Optimal Ultimate X strategies using the full action space

I'm planning on making this a long thread so here is a summary. I believe the currently published optimal returns for ultimate x by igt and wizard of odds aren't the true optimal returns. If one expands the strategy space to include the option to bet five coins per hand, then one can do better. I'm not sure whether my math is wrong so I need to submit these ideas to criticism if they are ever going to be useful.

Let me say one more thing before I begin. I believe these improvements are small enough that they won't negatively impact the current machines available, but if anyone has an objection to me discussing this, please let me know.

First a little background on the theory behind why ultimate x strategies are difficult and change as the current multiplier changes. In any given hand you have two goals. The first is to maximize the money you get from the current hand, just like standard video poker. Your next goal is to maximize the value of the next hand by generating multipliers. These two goals conflict because the hands that pay more on the current hand are not the hands that generate the largest multipliers. You have to balance these goals to maximize your total equity from the combined sources. In addition, the importance of each source changes as the average multiplier changes.

Now some basic math to show you what I'm thinking. Lets take as an example the strategy of switching from 10 to 5 every other hand. Definitions

BR- base return of the game on any five coin hand
M- the multiplier for a given hand
P- payout for a given hand on the current hand

Then the total payout for each hand in the alternating 10 5 strategy is

P+M*BR-1

For instance say a full house returns 9 and has a multiplier of 12 in 9/6 ddb. Then P=9 and M=12 while BR is 98.98%. We use this formula to figure out the optimal strategy for the ten coin hand. Now to figure out the return on the combined strategy of the ten coin and then five coin hand. The strategy for the five coin hand is obvious, it's just standard 9/6 ddb. This gives you the strategy for the ten coin hand. Together these give you the ability to calculate the total return of the strategy.

For 9/6 ddb I get 99.95%. This is compared to 99.86% as the upper bound that igt states for a continuous ten coin strategy. The strategy is also easier, you memorize two strategies instead of one potentially for every average multiplier.

However this is not the optimal return. Playing ten coins in any hand with no multiplier is better than five for every ultimate x game. This is built into the multipliers by igt. This work shows that at some point it becomes better to play five coins, if the average multiplier is large enough. However, one would expect some continuity, that playing ten coins for average multipliers close to zero is also better. How do I know the higher multipliers are the ones that you should play five coins for? That is where the conflict between the current and future hands is greatest as discussed above.

How did I get the 99.95%? I use

(P+M*BR-1)/2 and enter that into the wizard of odds calculator, giving 99.93 as the return for the ten coin hand, and then adjust for the fact that you are actually playing 15 coin over the entire strategy and subtract a third off the house edge. For instance say P was a deterministic 1 for all hands, BR was 99% and M was a deterministic 2 for al hands. Then one would play ten coins per hand and lose five coins, then play five coins per hand and make .99*2-1=.98*5. Total win is 14.9 coins and total investment is 15 coins for a return of 99.33%. One could get this by taking the formula above getting .99 and subtracting off a third of the house edge.

Anyway, I know that this is very likely wrong which is why it's important to subject it to public criticism. Have at me.

An alternating 10/5 strategy produces a lower EV than the single strategy that ignores the multiplier.

For the single strategy alternative, 10-hand variety of Ultimate X 9-6 DDB, the average multiplier is 2.027729 and the EV is 99.8253%. You can arrive at this iteratively by using the formula M-bar * P + (M-1)*EV for 1-coin payoffs. At the end, each final hand payoff = 2.027729*P + (M-1)*0.998253. And when you use those numbers to find the optimal strategy, you can find that the average multiplier and the EV that is produced are consistent with those "payoffs."

For the alternating 10/5 strategy, the formula should be P + M*BR in each case, including garbage (less than a pair of jacks or no pair) hands; generally for traditional vp software, you do NOT subtract out the bet, which would equate to the "-1" that is stated above. For garbage, that works out to 0 + 1 * 0.989808 = 0.989808. So, to find the optimal strategy (since most vp software requires the garbage payoff to be zero), you can subtract out the garbage payoff from each hand and use P + M * BR - BR or P + (M-1)*BR. After finding the optimal strategy, you can find the payoff by using P + M*BR to multiply by the probabilities of each final hand. That produces 99.7617%, which is LOWER than the EV of just using the single strategy.

