bonanova

If I repeatedly throw a fair 6-sided die, what is the probability that the running total will at one point equal 137?

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I'm wondering if the solution is unique. It will be interesting to see if another solution presents itself.

k-man, Great observation. Anza Power, very interesting puzzle.

OK I've worked on this for a while. Convinced it has no solution. BTW it's Hex, not binary.

This might help readability. Re-posting using Courier font and some added spaces. 5 A 3 2 6 D C D 0 5 F 9 3 B 2 2 4 B D 8 A 1 7 B 5 E 9 C 7 E F 7 B 6 3 F C A 0 6 1 9 D E 0 D 5 B 4 2 7 4 D F F D 6 B E 4 0 1 8 5 3 F 8 2 0 C 3 8 7 D A 8 C 7 6 F 5 0 4 D 5 A B 3 C E A 1 F 9

This puzzle asks for the strategy that maximizes the likelihood of choosing the babe with the greatest dowry. Real-life situations often ask for the strategy that yields the best result, on average - best result or not. Clearly a strategy that on average gets you into the top ten percent of the dowries would be preferred over one has a better shot at picking the highest dowry but on average gets you only into the top twenty percent. That said, and if someone has written the simulation for this puzzle, I wonder where the OP best strategy places you in the distribution, on average. And, is there a strategy that places you higher? A good sample set of dowries to find out would be simply 1, 2, 3, ... , 98, 99, 100. Run the algorithm 100,000 times without using knowledge of the distribution, and average the value of the result.

[spoiler=Nice puzzle.]The beauty of puzzles of this type is the presence of a simple and compelling analysis that leads to the wrong answer. I should say the beauty, for me, lies in the presence of a simple, and equally compelling analysis that leads to the right answer. And when considered side by side, the correct analysis identifies and explains the fallacy of the incorrect approach. The beauty is particularly enhanced when it obviates the need to crank through the appropriate Bayesian formulas. Gah! Here, the [erroneous] analysis springs from assigning forever a 1/40 probability to the first suitcase holding the money. How on earth could that value change, simply because we inspected 38 other suitcases? "It can't," is the compelling thought, so the other 39/40 probability certainly must lie with the remaining cases, finally devolving upon the 40th case. Switch! You're a fool not to. Just like in Monte Hall. But then one remembers that in the former puzzle Monte knew the intermediate cases [doors] were losing choices. Moreover, Monte was able to open a losing door for every distribution of hidden contents. Here, one does not know a priori they are losing choices. It just turns out that they are. And, suitcase 1 was chosen from a sample population of cases in which 2-38 [numbered by the order chosen] were losers. This was true even though we did not know it. An assignment of 1/40 is valid only when a choice is made from an unbiased distribution of equal a priori likelihoods. How do we see [intuitively] that the conditions of the problem make 1/40 the wrong measure of probability for the initial suitcase? Read on. I will change the problem slightly. In a way that [a] still adheres to the conditions of the puzzle and does not change the answer. Since events are independent, order does not matter. Defer the choice of the "initial" case until after 38 "intermediate" cases have been inspected. In ninety-five percent of the trials, money will be found in one of the "intermediate" suitcases. These instances do not fit the conditions of the puzzle, so when that happens we begin again. Eventually we will open 38 empty suitcases; now the puzzle conditions have been realized. Next, we choose an "initial" suitcase. Finally, we choose between the "initial" case and the 40th suitcase. Shall we switch, or not? Now intuition leads clearly to the correct answer.

Several solution have been posted. Mine is not greatly distinguished from any of them. It differs from k-man's (above) only in that it works with 12 chips.

First player always wins then?

I heard it given by Raymond Smullyan in a taped greeting to his close friend Martin Gardner. It was new to me.

Interesting puzzle!

I had in mind finding a method that did not seem to depend on knowing the value of x. There may be very many such ways. So let's amend the trick to make it seem more difficult, but actually providing a guide to the desired solution. Have Jamie construct three equal stacks. You do not know how many chips are in each stack. Instruct Ian to make three adjustments to the stacks. The words all, half, etc may not be used; at no point is any stack empty. Ask Davey to call out a number 0 < x < 13. Instruct Ian to make a fourth adjustment to the stacks: "Move __ chips from Stack _ to Stack _." Where "__" is a specific number different from x. The middle stack should now have x chips.

a. I'm asking whether that was the meaning. I don't think it has been previously posted.

It was amateur night at Morty's. Alex had finished his penultimate act of mysterious prestidigitation. Preparing for the finale, he handed a pile of poker chips to Jamie and, after turning his back to the table, asked Jamie to construct three equal stacks of at least four chips. Back still to the table, Alex next asked Davey to call out at random a number between 1 and 12. Suppose Davey calls out the number x. With the chips still out of his sight, and without knowing the number of chips in each stack, Alex finally gave instructions to Ian to shift chips among the piles, in a manner that left x chips in the center stack. What is the method? To clarify, neither the number x nor the word "all" were mentioned in the instructions. This rules out the obvious: move all the chips to stack 3, then move x of them to the middle.

Thinking of a puzzle that could be titled the Purloined Sirloin. Probably won't find one.

Very nice Cap'n!

Right. I meant to ask could not Lucy be insane.

Lucy's statement could not be true?