I'm going to try this and waste a lot of time if I don't do this right, but I should get close.
The first thing that I am going to do is eliminate, of the 22,100 possible hands, all hands that have a card higher in rank than a ten in them. Fortunately, that part is easy:
(36/52)*(35/51)*(34/50) = 0.32307692307
That's going to be very important because now I have to subtract certain other hand probabilities from that in order to arrive at how many combinations there are. In terms of how many possible hands that would be, we see:
22100 * .32307692307 = 7139.99999985, so, 7,140.
For the remainder of these calculations, we are going to be basing it on 7,140 and subtracting certain types of hands from that. We will use the same logic here:https://wizardofodds.com/games/three-card-poker/appendix/1/
With the remaining cards, which are essentially the same as a deck that only runs 2-10, which means that there are nine ranks and four suits. That tells us the number of combinations is:
(36*35*34)/(3*2*1) = 7,140
Which agrees with the above number, so that's fantastic.
The first thing that we will do is eliminate the straight-flush, the span of which can run 2-3-4 all the way to 8-9-10, which comprises seven total spans. (A-2-3 is gone because it would mean an Ace). 7 * 4 (suits) = 28
7,140 - 28 = 7,112
This is simple, the number of possible Trips equals the number of cards in the deck. There are 36 cards that remain in the deck, therefore there are 36 different Trips. Alternatively, nCr(4,3)= 4 which chooses three suits out of four and then you multiply by the number of ranks, which is nine, so 9 * 4 = 36
7,112 - 36 = 7,076
For the straights, from the straight-flush section we know that there are seven possible spans for a straight. A straight has three cards, each may be one of four suits. However if all three suits are the same then the player has a straight flush. So the number of suit combinations is (4)^3-4 = 64-4 = 60. So there are 7*60=420 possible straights.
7,076 - 420 = 6,656
There are four possible suits for the Flush. For each suit there are nCr (9,3) = 84 ways to draw three ranks out of nine. However, from the straight flush section we know that there are seven combinations which result in three connected ranks, giving the player a straight flush. So the combinations giving a straight, but not a straight flush, are:
nCr(9,3)-7 = 84 -7 = 77.
Therefore, the number of flushes is 77 * 4 = 308
6656-308 = 6,348
There are nine possible ranks for the pair and eight for the singleton, ergo, there are 72 ways to pick the ranks. Within the pair, there are nCr(4,2)=6 ways to pick two suits out of four. For the rank of singleton there are four possible suits. So the total suit combination is 6 *4 = 24. The total number of pair combinations is 72 * 24 = 1,728
6348-1728 = 4,620
4620/22100 = 0.2090497737556561
Thus, 0.2090497737556561 is the probability of the dealer getting a non-qualifying hand.
The probability of the dealer getting a qualifying hand is 1-0.2090497737556561 = 0.7909502262443439
In general, there are four game states:
Player Qualifies, Dealer Qualifies:
(17480/22100) * (17479/22099) = 0.6255947782490107
When that game state is met, the dealer and player will win, lose and tie equally. Therefore, no House Edge is derived from that game state.
Player Does Not Qualify, Dealer Qualifies:
(4620/22100) * (17480/22099) = 0.1653554479953332
The dealer would win in such a situation, anyway, so no additional House Edge is derived from that situation.
Player Does Not Qualify, Dealer Does Not Qualify:
(4620/22100)*(4619/22099) = 0.0436943257603229
Half of these situations actually benefit the player, because the dealer would then beat the player with a non-qualifying hand. Half of these situations actually benefit the dealer, because the player would beat the dealer when the dealer has a non-qualifying hand. Therefore, there is no effect on the House Edge.
Here is where the House Edge comes in:
Player Qualifies, Dealer Does Not Qualify:
(17480/22100)*(4620/22099) = 0.1653554479953332
The player would literally always win and never lose in this situation, but now the player always pushes.
NO HOUSE EDGE, NO QUALIFY RULE:
(0.6255947782490107 * 5 * .5) + (0.6255947782490107 * -5 * .5) = 0
(0.1653554479953332 * -5) = -0.82677723997
(0.0436943257603229 * 5 * .5) + (0.0436943257603229 * -5 * .5) = 0
(0.1653554479953332 * 5) = 0.82677723997
That would be a fair game, the House Edge would be zero.
However, with the, 'Dealer Qualify,' rule, the last part does not happen and the result of such an event is:
(0.1653554479953332 * 0) = 0
0 + -0.82677723997 + 0 + 0 = -0.82677723997
Which means that the player is expected to lose 82.677723997 cents for every five dollars bet, which makes the House Edge:
.82677723997/5 = 0.16535544799
Which is, of course, roughly the probability that the subject matter hand (Player Qualifies, Dealer Doesn't) comes up.
Vultures can't be choosers.