I know there are tons of math experts and I would like some help from you guys.

Ultimate X is a Markov decision problem. Suppose I play DDB 9/6 10-Play UX game, the current dealt hand is AAA33, with a total multiplier of 20 summing all 10 hands.

1. Holding AAA, return 61.364, possible outcomes with (# of combinations, multiplier) are

4AWK: 12, 4X

4ACES: 36, 4X

FH: 67, 12X

3OK: 968, 4X

2. Holding AAA3, return 58.511,

4AWK: 1, 4X

FH: 2, 12X

3OK: 44, 4X

3. AAA33, return 45.000,

FH: 45, 12X

4 to 32 ...

So what is the formula for aggregating all these numbers and expected multipliers and calculating the return of the current hand AAA33 with the total multiplier 20X?

I used this type of method on a three card poker game (where you get five cards and make three 3CP hands) to decide what hands the Dealer can make and whether you play or fold.

Personally I don't get into games where you have more than one decision point (except Blackjack which has a small number of possible hands). I can see you need to create a database at the second decision point before you can analyse the first decision point etc. That's too much hassle for me!!

To clarify, are you counting 1x multipliers in the 20x? This is the right way to do it. This is just above the average multiplier total.Quote:szaHello,

I know there are tons of math experts and I would like some help from you guys.

Ultimate X is a Markov decision problem. Suppose I play DDB 9/6 10-Play UX game, the current dealt hand is AAA33, with a total multiplier of 20 summing all 10 hands.

1. Holding AAA, return 61.364, possible outcomes with (# of combinations, multiplier) are

4AWK: 12, 4X

4ACES: 36, 4X

FH: 67, 12X

3OK: 968, 4X

2. Holding AAA3, return 58.511,

4AWK: 1, 4X

FH: 2, 12X

3OK: 44, 4X

3. AAA33, return 45.000,

FH: 45, 12X

4 to 32 ...

So what is the formula for aggregating all these numbers and expected multipliers and calculating the return of the current hand AAA33 with the total multiplier 20X?

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Also, are you asking how to decide whether to break up the full house or not?

If the multipliers are way above average, you can switch off the bonus for one hand and just keep the dealt full house. I know you cannot do this on Bonus Streak and some of the other new multiplier stream games. I proved that switching the bonus off increased the EV slightly for single-line UX, but switching off and on does slow down the throughput. If you have a good promo situation, this might be monetarily inferior strategy. I never worked on switching strategies for multiplay UX.

Let me assume you want to keep the bonus on. The short answer is that the next hand has a return of roughly 99% times the multiplier total / 20. You need to calculate the expected number of multipliers for the two decisions, multiply the difference by 0.99 / 20 and add the expected immediate return for the current hand under the two choices.

The long answer is to solve the Markov equations.

Quote:szaHello,

I know there are tons of math experts and I would like some help from you guys.

Ultimate X is a Markov decision problem. Suppose I play DDB 9/6 10-Play UX game, the current dealt hand is AAA33, with a total multiplier of 20 summing all 10 hands.

1. Holding AAA, return 61.364, possible outcomes with (# of combinations, multiplier) are

4AWK: 12, 4X

4ACES: 36, 4X

FH: 67, 12X

3OK: 968, 4X

2. Holding AAA3, return 58.511,

4AWK: 1, 4X

FH: 2, 12X

3OK: 44, 4X

3. AAA33, return 45.000,

FH: 45, 12X

4 to 32 ...

So what is the formula for aggregating all these numbers and expected multipliers and calculating the return of the current hand AAA33 with the total multiplier 20X?

link to original post

I cannot answer your question but here's a paper by Gary Koehler at the Wizard of Odds website on the analysis of Ultimate X: https://wizardofodds.com/pdf/ultimatex.pdf.

I am pretty busy today, but I will try to give more detail.Quote:szaHello,

I know there are tons of math experts and I would like some help from you guys.

Ultimate X is a Markov decision problem. Suppose I play DDB 9/6 10-Play UX game, the current dealt hand is AAA33, with a total multiplier of 20 summing all 10 hands.

