There are n Snakes on the Plane. There are n Passengers on the Plane. Each Snake will Bite One Passenger at Random. What is the Probability that The Target, Mr. X, does not get bit, as n tends to infinity?
- Suppose the Snakes are Labeled as S1, S2, S3…Sn
- Let’s look for a moment at Snake S1:
- P(Snake S1 Bites Mr. X) = 1/n
- In other words, there’s a 1/n chance that Mr. X will get bitten by that particular Snake, S1
- P(Snake S1 does not Bite Mr. X) = 1 – 1/n
- This is the complement of the bullet above
- P(Snake S1 Bites Mr. X) = 1/n
- Notice that the logic of [2] applies to each Snake. That is,
- P( Snake S1 does not Bite Mr. X) = 1 – 1/n
- P( Snake S2 does not Bite Mr. X) = 1 – 1/n
- P( Snake Sn does not Bite Mr. X) = 1 – 1/n
- Thus, P( Mr. X does not get Bitten ) = The Product of all the Probabilities in [3],
- (1 – 1/n) x (1 – 1/n) x … x (1 – 1/n)
- S1 no Bite & S2 no Bite & … & Sn no Bite
- So, the Final Solution for P(Mr. X Not Bit) = (1 – 1/n)n
- And note that (1 – 1/n)n -> 1/e as n -> ¥ (By Elementary Calculus)
- Also note that 1/e » 36.8 %
Note that we are assuming independence in Step [4], we’ll leave it to the Biologists to interpret whether this is a reasonable assumption
For more about the ending in Step 5, just email Dr.Canada – dcanada@mail.ewu.edu
