# Unexpected hanging paradox maxmin strategies

I have a question about strategies of the players of Unexpected hanging paradox (I am very sorry for a long topic, topic exist already for a while, during this time I try to develop idea how to solve the problem, recently I decided to seek solution in the field of probability, therefore I ask for anyone who has experience in solving problems in probability to take a look. For understanding essential is reading the Version of the Paradox paradox, Question, and may be few last addendums).

Version of the Paradox: Let's consider a slightly specific version of the paradox, where the players take actions in rounds (in terms of paradox rounds can be days). On every specific round (day) judges can take action $A$ - the prisoner will be hanged today, $B$ - the prisoner will not be hanged today, and the prisoner can have few guesses $A$ - I will be hanged today, $B$ - I will not be hanged today. The game is played simultaneously. If $A$ is played by one of the players then the judges win, if both players play $A$ simultaneously than the prisoner win, and if both of them play $B$ game goes to the next round. If as a result of four rounds the prisoner is not hanged than the prisoner win.

Question: what is max-min strategy for the judges and for the prisoner. If there are any solution concept (Nash equilibrium).

The following is a few I hope useful ideas to approach the solution.

The game should be represented in extensive from but with simultaneous play.

Let's start from judges. There is no dominant strategy for the judges and it looks like there is no a pure Nash equlibrium. What it's known is that the chances to win when judges play $A$ is 50% and the chances to win when judges play B is $>50%$. But every next round the chances to win by choosing $B$ is getting lower, however the prisoner should prefer $B$ when the number of rounds is getting higher.

In addition, we can try to do a backward induction, the last round is similar to the Matching pennies game, that has a mixed Nash equilibrium, with actions takes on probability of 50 %.

On the second backward induction step (using the fact that on the last step the mixed Nash equilibrium is 50/50 between playing A and B for both players), therefore judges would prefer playing $B$ on the second step with probability 75 % and A with 25%, however the prisoner would prefer playing $A$ with 75% and B with 25%.

Unfortunately I didn't get any closer to maxmin strategies, but at least I've shared few ideas.

Addendum: Instead of playing 4 rounds, let's try to play 2 rounds, when the last round is similar to the matching pennies game, we can try to convert the 2 round game to normal form (4 by 4 actions for prisoner and judges of the form {AA,AB,BA,BB}) and it will be apparent that the most effective strategy for judges is two columns (judges is column player, at the same time is the worsen strategies for the prisoner ) - B(don't hang) -B(don't hang); B(don't hang)-A(hang) for the prisoner the win strategy is the strategy of the judges(BA when judges played BA, and BB when judges played BB), then maxmin of the prisoner is to play B for the tree rounds and on the last round play A or B with probability 50%.

The worsen strategies for the judges is AA and AB (they actually don't yield the second turn because judges decide to hang on the first round, therefore consider only strategy A, in the normal form it's represent like AA and AB, maybe I get it wrong), therefore playing A is maxmin strategy of the judges.

Addendum: we can try to looks at the problem from the probability point of view. Let's consider few ideas.

Idea 1. Judges throughout the game should make five principal decisions, following is the decision with the probabilities.

$Pr(j_1=$hand the prisoner on the first day$) = 0.2, ..., Pr(j_4=$hand the prisoner on the last (4'th) day$) = 0.2, Pr(j_5=$don't hang the prisoner at all$) = 0.2$

the same for the prisoner,

$Pr(p_1=$I will be hanged on the first day$) = 0.2,..,Pr(p_4=$I will be hanged on the fourth day$) = 0.2, Pr(p_5=$I will not be hanged at all$) = 0.2$

$P($judges win$) = \sum_{1 \leq i \leq 4}^{} p_1(1-s_1)=0.8$

The above approach for me looks like not correct, just because taking an action on round $i$ depends on the decision on round $i-1$, therefore all the probabilities are not independent.

Idea 2 We can try to represent all the probabilities of the game in this form

$Pr(p_1=$prisoner wins on the first round$) = 0.5($prisoner:I will be hanged on the first day$) \cdot 0.5($judges : hang the prisoner on the first day$)$

$Pr(p_2=$prisoner wins on the second round$) = 0.5($prisoner:I will no be hanged on the first day$) \cdot 0.5($judges : don't hang the the prisoner on the first day$) \cdot 0.5($prisoner:I will be hanged on the second day$) \cdot 0.5($judges : hang the prisoner on the second day$).$

$Pr(p_4=$prisoner wins on the last round$) = (0.5 \cdot 0.5)^3 ($judges didn't hang the prisoner on the three previous round and the prisoner didn't make guess that he will be hanged during first three days$) \cdot (0.25($judges: hang on the last day, prisoner: I will be hanged today$) \cdot 0.25($judges: don't hang at all, prisoner: I will not be hanged at all$))$.

$Pr($prisoner wins the game$) = p_1+p_2+p_3+p_4 =0.25+0.25^2+0.25^3+0.25^3 \cdot 0.5 = 0.3359375$

We get some result, I don't really know what to do with it, and don't know if it's correct, if we consider the probabilities for the victory like in Idea 2, it's obvious that the prisoner should make decision as soon as possible because on the first round he has the maximum probability to win and it goes less with every consecutive round. If we consider probability like in Idea 1 in independent manner, the maximal probability to win is on the round 4.

What is the right way I still don't know.

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Each player can choose $A$ at most once, and they gain no relevant information in the course of the game – though on day $k$ they know that the game has not ended before day $k$, this does not put them in any better position to make their decision on day $k$ than if they'd fixed it beforehand. Thus if the game lasts $n$ days, each player has $n+1$ different strategies, either to choose $A$ on some specific day or never to choose $A$. The prisoner wins if and only if both players choose the same strategy. Then an obvious (subgame-perfect) Nash equilibrium is to choose uniformly among the $n+1$ strategies. The judges win with probability $n/(n+1)$, the prisoner with probability $1/(n+1)$.