(Introduction to Probability, Blitzstein and Nwang)

Player A chooses a random integer between 1 and 100, with probability pj of choosing j (for j = 1, 2, . . . , 100). Player B guesses the number that player A picked, and receives that amount in dollars if the guess is correct (and 0 otherwise).

(a) Suppose for this part that player B knows the values of pj . What is player B’s optimal strategy (to maximize expected earnings)?

(b) Show that if both players choose their numbers so that the probability of picking j is proportional to 1/j, then neither player has an incentive to change strategies, assuming the opponent’s strategy is fixed. (In game theory terminology, this says that we have found a Nash equilibrium.) 16.

(c) Find the expected earnings of player B when following the strategy from (b). Express your answer both as a sum of simple terms and as a numerical approximation. Does the value depend on what strategy player A uses?

a) Pick the integer for which $p_j \cdot j$ is highest.

b) I dont't get the question.

c) Let $X^A$ be the number that player A picks, and $X^B$ the one B picks. Let $W$ be the earning.Then

\begin{align} E(W) &= \sum_{i=1}^{100} \sum_{j=1}^{100} E(W|X^A=i, X^B=j) \cdot P(X^A=i, X^B=j)\\ &= \sum_{i=1}^{100} i \frac{c^2}{i^2} = c^2\sum_{i=1}^{100} \frac{1}{i} = \frac{1}{H_n} = 0.192775 \end{align}

Is this correct? How to answer part b) ? Maybe I don't understand the problem right.


For (b), notice that if A has the strategy of picking $i$ with probability $c/i$, then B's expected winnings is $c$ whatever strategy B uses, and so there is no incentive to change.

Similiarly, if B has the strategy of picking $i$ with probability $c/i$, then B's expected winnings is $c$ whatever strategy A uses, and so there is no incentive to change there either.

This also answers the last part of (c).

  • $\begingroup$ Doesn't this imply that if $B$ uses the aforementioned strategy, they are expected to win approximately $20$ cents no matter what strategy $A$ chooses? Seems like an unfair advantage. Is this because $B$ is under no risk of losing money? $\endgroup$ – heckeop Nov 17 '19 at 21:33
  • $\begingroup$ @heckeop Well, the game as stated $A$ can only lose money and $B$ can only gain money, it's not a sensible game to actually play! $\endgroup$ – Christopher Nov 19 '19 at 9:48

For (c) I would rather write $$E[x] = \sum_{j=1}^{100} j \cdot P(X_A = X_B) = \sum_{j=1}^{100} j \cdot \frac{c}{j}= 100 \cdot c$$ We get $c$ by normalization of the prob. \begin{align} 1 &= \sum_j p_j = \sum_j \frac{c}{j} \\ \Rightarrow c &= \sum_j \frac{1}{j} \end{align} Of course, this yields the same numeric solution as Dominik, however, this calculation is simpler.


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