The Last Dumpling: Another Game Of Probability

BBT? Sheldon? Leonard? Rings a bell?

Anyways, let's say, Sheldon and Leonard are two very good friends. While sharing a dinner, Sheldon notices that only one dumpling is remaining on the plate. Being his friend, he doesn't want to upset Leonard but also wants that delicious dumpling!

So, he proposes a game.  As an assertive character, he sets the rules. He and Leonard would roll two dices. If in both of the two dices, a 1, 2, 3 or 4 comes up, Leonard wins the last dumpling. If any of them contains a 5 or 6, Sheldon wins. Leonard greedily accepts the terms, thinking that he would have a much higher probability of winning it.

But we all know Sheldon wouldn't want to play if he doesn't have a more than half chance of winning. So what do you think? Who gets the last dumpling?

Ok, let's see what happens when two dices are rolled.

The probability that one of the dices would have the value between 1-4 is the summation of their individual probability of happening. Let's call the outcome of the roll X. So it would be:

P(Dice 1 being 1-4)  = P(1<=X<=4) = P(X=1) + P(X=2)  + P(X=3) + P(X=4) = 1/6 + 1/6 + 1/6 + 1/6 [as each of them has an equal chance of happening] = 4/6

The second dice would have the same probability of this happening. So both of them having this outcome is the product of the probabilities. It's simple counting principle. If one event can happen in 5 ways and another can happen in 6 ways, together they can happen in 5*6 = 30 ways.

P(Both Dices being 1-4) = P(Dice 1 being 1-4) * P( Dice 2 being 1-4) = 4/6 * 4/6 = 16/36 = 4/9.

So if the game is played 9 times, Leonard would win 4 of the times and rest of the times Sheldon would win. Meaning, probability of Leonard's winning is about 44% and Sheldon's is 56%.

Bad luck Leonard!

Chuck-a-Luck: A fine game involving probability

I stumbled onto this game while reading a fantastic book, 'Innumeracy' by John Allen Paulos. The rules of the game is simple: you play against a host. The  host has three dices with number 1-6 on them. Then you would be asked to choose a number in that range. Let's say you have chosen number 4. The host would then roll the three dices at the same time. Now there are three outcomes:

1. If all three dices have 4 on them, you get 3 dollars.
2. If two of the dices come up with 4, you get 2 dollars.
3. If only dice has 4 on it, you get 1 dollar.
4. Else(none of the dices has came up with 4) then you pay the host 1 dollar.

Now, what do you think about your chances are in this game? Obviously, as it is played in casinos, the casino owners won't have allowed this game if there were no profit for them. So even if this game seems quite winnable, you should re-calculate your chances.

Here is the expected outcome of this game:

expected outcome(EO) = sum(all the ways you can win * associated rewards) - sum(all the ways you can lose * associated cost)

EO = sum(P(all three dices has 4) * 3 + P(any two dices has 4) * 2 + P(any dices has 4) * 1) - P(no dice has 4) * 1

= (3C3 * (1/6)^3 * (5/6)^0)*3 + (3C2 * (1/6)^2 * (5/6)^1)*2 + (3C1 * (1/6)^1 * (5/6)^2)*1) - (3C3 * (5/6)^3 * (1/6)^0)*1

= 3/216 + 10/72 + 25/72 - 125/216 = -.08

As you can see, the outcome is quite counter intuitive. It seems that you would lose .08 dollars every time you play the game.

Following is a simulation of the game play. It shows how many times you can play before you lose all of your money in the cold-hearted probabilistic eventuality:

'''
code is also hosted at https://github.com/hasanIqbalAnik/probabilities/blob/master/monty_hall_simul.py
'''
import random
def dice_roll():
    return random.randint(1, 6)

balance = 10 #you start with 10 dollars
count_games = 0
choice = random.randint(1,6)

while(balance > 0):
    outcomes = [dice_roll(), dice_roll(), dice_roll()]
    if(outcomes.count(choice) == 3):
        balance = balance + 3
    elif(outcomes.count(choice) == 2):
        balance = balance + 2
    elif(outcomes.count(choice) == 1):
        balance = balance + 1
    else:
        balance = balance - 1
    count_games = count_games + 1

print balance # 0 :(
print count_games # sample output, 74 times

This also matches the probability calculation that you would be able to play around 80 times with 10 dollars, losing .08 dollars on average each time.