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This is a follow-up to "How to beat Count Dracula".


Abraham Van Helsing and Jonathan Harker play a game against Count Dracula. The three players agree on the following rules:

  1. Count Dracula and Van Helsing enter the crypt, while Jonathan Harker has to wait outside. The crypt contains three coffins that are respectively numbered $1$, $2$ and $3$, together with three golden lockets that are respectively engraved with the numbers $1$, $2$ and $3$.

  2. Count Dracula puts the lockets into the coffins, so that every coffin contains exactly one locket. Every possible assignment of the three lockets is used with equal probability. (That's important: the count does not pick the assignment.) Van Helsing observes this, and knows exactly which locket is in which coffin.

  3. Now it is Van Helsing's turn: Van Helsing may (but does not have to) pick a pair of coffins and switch the lockets in these two coffins.

  4. Van Helsing now leaves the crypt through the back door. Jonathan Harker (who has gained no additional knowledge over the choices of Dracula and Van Helsing in the first three steps) enters the crypt.

  5. Count Dracula uses a random generator to pick an integer $N$ with $1\le N\le3$, and announces $N$ to Jonathan. Every possible value for $N$ is picked with probability $1/3$.

  6. Jonathan is allowed to choose one coffin and to open it.

    • If Jonathan Harker finds the locket with number $N$ in the opened coffin, then he and Van Helsing have won the game. In this case, they are allowed to drive a wooden stake through Dracula's heart.
    • If Harker does not find the locket with number $N$, then Count Dracula is allowed to drink Van Helsing's and Harker's blood to the last remaining drop.

Harker and Van Helsing discuss their options and want to agree on a good strategy. If Harker chooses his coffins simply at random, then the team Harker & Van Helsing has a $1/3$ probability of winning the game. Is there a strategy that would guarantee them an even better probability of success?

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I'll start with an obvious improvement. Let $abc$ mean the locket $a$ in coffin $1$, locket $b$ in coffin $2$, locket $c$ in coffin $3$.

Switch the following.

$123\rightarrow123$

$132\rightarrow123$

$213\rightarrow123$

$231\rightarrow132$

$312\rightarrow132$

$321\rightarrow123$

Then, open the $i$th coffin when Count Dracula says $i$. The final probability is $\frac{7}{9}$.

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  • $\begingroup$ What "obvious improvement"?  This is exactly the same as Jon Mark Perry's answer (unless he edited it in the first five minutes?). P.S. Please use spoiler blocks. $\endgroup$ – Peregrine Rook May 28 '16 at 20:08
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    $\begingroup$ @PeregrineRook - Element785 posted only 7 seconds after Jon Mark Perry (checkable by mouseover text on the answer times). Element118 is not referring to Jon's post. His "obvious improvement" is over the 1/3 probability mentioned in the OP. $\endgroup$ – Paul Sinclair May 28 '16 at 21:09
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Van Helsing ensures Coffin $1$ has Locket $1$ in it. If this is already the case, he ensures Coffin's $2$ and $3$ are also correct. Now Jonathan picks Coffin $N$ which in all but the cases $231$ and $312$ works and $N=2,3$, so a $\dfrac{14}{18}=\dfrac{7}{9}$ chance of success.

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