Using the 5-coin payoffs, you get the following (rounded to 4 places):

Final Hand.........Payoff....Mult...........Freq........Eff Payoff....Subt Garbage

Royal Flush.........4000.......4...........48.9397....4019.7962...4014.8471
Str Flush...............250.......4.........282.7646.....269.7962.....264.8471
4 Aces + kicker....2000.......4.........131.6250....2019.7962...2014.8471
4 2-4 + kicker........800......4..........368.4901.....819.7962.....814.8471
4 Aces..................800.......4.........366.9608......819.7962.....814.8471
4 2-4....................400......4..........992.9830......419.7962.....414.8471
4 5-K....................250......3.........4157.2480......264.8471....259.8981
Full House..............45.....12.......29222.1520.....104.3885......99.4394
Flush.....................30......10......41294.8162.......79.4904......74.5414
Straight..................20.......8......40547.4984.......59.5923.......54.6433
3 of a Kind..............15.......4....185891.4317.......34.7962.......29.8471
Two Pair.................10.......3.....321305.8393.......19.8471......14.8981
Jacks or Better..........5........2....489100.9535........14.8981........9.9490
Nada.......................0........1...1485248.2976.........4.9490........0.0000

Multiplying the Effective Payouts times the frequencies and dividing by 2,598,960 produces 14.964259 coins for each pair of alternating hands. Dividing this by the 15-coin total bet, the EV is 99.7617%, which is lower than the single-strategy approximation with an EV of 99.8253%.

So the alternating strategy's EV is also not better than the published optimal return.

Can you guys just send me a strategy card.

Quote: AxelWolf

Can you guys just send me a strategy card.

I'd still just use JB's strategy for now. Even though they have a point, I think, it doesn't look to be a practical improvement of return.

Quote: drrock2

An alternating 10/5 strategy produces a lower EV than the single strategy that ignores the multiplier.

For the single strategy alternative, 10-hand variety of Ultimate X 9-6 DDB, the average multiplier is 2.027729 and the EV is 99.8253%. You can arrive at this iteratively by using the formula M-bar * P + (M-1)*EV for 1-coin payoffs. At the end, each final hand payoff = 2.027729*P + (M-1)*0.998253. And when you use those numbers to find the optimal strategy, you can find that the average multiplier and the EV that is produced are consistent with those "payoffs."

For the alternating 10/5 strategy, the formula should be P + M*BR in each case, including garbage (less than a pair of jacks or no pair) hands; generally for traditional vp software, you do NOT subtract out the bet, which would equate to the "-1" that is stated above. For garbage, that works out to 0 + 1 * 0.989808 = 0.989808. So, to find the optimal strategy (since most vp software requires the garbage payoff to be zero), you can subtract out the garbage payoff from each hand and use P + M * BR - BR or P + (M-1)*BR. After finding the optimal strategy, you can find the payoff by using P + M*BR to multiply by the probabilities of each final hand. That produces 99.7617%, which is LOWER than the EV of just using the single strategy.

Using the 5-coin payoffs, you get the following (rounded to 4 places):

Final Hand.........Payoff....Mult...........Freq........Eff Payoff....Subt Garbage

Royal Flush.........4000.......4...........48.9397....4019.7962...4014.8471
Str Flush...............250.......4.........282.7646.....269.7962.....264.8471
4 Aces + kicker....2000.......4.........131.6250....2019.7962...2014.8471
4 2-4 + kicker........800......4..........368.4901.....819.7962.....814.8471
4 Aces..................800.......4.........366.9608......819.7962.....814.8471
4 2-4....................400......4..........992.9830......419.7962.....414.8471
4 5-K....................250......3.........4157.2480......264.8471....259.8981
Full House..............45.....12.......29222.1520.....104.3885......99.4394
Flush.....................30......10......41294.8162.......79.4904......74.5414
Straight..................20.......8......40547.4984.......59.5923.......54.6433
3 of a Kind..............15.......4....185891.4317.......34.7962.......29.8471
Two Pair.................10.......3.....321305.8393.......19.8471......14.8981
Jacks or Better..........5........2....489100.9535........14.8981........9.9490
Nada.......................0........1...1485248.2976.........4.9490........0.0000

Multiplying the Effective Payouts times the frequencies and dividing by 2,598,960 produces 14.964259 coins for each pair of alternating hands. Dividing this by the 15-coin total bet, the EV is 99.7617%, which is lower than the single-strategy approximation with an EV of 99.8253%.

So the alternating strategy's EV is also not better than the published optimal return.

Thanks for this very thoughtful response. This disagreement comes down to what happens when you brick. My way of calculating the optimal strategy and return assumes that you don't play the five coin hand if you don't generate a multiplier, while yours assumes that you always play the five coin hand even if you brick. We both end up doing kind of silly things as a result. You are assuming that someone plays a five coin hand even with no multipliers. I am essentially assuming a one handed ultimate x game, but my return calculation is correct for that non-existent game (the 99.93% part at least, the adjustment up would be less).

That being said, I am trying to stimulate discussion. In reality, the optimal strategy is likely to be ten coins for low multipliers and five coins for high multipliers. The limitted analysis I did that applies to a game that doesn't exist still proves that improvements are possible, once we realize that you really aren't going to be playing those five coin hands with bad multipliers anyway.

Quote: randomperson

I am essentially assuming a one handed ultimate x game, but my return calculation is correct for that non-existent game (the 99.93% part at least, the adjustment up would be less).

FWIW, 1-hand Ultimate X does exist on "Five Star Poker" machines, but the key multipliers are often downgraded so they aren't worth noting.