The last banana: A thought experiment in probability - Leonardo Barichello 

Just eat the banana she's so hungry that she thinks that the cat is a person

how do you find dice on a deserted island yet no food

i'd just suggest we split the banana.

Here is a thought experiment: This is a coin toss game. It's called "Heads I win; Tails you lose." Enjoy.

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.

Real solution: Don't buy odd number of bananas

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.

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)) + '%'

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!

Or player 1 and player 2 could not be greedy assholes and just split the banana in half...

CATS ARE CARNIVORES THEY CAN'T EVEN PROPERLY DIGEST BANANAS

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?

I just wanted to be the cat

You know, just devide the banana into two

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! :)

Seems like all these probabilities go out the window when I'm playing DnD

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

If it were me I would give the banana to the cat so it will be healthy once I ate it. YUMYUM CAT STEW!!

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.

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I'm not greedy, I would just share the Banana.

Why play dice over a banana? If they have 2 dice they could have been playing RPG instead!

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.

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

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!

Now if your playing first to 10 points wins, what would the probability that player 2 wins the game overall?

Nice video, explaining both the methods ....and the difference between theoretical and experimental probability...

Cat:I will take 5 and 6 Girl:hah sucker. 5 minutes later Cat:num num I eet

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.

Ted ed for the million time you are my best RU-vid channel

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.

How come the Cat is rolling dice?😂 My Cat only knows how to Make scars on my face😂

Could you guys also try explaining Bernoulli's trials?

So player two won. Player two is a cat. Ted-Ed just gave the last banana on Earth to a cat. Congratulations

I did 2/3 * 2/3 = 4/9, and 1-4/9 = 5/9

DO YOU LIKE BANANAS? Cuz I know where you can find bananas! ON THE MOOOOOOOON!!!!

I know what I'm doing now during the apocalypse!

The comments section here signifies the end of progress of humans

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!!

I think each one of them should have 50% of the banana so that way they don't have to do all those calculations.

Nine people must have lost their last banana to a smarter cat! Seriously, how can anyone dislike a video like this?

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.

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.

That made both my brain and banana hurt.

put it on playback speed 0.5 and thats how fast my teacher speaks

Why should a cat in an island playing dices for banana?! Cat eat fish. LOL!

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 : )

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 wish I had seen this video when I was studying probability in college.

Wait wait wait... So You like... *mmhmm* Bananas?

God forbid I be stuck on a deserted island with a cat rolling dice over the last banana.

Can anybody stop arguing about what cats eat? It's an example, (FICTION)

First ever Ted-Ed video I didn't need to watch beyond the first 50 seconds.

Thanks for giving me more ideas about how to rig my casino.

why the hell are you stranded on an island with a giant cat that can play dice and like banana...

Why would you have a dice when u are stuck in an island....

This awesome channel deserves more subscribers

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 Last Banana* Conning to a theater near you

100% chance of getting the banana: Plant the banana seeds

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

Split the banana god dammit!

it's a cat, dude! Forget the complex math and roast the kitty for burgers with banana desert.

Fascinating stuff! ☺

This just took my RISK(tm) game to a whole new level

banana will make your teeth go gray.

Never seen a cat playing dice on a deserted island!

We have the same last name, lol

Why argue over the banana when you can have roasted turtle?

or just split the... GAWD DAMN IT SOMEONE ALREADY SAID THAT

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

why don't they just break the banana in half?

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 split the FUKING BANANA

I didn't really understand the joint probability explanation.If you calculate the probability for the cat, it should win if the highest is 5 or six, so that would be 1 in 3 for each dye. The joint probability should then be 1/9, while the probability of the girl winning is 4/9.the player 1 has a higher chance of winning an the probabilities don't add up to 1.Can anyone explain?

You want to be the cat and pray she is a furry.

Lol the turtle in the end

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.

The problem was fairly easy even before the explanation, but great video. Haven't heard of probability in a while

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.

They could've just made a phone out of that banana to call for help... *BA-BA-BA- BANANA PHONE DOODOPOPDDILLIDYDOOP*

Step one, realize you are playing dice with a cat. Step two, eat banana so you stop hallucinating.

Probability admits a lack of information. It's a best guess based on what you know at the moment. However, how a player rolls the dice, how they land, how the edges of the die catch the table's surface will absolutely determine the result. Actually the result is determined before the die is cast. If one could know 100% of the information of all the forces, vectors, mass, etc. involved before and during the cast of a die, one could predict exactly the result. "Odds" of something happening boils down to a best guess. Take a horse race for example. If one could know the background of every horse, it's genetics and lineage, how it feels [health wise before and during the race, the jockey's condition past and present, how each horse might react to its post position, track conditions, etc. and know all these variables to 100%, then the results of the race is a fait accompli. Conclusion: Hard determinism.

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

Here's the way I look at: Player one wins if a 5 or a 6 doesn't show on either dice The odds of a 5 or 6 NOT being on the dice is 2/3 Since there are two dice you have to multiply 2/3 by 2/3. That is the odds of neither a 5 nor 6 showing on either die. So the odds of the first player winning is 2/3 x 2/3 = 4/9 = 0.44 And the odds of the second player winning is 5/9 = 0.56

If they just used 1 die, they'd both have 50% chance of winning. If they added the rule that player 1 wins if both dice are 5s or 6s, it's also equal.

or you can just freaking split the banana this ain't gotta be no hunger games (literally)

Me and my buddies took notes of the dice rolls whilst playing the boardgame Risk many many times. Perfect Gaussian distribution.

*split the banana.* *_Modern problems require modern solutions._*

This is a paradox that I thought of: Space is infinite, but by logic, infinity doesn't exist, everything must have an end, but what's behind that end? Maybe another Space, let's call it Space 2, but what's behind the end of Space 2? Space 3, but what's behind Space 3? Space 4, and that goes on to infinity, which makes the statement that infinity doesn't exist meaningless. The infinite paradox.

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.

2 things 1. why a banana 2. just split the blooming banana

just eat the banana shes so hungry that she thinks that the cat is a person

Say you go only one round of dice. Which would be better to go first or second?

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.?

I am so confused. Not by the video, but by the comments. Some say "one person eats the other and banana". Others are saying "the cat doesn't eat bananas". And 50% of the comments are arguing to split the banana. Still, others are stuck how they got the dice in the first place, or why there's only 1 banana. Or why they can't go fishing. (If this comment seemed rude, then no offense to anyone.)

How have I never seen this one! Love it!

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?

break the banana in half and if you're breaking it give yourself more