
BONUS QUESTION:
Three people roll one die each. if the smallest number rolled is one - player 1 wins. If the smallest number rolled is two or three - player 2 wins. If the smallest number rolled is four, five or six - player 3 wins.
Who has the best probability of winning the game?
+scarletovergods Player 2 has the highest probability; his probability of winning is 4/9. The probability for player 1 is 31/72, which is very close to that of player 2, but for player 3 it is only 1/8.
+scarletovergods Player 2 has the highest probability of winning. Player 3 has the lowest.
+scarletovergods
SOLUTION
Player 1: 1 - (5/6)^3 = 91/216
Player 2: (5/6)^3 - (3/6)^3 = 98/216
Player 3: (3/6)^3 = 1/8 = 27/216
+Mark Ten
DETAILED SOLUTION
For player 1 to win, there must be *at least* a 1. The probability of getting at least a 1 is complementary to the probability of not getting a one.
Probability of not getting a one = probability of getting all 2-6s = (5/6)^3
Hence, probability of getting at least a one = 1 - (5/6)^3 = 91/216
For player 3 to win, there can only be 4, 5 and 6. Thus the probability is simply (3/6)^3 = 1/8.
Since the probabilities of each player winning all add up to 1, the probability of player 2 winning is simply = 1 - 91/216 - 1/8 = 98/216.
Mark Ten You are right. I must have made a small mistake.
CATS ARE CARNIVORES THEY CAN'T EVEN PROPERLY DIGEST BANANAS
haha lol
My cat eats pazzeles.
They are omnivores, you probably never seen a cat eating grass just like my neighbor's
@@matttejada7381 Actually no, Cats are obligate carnivores
If they eat grass, it is most likely because they are trying to induce vomiting to clear out something irritating it's stomach (fur, parasites, etc.)
That said, even obligate carnivores can consume plant based matter in small quantities, most cat foods are a blend of animal protein and starch (the plant matter in cat food is processed to be more readily digested, if that same plant matter were encountered in the wild, the cat's stomach would have a hard time breaking it down)
@@Deathnotefan97 thanks for this information
Soo satisfying to watch these videos after you learn it's topic. Just had some lessons in college about probability and that's really accurate!
You know, just devide the banana into two
+✏Hello_Random✏ i was gonna say to :D
+Turtles and then eat the person
A banana is a banana, you can't say it's only a half.
well TJ "Timestoppa" Yoshi!
luigisghost227 but first, we need to talk about parallel universes.
Nice video, explaining both the methods ....and the difference between theoretical and experimental probability...
Seems like all these probabilities go out the window when I'm playing DnD
Ted ed for the million time you are my best TH-cam channel
Okay. If number 1-4 comes up on the first dice, there's a 2/6 chance that it will be 5 or 6 on the second one, but if number 5-6 comes out on the first dice, player two have already won.
So the total probability P(player one winning) = (1/6 * 4/6) + (1/6 * 4/6) + (1/6 * 4/6) + (1/6 * 4/6) + (1/6 * 0) + (1/6 * 0) = 4(1/6 * 4/6) + 2(1/6 * 0) = 4/6 * 4/6 + 0 = 16/36 = 4/9 ≈ 0.44.
And P(player two winning) = 1 - P(player one winning) = 1 - 4/9 = 5/9 ≈ 0.56.
We can also count the numbers of successful outcomes for player one and divide that by the total amount of outcomes, like so: P(player one winning) = (4 + 4 + 4 + 4)/36 = 16/36 = 4/9.
I'm not greedy, I would just share the Banana.
Nope *takes your half and my half of the banana and eats it in one bite* meh benana
If it were me I would give the banana to the cat so it will be healthy once I ate it. YUMYUM CAT STEW!!
Franco Earl Lacno Although, that is a bit inefficient. I think you can see this in the Food Pyramid or something like that.
+Franco Earl Lacno I don't think a saltwater cat is yummy.
+The Gaming Oracles Faith in humanity vanquished. Congratulations.
No the cat is too huge. Even if I beat it, it will scratch me and I will get infected, andjust recovering from that would take a ton of energy.
Not if it's asleep
Another bonus answer: If the biggest number is 1, there is 1 possibility to win for player one; if it's 2, there are two; if three, three, etc. So, player one's possibilitys are ((1+2+3+4)*2)-4=10*2-4=20-4=16 and player two's are ((5+6)*2)-2=11*2-2=22-2=20. The *2 part is for the cases where whats explained before happens with the other dice. The -4 is for the ones that are repeated (1 and 1, 2 and 2, 3 and 3, 4 and 4), and -2 is for 5 and 5, 6 and 6
Good Morning! watched this last night and slept on it, this morning it hit me on how one might even the odds... add that all doubles go to the first player. The only change will be the 2 matches which would have been going to the second player so 16-20 become 18-18.
