All of the original cards were removed and replaced with 4 similar cards (8 of diamonds -> 8 of hearts, etc.). This fools your brain into thinking that Simon guessed your selected card...
Im an amateur magician, and although this concept seems pretty simple to spot, a great number of card tricks rely on this very technique. Takes balls to try it with a live audience, (because it seems SOOOO SIMPLE) but I've yet to have someone (who doesn't have the luxury of a "back button" on their brain) burn such a trick...
Matt: I just beat the gemstone game, so it is possible.
The way I look at it is...the key to winning is be sure YOU are the one that gets it down to 13 gemstones. On your next turn, get it down to 9, and you're all set.
I will elaborate a little more than Greg, in case someone is still in doubt. we want Simeon to have 1 left to make him lose. Suppose we are left with 2, 3, or 4 stones left. Then take away 1, 2, or 3 stones, respectively to leave Simon with just 1 stone. In other words, we win the game if we have 2, 3, or 4 stones left. If we are left with 5, then we lose, since any choice we make will leave Simon with 2, 3, or 4 stones left.
So now the objective is to force Simeon to be left with 5 stones. Similar to the above argument, we can force Simeon to lose if Simeon was left with 5+4=9 stones earlier. Continuing this argument, it's not too hard to see that we can always force Simeon to lose if Simeon is left with 1, 5, 9, 13, 17, or 21 stones (remainder of 1 when divided by 4).
So when the game starts with 23, just take two and leave Simeon with 21 stones. Then always take away appropriate number of stones so that Simeon will be left with the appropriate number of stones. For example, if he takes away 3, we then take away 1. So the net change is always 4, without changing the number's remainder when dividing by 4 (This also explains why Simeon wins if he starts).