Why is a Pair of Tens better than a Pair of Aces in Texas Hold 'Em? A couple years ago I developed a program to calculate the optimum betting amount for a round of Texas Hold 'em by using the Kelly criterion. In the process of computing the probability of winning for certain poker hands, I discovered an interesting fact regarding hole cards: a Pair of Tens is a better than a Pair of Aces.
I discovered this by iterating through each possible hand for a given set of hole cards, and comparing it to every possible opponent hand combination; I recorded the number of hand showdown scenarios that resulted in a win for each set of hole cards.

Side note: In Texas Hold 'Em, player hands are played off a 5 card community pile on the table; each player has two unique hole cards, but create their hands using the community cards on the table. It's therefore possible for player and opponent to have the same hand despite different hole cards.

Since I knew that there are $2097572400$ possible hand combinations given a set of hole cards, I was able to derive the probability of winning a single opponent showdown, given a set of hole cards:
$P(win) = \frac{number\ of\ wins}{2097572400}$
Regarding hole cards against a single opponent, a Pair of Tens results in an approximate $61.09$% showdown win probability, while a Pair of Aces only results in an approximate $59.97$% showdown win probability. I've posted the entire probability table online for those who are curious.
Although my bruteforce computation proves that starting out with a Pair of Tens in Texas Hold 'Em is better than starting out with a Pair of Aces, I would like to understand why.

I suspect that the answer lies in how a straight can be formed. For a Pair of Tens, high-ranking straights can be formed both above and below the Ten card:
+------------------+
|6|x|x|x|10|y|y|y|A|
+------------------+

In this case, Ten is right in the middle between a Ten-High Straight and an Ace-High Straight; all straights between may also be formed.
Take a lot at the disadvantage of an Ace:
+------------------+
|10|x|x|x|A|y|y|y|5|
+------------------+

Since Ace is right at the straight cutoff border, Ace-High Straight and 5-High Straight are the only possible straights.

Aside from my bruteforce method, is there a succinct mathematical explanation of why a Pair of Tens results in more winning hand combinations than a Pair of Aces?
 A: My showdown simulation was implemented incorrectly; hands were incorrectly ranked, resulting in the wrong number of wins for certain hole cards. As such, the premise of this question is false: a Pair of Tens is not better than a Pair of Aces in Texas Hold 'Em.
Here's a new table I generated, correctly showing a Pair of Aces as the most likely hand to win a single opponent showdown:
Win prob.       Split prob.     Hole cards
---------       -----------     ----------
0.849332        0.005433        Ace of Clubs, Ace of Diamonds, 
0.821185        0.005564        King of Clubs, King of Diamonds, 
0.796330        0.005861        Queen of Clubs, Queen of Diamonds, 
0.771539        0.006328        Jack of Clubs, Jack of Diamonds, 
0.746610        0.007028        Ten of Clubs, Ten of Diamonds,
0.716662        0.007830        Nine of Clubs, Nine of Diamonds, 
0.687178        0.008911        Eight of Clubs, Eight of Diamonds, 
0.662206        0.016497        Ace of Clubs, King of Clubs, 
0.657255        0.010211        Seven of Clubs, Seven of Diamonds, 
0.653146        0.017899        Ace of Clubs, Queen of Clubs, 
...

Note: I've removed equivalent hands from the table to simplify it.