1. Holding AAA, return 61.364, possible outcomes with (# of combinations, multiplier) are

4AWK: 12, 4X

4ACES: 36, 4X

FH: 67, 12X

3OK: 968, 4X

2. Holding AAA3, return 58.511,

4AWK: 1, 4X

FH: 2, 12X

3OK: 44, 4X

3. AAA33, return 45.000,

FH: 45, 12X

4 to 32 ...

So what is the formula for aggregating all these numbers and expected multipliers and calculating the return of the current hand AAA33 with the total multiplier 20X?

link to original post

Your numbers are wrong, for starters. 4AWK: 12, 4X should be 4AWK: 10, 4X, since two of your kickers are used up.

Your next hand will have a 12x multiplier if you go for option #3. Since the multiplier is so high, you will mostly play this next game like normal DDB. You lose a lot for any deviations trying to get better multipliers for the third game. Your immediate return will be (20/10)*10*9 and the return for the next hand will be roughly (120/10)*10*0.99. The item in parentheses is the total multipliers divided by the number of hands, that is, the average multiplier.

If you hold AAA (option #1), your average multiplier will be just 4.5 and your average total will be 45. Your strategy changes will reduce the return of the next hand. Let me just guess 0.98 because I don't care to calculate it now. Your immediate return will be (20/10)*10*(61.365/5) and the return for the next hand will be roughly (45/10)*10*0.98

Option #1: 245.46 + 44.10 = 289.56

Option #3: 180.00 + 118.80 = 298.80

It turns out that taking the full house now is better than taking the higher current EV and then losing out on the value of the next hand.

Quote:MentalI am pretty busy today, but I will try to give more detail.Quote:sza

I know there are tons of math experts and I would like some help from you guys.

Ultimate X is a Markov decision problem. Suppose I play DDB 9/6 10-Play UX game, the current dealt hand is AAA33, with a total multiplier of 20 summing all 10 hands.

1. Holding AAA, return 61.364, possible outcomes with (# of combinations, multiplier) are

4AWK: 12, 4X

4ACES: 36, 4X

FH: 67, 12X

3OK: 968, 4X

2. Holding AAA3, return 58.511,

4AWK: 1, 4X

FH: 2, 12X

3OK: 44, 4X

3. AAA33, return 45.000,

FH: 45, 12X

4 to 32 ...

So what is the formula for aggregating all these numbers and expected multipliers and calculating the return of the current hand AAA33 with the total multiplier 20X?

link to original post

Your numbers are wrong, for starters. 4AWK: 12, 4X should be 4AWK: 10, 4X, since two of your kickers are used up.

Your next hand will have a 12x multiplier if you go for option #3. Since the multiplier is so high, you will mostly play this next game like normal DDB. You lose a lot for any deviations trying to get better multipliers for the third game. Your immediate return will be (20/10)*10*9 and the return for the next hand will be roughly (120/10)*10*0.99. The item in parentheses is the total multipliers divided by the number of hands, that is, the average multiplier.

If you hold AAA (option #1), your average multiplier will be just 4.5 and your average total will be 45. Your strategy changes will reduce the return of the next hand. Let me just guess 0.98 because I don't care to calculate it now. Your immediate return will be (20/10)*10*(61.365/5) and the return for the next hand will be roughly (45/10)*10*0.98

Option #1: 245.46 + 44.10 = 289.56

Option #3: 180.00 + 118.80 = 298.80

It turns out that taking the full house now is better than taking the higher current EV and then losing out on the value of the next hand.

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Mental, thank you very much. Yes, that's a typo, it should be 10 for 4AWK.

Option #1, I looked it up in the EV table of VP.com

EV: 1399.548, which is Win + XEV = 1227.29 + 172.259

I got the number 1227.29, which is 20 x 61.364, but why the XEV is 172.259?

