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

Imagine a game of dice: if the biggest number rolled is one, two, three, or four, player 1 wins. If the biggest number rolled is five or six, player 2 wins. Who has the best probability of winning the game? Leonardo Barichello explains how probability holds the answer to this seemingly counterintuitive puzzle.

Lesson by Leonardo Barichello, animation by Ace & Son Moving Picture Co, LLC.

[StudentPursuit]

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

[bored_person]

i'd just suggest we split the banana.

[KeaganKenalty]

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

[djsyntic]

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.

[anitkaurbagga]

Real solution: Don't buy odd number of bananas

[amritpalsingh]

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.

[KJKP]

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

[ergogray]

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!

[lprocks]

CATS ARE CARNIVORES THEY CAN'T EVEN PROPERLY DIGEST BANANAS

[iso]

Using a computer simulation in python, I tried this out:
Results:
Player A:  44.4457644446% (expected:
Player B:  55.5542355554% (expected:

from random import randint
pA = 0
pB = 0
trials =
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)) + '%'

[daddywambat]

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

[backattack]

You know, just devide the banana into two

[scarletovergods]

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?

[rebeccarich]

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! :)

[lopsidedhead]

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

[joeybiegel]

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

[luckingk]

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!

[Markus]

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.

[XDDISTURBEDXD]

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

[francoearllacno]

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