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?
(In reply to
re: computer discovery by Jer)
I am going to try to modify the program for non-simultaneous card flips during one round to see what I get.
Also, for small values of n, even with the simultaneity rule in effect, the strategy I was using isn't optimal, at least for n=2 (four cards). If one does not get a match looking at the first two cards, then if one selects the remaining two cards there's no chance of getting a match then either, whereas if you choose one of the known cards and one of the unknown cards, there's a 50% chance of getting a match, and then an assured match next round. If this modification were made for the strategy in the simultaneous flip version, there is:
1/3 probability of an immediate first match, for a total of 2 rounds.
2/3 * 1/2 = 1/3 probability of no match on first try, but a match on the second try as described in the preceding paragraph, for a total of 3 rounds.
1/3 probability of 4 rounds (no match, no match, match, match).
That brings the expected number of rounds down to 3.
I would think for larger numbers of cards, the exploratory value of selecting solely unknown cards (if no pairs are known) outweighs the probability of getting a match by chance.
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Posted by Charlie
on 2017-06-01 08:58:32 |
re(2): computer discovery