This awesome channel deserves more subscribers
It doesn't matter who wins because in the end both are gonna die if that's the last piece of food left. Unless help arrives ofc.
Could you guys also try explaining Bernoulli's trials?
DO YOU LIKE BANANAS? Cuz I know where you can find bananas!
ON THE MOOOOOOOON!!!!
How have I never seen this one! Love it!
on a beach...you cant be left with just one banana...i'd rather go fishing !!
but the probability is very well explained!! TEDed
rocks!!
First ever Ted-Ed video I didn't need to watch beyond the first 50 seconds.
How come the Cat is rolling dice?😂
My Cat only knows how to Make scars on my face😂
It is not really there and the game is all in the girls head
@@rattriangle agree✌
Me and my buddies took notes of the dice rolls whilst playing the boardgame Risk many many times. Perfect Gaussian distribution.
And this is why I have rolled quite a few critical misses in the past few weeks of D&D... Nothing like having three critical misses in a row with a 5% chance that I am going to roll a 1 on a D20 each time I roll. It's almost to the point that I have a huge amount of anxiety come up whenever I am rolling my attacks.
This just took my RISK(tm) game to a whole new level
I did 2/3 * 2/3 = 4/9, and 1-4/9 = 5/9
amazing work on all of your videos
Fascinating stuff! ☺
hello
Cat:I will take 5 and 6
Girl:hah sucker.
5 minutes later
Cat:num num I eet
Why play dice over a banana? If they have 2 dice they could have been playing RPG instead!
DICE IS ALREADY PLURAL YOU DICKWAD unless English isn't your first language in which case I politely inform you on this matter without swearing
No, English is not my primary language and I never been on an English school. Thank you very much for the advice.
maximuscesar The singular would be die, like in the video. Although this form is rarely used.
maximuscesar Great! I'm glad we could have this intelligent, mature conversation.
Mauri .tanga I can't argue with that. Excellent point.
If you make it so player 1 wins if the highest number is 1, 2, 3, or 5 and player to wins if it's 4 or 6, they'll each have a 50% chance of winning.
Also, if you use dice with sides that are divisible by 4, you can give the lower 4th and the upper 4th to one player and the middle half to the other and they'll both have a 50% chance of winning.
I know what I'm doing now during the apocalypse!
Little did you know...
First one of these videos I could figure out.
Wait wait wait... So You like... *mmhmm* Bananas?
OH MY FUCKING GOD
Lol
DO YOU LIKE BANANAS????
Nope but I'm willing to share mine if anyone wants some ;-)
Now if your playing first to 10 points wins, what would the probability that player 2 wins the game overall?
Thanks for giving me more ideas about how to rig my casino.
Thanks. Simply great. I love you people.
That made both my brain and banana hurt.
You should reaallly go back to school.
I wish I had seen this video when I was studying probability in college.
+Alvin Lee I'm glad I didn't have too much college, I've been learning from this guy: dfpolis R-9 Have Reason Will Travel. That's his 13th response video to naturalists who learned that the brain produces the mind (they learned it in college), but that doesn't make any sense, we are aware before our brain could process awareness!
Nine people must have lost their last banana to a smarter cat! Seriously, how can anyone dislike a video like this?
"probability? I HATE PROBABILITY!!!" *dislikes*
henriandco (Even though the title states clearly that this is a video about probability)
You can please ALL of the people SOME of the time. You can please SOME of the people ALL of the time. But there are SOME people that you just CAN'T please!
KrisBlueNZ Doesn't the statement "There are some people you can't please" directly contradict the statement "You can please all of the people some of the time"?
Yes. Yes it does.
The problem was fairly easy even before the explanation, but great video. Haven't heard of probability in a while
Why would I "at first glance" think that player one has an advantage? Player two could win on either die. Even if I didn't sit down and work out the odds like twenty different ways (really, did it take this long to explain?), it would seem like player two had the clear advantage.
I was thinking the same!
Shawn Ravenfire The reason it might look that way on first glance is that Player One has more sides in favor; any of the first 4 numbers are more likely to turn up than any of the last two, appearing to put the odds in Player One's favor with 2/3 chance of winning.
Of course, when one looks at it more carefully, the latter two numbers actually hold a stronger chance for victory when they do turn up; any of Player One's numbers can be overpowered by Player Two's, while Player Two can never be overpowered (Even if Player One gets a 6, it counts as a win for Player Two).
