10C, 10S, JS, QS, KS Draw Poker

I'm developing a program that will tell you what cards to hold when playing Draw Poker. I will eventually figure this out on my own but I'm looking for a shortcut. My program (either correctly or incorrectly) is telling me that when I am dealt 10C, 10S, JS, QS, KS that my best chance of improving this hand is to hold onto the pair AND the king. This seems counterintuitive. I would expect that you have a better shot of improving the hand if you toss the king as well. Is there a free program available that can confirm or reject this?

How about tossing the 10C?

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I guess it depends on how the hands are ranked and the ultimate objective. If the hand are ranked as "one pair" as opposed to "pair of tens" then "one pair with a king" beats "one pair with a queen (or lower)". In that case it may make sense to keep the king as a high kicker.

if you're talking VP, toss the 10C and go for the straight, flush, SF, or RF as these will have high enough payouts relative to the low pair (unless it's some crazy version of VP with wild 10s or something).

Keeping the royal draw, your outs are:
1 to RF
1 to SF
7 to a flush
6 to a straight
9 to a face pair

This totals 24, of them 15 are high. Another 2 keep you par, and the remaining 21 (or less depending on burn cards) give nothing.

Keeping the pair and a king:
2 to trips
3 to two-pair

Each of these is drawn twice. There is exact math for this, but it's harder, and, approximately, it's equivalent to 10 outs.
Another 42 cards, however, keep you with a middle pair and a king or ace kicker.

The average winning hand in draw poker is a two pair. A middle pair is weak. Considering that the first option gives you 15 or 24 outs and the second 10 outs, though leaves you with a better hand on missing, it seems like the program is in error. This will depend somewhat on the table, but even on a perfectly loose table, the second option might still not be better.

The OP seems to have made it pretty clear that this is Draw Poker, which is NOT video poker and rather a live multiple player game.

I should have been more specific. This isn't relative to a any game. It's a math exercise. All 7462 possible poker hand rankings are considered, from unsuited 2,3,4,5,7 (of which there are 1020 combinations) to Suited 10,J,Q,K,A (of which there are 4 combinations). The exercise is to take a set of five cards (like the ones in the title of this post) and discard zero, one, two, three, four and all five cards and draw from what remains in the deck to determine which action produces the highest probable result on average.

The methodology is to build five tables. The first has a single card and then all of the possible combinations for all of the other remaining cards making a five card hand are calculated and summed up. Take the 2 of clubs for example, there are 255 ways (out of 249,900) for the 2 of clubs to wind up in an unsuited hand of 2,3,4,5,7. There in only one combination for the 2 of clubs highest hand, a straight flush to the 6. The second table calculates two cards held then three then four. The last table is the relative strengths of all 7462 rankings that five card poker allows ranked from 1 to 7462 (it also serves as the "Discard All" basis).

By joining these tables, I think I can create a quick calculator for determining what to do when playing 5 card draw. The tricky part is eliminating the discarded cards from the overall calculation. I think that's where my biggest bug is. I have no issue with the 'four cards held' because all you have to eliminate is the one possible starting hand. In other words, if you start with a 10c, 10s, Js, Qs, Ks and discard one card, you can't end up having 10c, 10s, Js, Qs, Ks. So you subtract one when that becomes a possibility. There are 48 possible draws for 10s, Js, Qs, Ks but 1 contains an instance of your original hand which must be subtracted leaving 47. This part I have correct. but I'm having a little difficulty removing the discarded cards from the three card combinations on down to zero.

Is there a free calculator already available that does this? I googled but came up empty.

Small victory.

I got it to work for discarding None, One, Two and Three cards. There is a wrinkle in discarding Four or All cards. The wrinkle is that some of the used card combinations will double or quadruple count and since I didn't bother setting up the tables with suit identifiers, there is no way to correct that problem. I'll have to be happy with this.