10 cards with identical backs are face down on a table. Each card face matches exactly one of the other card faces. The cards are in a random sequence. A turn consists of choosing 2 cards, simultaneously reversing them so that they are face up, discarding them if they match, and turning them face down if they do not match. The game ends when all cards are discarded.

a) If you have perfect memory, and an efficient strategy, then what is the expected number of turns in the 10 card game?

b) What is the expected number of turns if instead there are 2n cards in the starting tableaux, with each card matching exactly one other?

Is the player always going for the least possible draws?  Or is he simply trying to keep his draws to a minimum?

For example, let's say the cards are marked 1A thru 5A and 1B through 5B, with 1A-1B  being a match, and so on.

On the first draw, the player draws 5A and 5B, a match.  On the second draw, he draws 1A and 2A.  On the next draw, he draws 3A and 4A.  At this point, he has a decision to make.  He can risk the fewest draws possible by selecting a known card with an unknown one:

Round 4 match - 25% probability
Round 5 match - 33% probability
Round 6 match - 50% probability
Round 7 match - 100% probability

But if he misses, he could default to 9 draws:

Round 4 match - 25% probability
Round 5 No match - 67% probability
Round 6 match - 100% probability
Round 7 No match - 50% probability
Round 8 match - 100% probability
Round 9 match - 100% probability

In order to avoid having 9 draws, the player could select two unknowns in Round 4 with 0% probability of getting a match.  But in doing so, the player guarantees a number of 8 draws - one more than the best-case, but one less than the worst case.

So, some rules are in order here.  What is the objective of the player?  Is he risk-loving or risk adverse?  Will he always attempt the fewest draws possible at any stage, or will he accept more draws to minimize risk of taking even more?

Posted by hoodat on 2017-06-01 19:46:32