What is the probability of choosing the bomb?

Question

Five crackers are put in a bowl full of bran. In four of them, there is a check for $ 2 million, one contains a bomb, strong enough to kill the person who pulls the cracker.

1. Mr. Adam and Mr. Bill reach for the bowl together. Mr. Adam opens up the lid and picks up one of the crackers. What is the probability of choosing the bomb for Mr. Adam and Mr. Bill? (It should be 1/5 for Mr. Adam and 1/5 for Mr. Bill)

$$P(X_1) =1/5$$

$$P(X_2) =P(X_2|X_1)*P(X_1) + P(X_2|¬X_1)*P(¬X_1)$$

where $X_1$ means choosing the bomb on the first turn and $X_2$ means choosing the bomb on the second turn

$$P(X_2) =0*(1/5) + (1/4)*(4/5)$$ $$P(X_2) =1/5$$ 2. Mr. Adam and Mr. Bill decide to draw for turns to pick from the bowl. What are the probabilities of choosing the bomb in this case? Let's say they toss a coin and if it comes head then Mr. Adam goes and if it comes tails then Mr. Bill. So, they could choose a bomb if they get heads and also if they get tails and the other (first mover) does not choose the bomb. $$P(X) =P(X|H)*P(H) + P(X|T,¬X_1)*P(T)*P(¬X_1)$$ $$P(X) =(1/5)*(1/2) + (1/4)*(1/2)*(4/5)$$ $$P(X) =(1/5)$$

So, there is no difference in the probabilities of choosing the bomb whether they draw for turns or not. Am I getting it right?

Answer 1

Say we have two persons A and B. If person A starts first, the probability of A choosing a specific item should be higher than that of person B.

Probability of person A (first person) choosing a specific box $ = \frac{1}{5} + \frac{4}{5} \cdot \frac{3}{4} \cdot \frac{1}{3} + \frac{4}{5} \cdot \frac{3}{4} \cdot \frac{2}{3} \cdot \frac{1}{2} = \frac{3}{5}$

Then probability of person B choosing the specific item $ = 1 - \frac{3}{5} = \frac{2}{5}$

In the above working, person A has probability of choosing a specific item as follows -

i) in the first chance, it is $\frac{1}{5}$

ii) gets a second chance only if fails to choose the specific item in the first chance (probability of $\frac{4}{5}$) and person B also fails to choose the item next, probability of which is $\frac{3}{4}$.

iii) is certain to choose the item if both of them fail to choose the item in their first two chances.

Now if the first person can be either of them with probability $\frac{1}{2}$ based on toss of a coin, then it is easy to see that they have equal probability of picking a specific item: $\displaystyle \frac{1}{2}$.

$\big(\frac{1}{2} \cdot \frac{3}{5} + \frac{1}{2} \cdot \frac{2}{5} = \frac{1}{2} \big)$.