P1 is in a better position to benefit from relatively lucky shuffling, while P2 is more vulnerable to relatively unlucky shuffling. If P1 and P2 have similar shuffle luck, then P1 has no advantage when it comes to splitting key cards.

This is just not mathematically sound. One player benefitting more from lucky shuffling and the other being more vulnerable to unlucky shuffling aren't effects that cancel each other out but in fact

*compound* each other.

Let's assume that each player has an equal chance, each turn, of missing an opportunity to buy a key card. We'll count the number of missed opportunities for each player over the course of, let's say, 8 turns. (It doesn't matter how many, really.) Let's say each player has a 20% chance of a turn being a missed opportunity. (The exact percentage affects the magnitude of the resulting advantage but not whether or not there is one.) Let's let M1 = the number of P1's missed opportunities. Let M2 = the number of P2's missed opportunities.

If we simulate this, then the values of M1 and M2 we come up with will determine how the key card splits. If M1 < M2, then P1 wins the split. If M1 is

*equal to or one more than* M2, then the split ties. For P2 to win the split, M2 must be

*at most two less than* M1.

The result of averaging this calculation over many simulations should be apparent, but just for fun I simulated this 10,000 times and got the following results:

* P1 wins the split: 3789 times

* Split is even: 4494 times

* P2 wins the split: 1717 times

In other words, P1 is more than twice as likely to win the split as P2 is. Not that I recommend reading too much into these specific numbers! Obviously I've made a lot of assumptions on how many turns it will take to exhaust the supply pile of the key card, and the likelihood of missing an opportunity to purchase one. Moreover, the chance of missing any one opportunity is correlated with the chances of missing any of the others, because having a bad hand means possibly having a good hand next. Still, you can finagle these numbers any way you want to: ultimately, the odds MUST favor P1, simply because P1 can win the race after making more mistakes than P2 can make in order for him to.