The following problem is taken from Mark Joshi's quant interview book.

"I pick a number n from $1$ to $100$. If you guess correctly, I pay you $\$n$ and zero otherwise. How much would you pay to play this game?"

The question was answered in this post. I reproduced a part of the answer there.

The intuition is that in an optimal strategy, the picker should be indifferent to what the guesser chooses.

Suppose we just take $n=3$ for simplicity. Suppose the picker chooses $1$ with probability $p_1$, chooses $2$ with probability $p_2$, and $3$ with probability $p_3$. The selection of $p_1, p_2, p_3$ constitutes the picker's strategy.

The indifference criterion means that $1p_1=2p_2=3p_3$.

I do not understand the last bold sentence. Why would indifference criterion implies that $1p_1=2p_2=3p_3$? Is this a definition of being indifference?


In a sense, that is the definition. Or to be more precise, the indifference principle states that:

Indifference principle: The picker shall randomize their choice (in this game: the hidden number $n$) s.t. no matter what the guesser does (in this game: guess value $G \in \{1,2,\dots,100\}$), the expected payoff is the same (in this game: the expected payoff is independent of $G$).

Then for this specific game, since the expected payoff for guessing $G$ is $G\times Prob(G=n) = G \times p_G$, the indifference principle becomes the formula $1 p_1 = 2 p_2 = 3 p_3 = \cdots$

So that answers your question: Yes it is the definition, or rather, a direct, immediate consequence of the definition.

The underlying question, which you didn't ask, is: Why should the picker adopt the indifference principle? I.e., why is indifference the "optimal" strategy?

If the picker has some insight into the personal tendencies of the guesser, then the picker might do something different to foil the guesser. E.g. if the picker knows the guesser is bad at math and will most likely guess $100$ thinking all numbers are equally likely and that number pays the most when correct, then the picker can foil the guesser by picking $1$. But of course, the guesser, knowing that, would have picked $1$, and the picker, knowing that, might have picked $2$ instead, etc. This turns into a game of think and double-think.

The indifference principle avoids all this. Essentially, the picker is assuming the worst case "meta" scenario: that the guesser knows the picker's probability numbers $p_i$ for all $i$. Perhaps the guesser has a spy, or can read mind, or is just really really good at math and logic. Under this worst case scenario, the best (optimal) the picker can do is to adopt the indifference principle. If the picker did anything else, the guesser (knowing all $p_i$ values) can foil it and obtain a higher expected payoff by guessing $G= \arg\max_i i p_i$.

The indifference principle is even easier to explain in Rock-Paper-Scissors. One can certainly play it as think-vs-double-think, but most people(?) try to adopt the indifference principle $p_R=p_P=p_S = 1/3$ s.t. no opponent can have any extra benefit.

Note that the indifference principle does give up something: in return for protection against a really good opponent, you give up the opportunity to exploit a really bad opponent. E.g. if your opponent always plays Rock (or always guesses $100$), your indifference principle would still give the same expected payoff - and in a sense, let your opponent "off the hook". In other words, indifference is optimal under the "worst case meta scenario" I described above, but it may not be optimal if you know in advance your opponent's (probabilistic) strategy.

You can read much more about this in the context of Nash equilibrium

  • $\begingroup$ Thanks for your detailed explanations. Regarding the statement for indifferent principle which involves probability, can you give a reference to it? I would like to read more about it. $\endgroup$ – Idonknow Dec 29 '19 at 2:09
  • $\begingroup$ Not sure I can give a reference that's different from the Nash equilibrium. In particular, the indifference principle here is a mixed (i.e. probabilistic) strategy that is one half of a Nash equilibrium for this game. (It is one half because the Nash Eq. consists of a pair of (usually mixed) strategies, i.e. what each player would do. This question deals mostly with what the picker would do.) $\endgroup$ – antkam Dec 29 '19 at 3:43
  • $\begingroup$ 'Then for this specific game, since the expected payoff for guessing $G$ is $G\times Prob(G=n) = G \times p_G$' Why would the expected payoff for guessing is $G\times Prob(G=n)$? $\endgroup$ – Idonknow Feb 12 '20 at 2:06
  • $\begingroup$ If you guess $G$, you win with prob $P(G=n)$, and if you win, you gain $n (= G)$ dollars. So your expected payoff is $G \times P(G=n)$. $\endgroup$ – antkam Feb 12 '20 at 5:21
  • $\begingroup$ I see. Thanks so much. $\endgroup$ – Idonknow Feb 12 '20 at 5:23

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