12 replies to this topic

[witzar]

Spoiler for Another way to look at it

1. We form all possible sequences of 40 cases (39 empty and 1 with the treasure).
2. Then we remove all sequences, where treasure is in the middle (1 < case# < 40). In other words we retain only sequences with treasure in case #1 or case #40.
3. Finally we ask what kind of sequences are more frequent among retained: those with treasure in case #1 or those with treasure in case #40.

It should be obvious that both types of sequences are equally frequent, since there is an obvious 1-1 correspondence between those two sets.

(k-man nicely proved the same by counting cardinality of both sets and showing their equality.)

[k-man]

Spoiler for Yet another way...

I thought of a better way to show that the probability to win is the same whether the player chooses to keep the chosen case or chooses to switch.

 

Let's consider 2 players with 2 different strategies. Player A keeps his chosen case till the end while Player B sticks with his case until there is one last case remaining and then switches. Both players always follow their strategy, so if we find the probability of winning by each player we can determine which strategy is better or if they are the same.

 

For Player A to win he must pick the winning case from the start, so he obviously has 1/40 chance of winning the game.

For Player B to win he must sequentially pick 39 empty cases leaving the winning case as the last case, so that he wins by switching to it. Probability of picking 39 empty cases is 39/40 * 38/39 * 37/38 * ... * 2/3 * 1/2, which equals 1/40.

 

Since both strategies have exactly the same winning probability it doesn't matter which strategy to follow. 

 

Nice puzzle, Bushindo.

[bonanova]

Spoiler for Addendum

If [with knowledge of their contents] Monte had opened 38 empty suitcases after my initial choice, then I should switch.

What makes the difference? For every distribution of which suitcase holds the money, there will be 38 empty suitcases for Monte to disclose. Thus no distributions have been ruled out a priori. My initial choice in this case therefore has a true 1/40 probability. The remaining suitcases comprise the remaining 39/40 probability of having the money. By exposing 38 empty ones, Monte has directed me to the single suitcase that has that 39/40 chance.

I switch; and by doing so I multiply my winning chances by 39.