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Two persons $A, B$ roll a fair $n$-face dice separately and get $1 \le x,y \le n$ points. Then the third party will put $x + y$ dollars in a black box. $A$ and $B$ only know the point they roll and don't know the other's.

Then they bid on the black box. $A$ bids a integer price $p_1$, then $B$ can only bid at least $1$-dollar higher integer price $p_2$ or give up. If $B$ gives up, then $A$ must buy the black box with price $p_1$. If $B$ bids $p_2$, then $A$ can bid at least $1$-dollar higher integer price $p_3$ or give up, etc. Until one gives up, the other one should buy the black box with the latest price.

Question:

  1. Assuming that $A$ and $B$ are rational, what's the optimal strategy of first player $A$ and second player $B$? How can, from the other one's bidding price, infer the range of points the other one has?

  2. What if the other one should have to bid at least $k$ dollars higher ($k$ is integer)?

  3. What if when there are $m$ players? (same question as Q1 and Q2).

Is there any terminology of this problem? Is there any reference or literature which thoroughly discusses this problem?

maplemaple
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  • What is the source of the question? – user2661923 Oct 28 '22 at 19:57
  • @user2661923 course homework with bonus points – maplemaple Oct 28 '22 at 20:07
  • What is the name of the class? What is the name of the corresponding textbook? – user2661923 Oct 28 '22 at 20:08
  • @user2661923 Game theory, recent topic is about Auction Theory. No official textbook – maplemaple Oct 28 '22 at 20:11
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    This question combines maths and psychology.

    Strategy n°1 for A : he bids x+(n-1)/2, with objective to win 1pt. But in this case, if B knows that A applys this strategy, he know exactly what to do, so he will win 1pt, or A will loose 1pt or even more if y is very small. B is the winner.

    A needs to mix this strategy, and some bluff.

    So there is no mathematics answer.

     – Lourrran Oct 28 '22 at 20:13
  • Were there any previously solved problems, worked examples, or theorems, that led up to this problem, that might reasonably be regarded as training (AKA tools) for the problem? It would help improve the quality of your posting if you could edit your posting with a discussion of what tools that you think that you are supposed to use, along with an attempt to employ the tools to solve the problem. – user2661923 Oct 28 '22 at 20:15
  • @user2661923 No warm-up questions. It's a separate bonus question. – maplemaple Oct 28 '22 at 20:17
  • @Lourrran You may be right. My instinct is that this is a chess situation rather than a Poker situation. That is, the player is supposed to assume that the other player will play perfectly, and that there are $n$ possible outcomes that the other player will be faced with. Perhaps I am mistaken, but I suspect that the intent of the problem composer was to treat this as a Kriegspiel type Chess situation, where the opponent's moves are unknown. – user2661923 Oct 28 '22 at 20:18
  • @maplemaple I question the assertion in your last comment. Math courses (below the graduate level) don't normally ask the student to attack a problem without giving the student relevant prior training. Otherwise, what is the educational value of such a question? It would be like asking a high school student to prove that $~e^{\pi}~$ is a transcendental number. – user2661923 Oct 28 '22 at 20:21
  • @user2661923 It's a graduate course. This prof will ask some unsolved problem in homework as bonus without pointing it out. Last semester he asked this question in another game theory I course https://math.stackexchange.com/q/4402548/481713 without pointing out the terminology and it's an unsolved problem. – maplemaple Oct 28 '22 at 20:28
  • @user2661923 Therefore I post here to see whether it's the same case. If so, what's the current stage of this problem in research area. – maplemaple Oct 28 '22 at 20:31
  • Very nice response. That changes things remarkably, with respect to how MathSE reviewers should react to your posting. Based on your comments, it is certainly plausible that you have had no prior training. This implies that it is not reasonable to expect you to either detail pertinent tools or to show work. – user2661923 Oct 28 '22 at 20:31
  • To make sure I'm understanding everything correctly, the payoff to whoever wins the auction is $x+y$ minus the price they paid for it, and the payoff to the other is $0$? And when you say "Assuming $A$ and $B$ are rational," you mean that they only care about maximizing their expected payoff (i.e. no risk-aversion) and you are looking for a Nash equilibrium? – user6247850 Oct 28 '22 at 22:06
  • @user6247850 Yes. – maplemaple Oct 28 '22 at 22:07
  • It is similar to poker rather than chess. In games, there are 2 categories : When all players have all informations (chess), and when some players don't have all informations (cards game especially). – Lourrran Oct 30 '22 at 23:42

1 Answers1

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Assume strategies are known by the other player.

n=1. Always bid 1. You win 1 dollar or the other player bids 2 and no one wins.

n=2. When $x$ is rolled, the pot is $x+1$ or $x+2$, so bidding $x+1$ nets 0 or 1. Bidding 2 each time gives no information to player 2, so if player 2 rolls a 1, then player 2 is risking a loss by bidding 3, so player 2 will only bid 3 if they roll a 2. Thus bidding 2 each time will net a profit of 1 half the time (and 0 otherwise). The second strategy (bidding 2 every time) has the advantage that your opponent cannot cut you out of your profits by responding to a bid of 3 with 4. Though a bid of 4 may be considered irrational as it will never net a profit, as it is possible that your opponent's strategy may change. If you don't want your opponent to win money, bidding $x+1$ each time ensures he cannot profit. If you only care about your own winnings, then either strategy is suitable. Player 2 should always bid $y+1$. As you can see from this example, the problem is not specific enough to have a single answer, even in this very simple case. So the problem needs more specification around preferences, though more could be said in the instance where $k > 1$ .

Mark
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