Necktie paradox
The necktie paradox is a puzzle and paradox with a subjective interpretation of probability theory describing a paradoxical bet advantageous to both involved parties. The two-envelope paradox is a variation of the necktie paradox.
Statement of paradox
Two persons, each given a necktie, start arguing over who has the cheaper one. The person with the more expensive necktie must give it to the other person.
The first person reasons as follows: winning and losing are equally likely. If I lose, then I will lose the value of my necktie. But if I win, then I will win more than the value of my necktie. Therefore, the wager is to my advantage. The second person can consider the wager in exactly the same way; thus, paradoxically, it seems both persons have the advantage in the bet.
Resolution
The paradox can be resolved by giving more careful consideration to what is lost in one scenario ("the value of my necktie") and what is won in the other ("more than the value of my necktie"). If one assumes for simplicity that the only possible necktie prices are $20 and $40, and that a person has equal chances of having a $20 or $40 necktie, then four outcomes (all equally likely) are possible:
| Price of 1st person's tie | Price of 2nd person's tie | 1st person's gain/loss |
|---|---|---|
| $20 | $20 | 0 |
| $20 | $40 | Gain $40 |
| $40 | $20 | Lose $40 |
| $40 | $40 | 0 |
The first person has a 50% chance of a neutral outcome, a 25% chance of gaining a necktie worth $40, and a 25% chance of losing a necktie worth $40. Turning to the losing and winning scenarios: if the person loses $40, then it is true that they have lost the value of their necktie; and if they gain $40, then it is true that they have gained more than the value of their necktie. The win and the loss are equally likely, but what we call "the value of the necktie" in the losing scenario is the same amount as what we call "more than the value of the necktie" in the winning scenario. Accordingly, neither person has the advantage in the wager.[1]
This paradox is a rephrasing of the simplest case of the two envelopes problem, and the explanation of the resolution is essentially the same.
See also
- Bayesian probability
- Bertrand paradox
- Decision theory
- Dutch book
- Monty Hall problem
- Two envelopes problem
- Newcomb's paradox
- St. Petersburg paradox
References
- ↑ Brown, Aaron C. (1995). "Neckties, Wallets, and Money for Nothing". Journal of Recreational Mathematics 27 (2): 116–122.
- Kraitchik, Maurice (1943). "Mathematical Recreations". London: George Allen & Unwin.
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