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!
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!
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