Assume XEV is 172.259, since the average X for next every hand is 4.495837188, this equation should hold:

4.495837188 X per hand * 10 wager per hand * 10 hand * ??? = 172.259

What would be the ??? in the formula? How to decide it?

Sorry, I am not going to look at VP.com and see what they are doing. I just know how I did it 10 years ago. I have not played UX in 8 years. The games I had available in good pay tables were all single line. I played by playing all multipliers off at 5 coins and then playing 10 coins on the next hand. UX had to be designed very carefully to make 10-coin play come out about the same EV as 5-coin play. IGT is giving the player an option to play the multipliers two ways, and they had to be careful that that option could not be exploited for a large EV boost.Quote:szaOption #1, I looked it up in the EV table of VP.com

EV: 1399.548, which is Win + XEV = 1227.29 + 172.259

I got the number 1227.29, which is 20 x 61.364, but why the XEV is 172.259?

Assume XEV is 172.259, since the average X for next every hand is 4.495837188, this equation should hold:

4.495837188 X per hand * 10 wager per hand * 10 hand * ??? = 172.259

What would be the ??? in the formula? How to decide it?

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You can look at UX as a game where you play infinite sessions and therefore need to solve an infinite Markov chain to play perfectly. Gary just calculates until the Markov chain converges on a good enough strategy. I always looked at it as a finite chain. You could decide that you were always going to play off 12X multiplier at 5 coins, for example. Then the chain is finite but the length is determined by luck in getting a 12X multiplier. Since these are rare, the initial strategy would be almost the same as the infinite chain. The EV would also be virtually the same.

I went further and asked what would be the EV if I terminated the chain by playing at 5 coins whenever the multiplier was greater than 1X. It turns out that the EV is better than the infinite game. If you go back to your original question, you have already done the math to answer which option would be better if you play off the hand at 5 coins. In that case, you just hold AAA.

The answer is different because we are saving ourselves 5 coins to play this way and we are avoiding the need to make bad compromise. When you play any hand at 10 coins, you have to compromise between a strategy that maximizes current payoff and a strategy that maximizes next-hand multipliers.

By switching back and forth between 5 and 10 coin play, I also was able to play optimally after memorizing only two strategies. Since I rarely played UX, I was never going to memorized seven different strategies or carry around a cheat sheet.

I try not to speculate about calculations that I have not done, but I always supposed that the optimal EV for multi-play UX is to often play the larger multipliers out at 5 coins per hand. I am just guessing that a total of 14x on 10 play would be enough to warrant playing off at 5 coins. This would avoid the need for complex strategy charts for every total.

I am working on some really difficult calculations right now, and I am not going to take time right now to prove my conjecture. I have all the code to do the calculations, but I would need to create a Monte Carlo simulation to test different total cutoffs.

The Markov decision problem requires the calculation of the probabilities for each transition from the current state to next state, and then calculate the weighed average of the expected value of all next states according to each transitional probability.

For 10-Play, the current state has 20X, the next state for the No.1 scenario is {2000-(4X), 800-(4X), 45-(12X), 15-(4X)}, so basically there are 4^10 transitions from current state to the next states. And I have to calculate all of them to decide the XEV??

Or do I just over-complicated it??

The only thing that changes the value of the next hand is the total of all the multipliers. From any one hand like the example you gave, the different possible totals are quite restricted. You can get 40X, 48x, ..., 120x or 11 different states with the last one being extremely unlikely.Quote:szaI thought about the Markov Decision calculation again.

The Markov decision problem requires the calculation of the probabilities for each transition from the current state to next state, and then calculate the weighed average of the expected value of all next states according to each transitional probability.

For 10-Play, the current state has 20X, the next state for the No.1 scenario is {2000-(4X), 800-(4X), 45-(12X), 15-(4X)}, so basically there are 4^10 transitions from current state to the next states. And I have to calculate all of them to decide the XEV??

Or do I just over-complicated it??

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For the UX game overall, you can get more different totals. So, single-line UX might have 7 different states and 10-line UX might has less than 100 states. (I didn't actually could how many multiplier totals are possible).