George Dolley If you need to explain why it immediately looks like player one has the advantage, then it doesn't immediately look like player one has the advantage.
So player two won. Player two is a cat. Ted-Ed just gave the last banana on Earth to a cat. Congratulations
He's a good boy, he deserves it.
The comments section here signifies the end of progress of humans
But aren't you a part of that too?
Accept for the top comment which suggest they split the banana.
@@lucaspawprint1888 Lol. Exhibit A. (That is, if you are a native English speaker)
Meh. I'm allergic to bananas 🙀
You can simply do these calculations in your head if you calculate the chance of NOT rolling a winning number.
Each die has a 2/3 chance of not rolling a winning number, multiply them and you have a 4/9 chance of not winning, meaning you have a 5/9 chance of winning. This works for any number of winning numbers and any number of dice.
How can player one automatically win if it's a tie ( 1 and 1, 2 and 2, 3 and 3, 4,and 4)? That doesn't make sence, neither number is higher.
+robert Butcher Its not who gets the 5 or 6 its if the 5 or 6 gets rolled by anyone player 2 wins.
+robert Butcher and your sence doesn't make sense
+mrsCrab Neither does yours so why are you talking.
confederate flag, nice touch
I love Ted ed
Why should a cat in an island playing dices for banana?!
Cat eat fish. LOL!
God forbid I be stuck on a deserted island with a cat rolling dice over the last banana.
Why would you have a dice when u are stuck in an island....
Why not?
Actually the coin toss is also unfair when you catch the coin after tossed. You can experiment by tossing 3 times and results are the side which is up when hold 2 times is the result. So to be fair it is wiser to let coin dropped on the ground in cricket games.
It is dropped on the ground
Can anybody stop arguing about what cats eat?
It's an example, (FICTION)
really good video, like the cartoons and makes sense.
100% chance of getting the banana:
Plant the banana seeds
There are no banana seeds in the modern bananas, they are mostly heavily modified so you cannot plant them
Luke banana seeds? Wtf dude? Have you ever eaten a damn banana?
Luke They arent full grown seeds, your banana has been genetically modified... And the chance of your banana to be real is about 0%
100% chance of getting the banana:
Drop the soap in prison.
Cheapskate X Videos key and peele for the win, my man!
*The Last Banana*
Conning to a theater near you
Split the banana god dammit!
banana split
*eats the whole thing, don't share any then walks away and rides boat away standing you on the island again*
player 2 gets 56% of the banana and player 1 gets 44% of the banana
put it on playback speed 0.5 and thats how fast my teacher speaks
We have the same last name, lol
Yep :)
Doppler gangers? Imposter?
I think each one of them should have 50% of the banana so that way they don't have to do all those calculations.
yeah but why the fuck would you do it like that ...why not whoever rolled the highesst number wins... is that not the normal way
The problem there is you could tie. (Theoretically, long enough for both of you to starve.)
Winning the banana is not the objective of this video, the objective here is to teach you something about how probabilities work.
GuyWithAnAmazingHat I disagree about the unimportance of a plausible demonstrative scenario, but you most definitely have an amazing hat.
Just to point out the math for Player 2:
*Player 2* has two winning numbers, but he only needs *one* of the dice to be a winning number to win. So his chances are:
(5-6 on the first die) OR (1-4 on the first die AND 5-6 on the second die)
Pr(Player 2 wins) = 2/6 + (2/6 * 4/6) = 5/9
: )
+Copydot
........just saw this after posting my comment up there lol.
How come the win requirement for player 2 is not just % of first roll >4 OR % of second roll >4 = 2/6 + 2/6 = 2/3.
Is this 2/6 * 4/6 part of the win requirement for player 2?
+dzpak Well, think of it like this: if you roll a 5 or 6 on either die, you can stop rolling right? So the odds of rolling the second die at all are dependent on getting a 4 or less on the first. So even though your odds of rolling a 5 or 6 on the second die are the same, there's only a 4/6 chance that you'll *have to* roll that second die.
And if you could just add 2/6+2/6... for each die, then for 4 dice you'd end up with an 8/6 chance of rolling 5 or 6. That doesn't make much sense ;)
+Copydot
Cool, that does make sense. We did probability when I was a kid, but can't rmb this type of math coming up lol. Thx for the explanation :)
banana will make your teeth go gray.
thanks, doing this for halloween.
+г-н зонт lik
my teeth r grey, best costume pls rate
11/10 ign
I think he was talking about dhmis...
Lol the turtle in the end
or just split the... GAWD DAMN IT SOMEONE ALREADY SAID THAT
how do I split that? :/
You can launch a dice in a centrifuge with rockets from the moon, make it land in a fast flowing river, picking up the centrifuge after it has fallen from a waterfall and see what the dice has "rolled", and it still wouldnt be random, as long as you calculate all the variables. Now compared to a flat surface and a simple hand throw to roll some dices wouldnt be hard to calculate.
why don't they just break the banana in half?
+Dragons' Majesty She doesn't want it for eating.
+Lugh Summerson DUDE
Lugh Summerson 😂😂😂😂
Excelente ideia!
just split the FUKING BANANA
hahahaha xd
Why argue over the banana when you can have roasted turtle?
You want to be the cat and pray she is a furry.
Yep hopefully
Hope you boys learned some hustling tricks here this is gold!
this one was pretty easy, lobe these thoughts provoking videos tho
Say you go only one round of dice. Which would be better to go first or second?
They should share it
Besides making a table is there an easy way to see right away that it was the case player 2 had the better probability? At first when asked I mistakenly thought player 1 but now it's obvious. How can you calculate the odds right away with no table or listing.?
+Chandler C You can calculate in your head the probability that the girl wins, which is 4/6 * 4/6 = 16/36. From this you can quickly see that it is less than half, so the cat has a better chance.
I have a question for very smart math nuts, what is the probability of getting any 7 random numbers from 0-50 vs the probability of getting a single number 7 times in a row??? Intuition says 7 of the same number is less probable however, aren't all numbers equally as improbable even if it's repeated? Ahhh it's killing me, which has a lower probability? Or are they equal?
+95TurboSol The reason getting "any 7 random numbers" seems more probable is that you aren't thinking of any specific numbers. For instance, if you were thinking of the numbers "1, 2, 3, 4, 5, 6, 7" vs "7, 7, 7, 7, 7, 7, 7" it would be at least a little more obvious that they have the same probability of showing up. We just tend to think of certain strings of numbers as "special," whereas other number combinations are boring and unremarkable, so the boring numbers SEEM more likely to show up even though they aren't, because that category of numbers is much larger. Does that help?
eljazeccle19
Exactly what I was thinking. I would love to see someone do the math on it though just to be sure, sometimes intuitions can be misleading.
+95TurboSol The first one is 100% because saying 'any' implies there's no limitations on the numbers you can receive, so you can receive 7 0-50 numbers, unless you mean receiving up to 7, then that would be (8/51)^7 which equals about 0.00023%. The second one would just simply be (1/51)^6 (6 because all the numbers are based off the first number you receive, so you don't take the first number into account) which equals about 0.000000006%; you'd have a better chance at the lottery at 0.000007%.
Chicken Surprise
Thanks, yeah I meant for instance if I did an auto pick of any 7 random numbers up to 50 and allowing repeat numbers. Now suppose I got 2 19 46 31 28 10 39 on one random pick, then doing it again suppose I got 5 5 5 5 5 5 5, which is less probable? Or are they the same?
+95TurboSol Getting straight 5's would be less than the random pick.
player 2 also wins because each side is slightly heavier or lighter than another, like 6 is the lightest because it has 6 holes, unlike 1 has only 1 hole
They could've just made a phone out of that banana to call for help...
*BA-BA-BA- BANANA PHONE DOODOPOPDDILLIDYDOOP*
One fair way to set up the game: Player 1 wins if the total of the numbers on the dice are odd, and Player 2 wins id the total is even.
good job
Took this to VBA for a test of 10.000.000. Giving the following wins
Ta = 5555139
Tb = 4444861
So the Rnd of VBA seems to work just as expected.
Sub banana()
Dim Ta, Tb, count, A, B As Long
For count = 1 To 100
Ta = Int(6 * Rnd) + 1
Tb = Int(6 * Rnd) + 1
If Ta >= 5 Or Tb >= 5 Then A = A + 1 Else B = B + 1
Next count
Debug.Print A, B
End Sub
it is permutation and combination! a helpful way to win board games haha
why the hell are you stranded on an island with a giant cat that can play dice and like banana...
Great!
But I've always thought that the number 1 side is slightly heavier because they indent holes into the surface for the number of spots, if this is the case number 6 (the opposite side) would come up more frequently than its 'probability' would claim?
Time to pull this on my classmates and won those burgers!
long gone our fellow banana ladies and gentlemen
Plot twist...
The cat wins and eats the banana.
But, alas it is allergic and dies.
The girl knew this all along and make stew out of the cat
You can also say that if they play forever then eventually player 1 would win 99% of the time since if you have an infinite amount of time then you can just keep playing until player one gets insanely lucky, but by that time they may need to win 1000s of games in a row, yet it is theoretically possible given enough time.
Corporal_Cake it would also be cherry picking. I watched those other vids.