Ultimate X strategy formulas

Wizard,

Regarding the Strategy section on the page:

Ultimate X (single line)

Is the formula win+multiplier-1 just an estimate for strategy generation purposes? I can’t seem to get the EVs for each of the multiplier strategies provided for 8/6 JoB to come up to the total EV for the game, using the probability of each multiplier provided:

Multiplier	EV	Probability	P*EV/2
1	1.990482	56.29%	0.5602211589
2	2.970095	19.93%	0.2959699668
3	3.951610	12.53%	0.2475683665
4	4.933877	7.24%	0.1786063474
7	7.882715	1.38%	0.0543907335
10	10.833378	1.50%	0.0812503350
12	12.801031	1.13%	0.0723258251
Total		100.00%	1.4903327330

Total EV is 149.03%, supposed to by 98.57%.

(Note: when I use the analyzer for 8/6 jacks with the standard multipliers it comes up with 98.4971%. I assume this is because the text says for Ultimate X it uses “the results of the optimal single strategy” rather perfect play)

What am I missing? Is the formula simplified by a factor of 150% for the sake of using integers? If so is there an exact formula? I would think for single line it would be relatively straightforward(?) to calculate the exact EV for the various multipliers that would sum up to the EV of the game? Or is it still an iterative process like multiline? There is only 1 hand and 7 multipliers and 10 paytable states including zero win so only 70 total states instead of the very large numbers of states in multiline. I see in your analysis tables it shows the 70 states in a table to calculate the total EV. I’m not sure how you calculate the probability for each state tho. I can get close by multiplying the probabilities of the 2 hands eg royal flush x fill house (12x), but it doesn’t quite match. Do you have to iterate until the change is within a certain delta like Koehler discussed?

For multiline Ultimate X Strategy section on:

Ultimate X -- Multi-Line

Why is the formula different from single line? It is 2 x win + multiplier - 1, which is equivalent to win + (multiplier - 1)/2. So why is the effect of the multiplier on strategy watered down by 1/2? I would expect the same formula as single line. Either way I’m sure this is a rough estimate because calculating the exact EV and strategy for multiline UX is quite complicated, per the Koehler paper linked to on the above page.

JD

Where are you getting your EV values from? How can any of the multipliers have an EV > the multiplier if the base EV < 100% ?

When I subtract 1 from each value in the EV, I get P EV / 2 = 99.033%

The EV values come from the WOO strategy page links for each multiplier:

1x: /games/video-poker/strategy/a-1-b-74-c-1-d-0-d-2-d-4-d-6-d-10-d-15-d-19-d-26-d-51-d-801/

2x: /games/video-poker/strategy/a-1-b-74-c-1-d-0-d-3-d-6-d-9-d-14-d-21-d-27-d-51-d-101-d-1601/

3x: /games/video-poker/strategy/a-1-b-74-c-1-d-0-d-4-d-8-d-12-d-18-d-27-d-35-d-76-d-151-d-2401/]

4x: /games/video-poker/strategy/a-1-b-74-c-1-d-0-d-5-d-10-d-15-d-22-d-33-d-43-d-101-d-201-d-3201/

7x: /games/video-poker/strategy/a-1-b-74-c-1-d-0-d-8-d-16-d-24-d-34-d-51-d-67-d-176-d-351-d-5601/

10x: /games/video-poker/strategy/a-1-b-74-c-1-d-0-d-11-d-22-d-33-d-46-d-69-d-91-d-251-d-501-d-8001/

12x: /games/video-poker/strategy/a-1-b-74-c-1-d-0-d-13-d-26-d-39-d-54-d-81-d-107-d-301-d-601-d-9601/

I attempted to provide the actual links, but Preview seems to strip out wizardofoddsDOTcom from the front of each link, making them invalid. The actual links can be found on the Single Line Ultimate X page.

The EVs appear greater than the multiplier because they have to be divided by 2. They are based on a 1 unit bet instead of 2.

Seeing these links made me realize that although I had replicated the values I had not connected it back to the original formula, which is given as win + multiplier - 1. This is after the currently active multiplier has been applied. So the formula could more accurately be stated as current_multiplier*win + win_multiplier - 1, which is what my code does except it also adjusts for 10 coins and divides by 2.

JD

After I posted this I searched thru old posts and found a couple on Ultimate X with one of the same questions.

It appears that for multiline, the formula multiplying by 2 is due to the fact that the weighted average (current) multiplier for multiple hands is close to 2. It is actually 2.0092 by my calculation.

So this also supports having the single line formula multiply by the current multiplier, which is known and doesn’t have to be averaged. This is in fact what is done in the provided strategies but it is implied before doing the formula win + multiplier - 1.

JD

You are betting twice as much per hand so you should divide everything by two. In common sense you bet ten credits and if you get a jacks or better, that pays five credits plus it multiplies a future five credit hand from 1x to 2x, so your jacks or better payout should be around 1 in the calculator. But I was wondering where are the four of a kind, straight flush and royal flush in your chart?

Also you should understand that Wizards way of generating a strategy will not and should not generate exact EVs. It’s an approximation useful only for generating strategies and will not give you exact EVs. It was not designed to do that and his language should be more explicit about what it can do.

Yes I did divide by 2 in the last column. The chart is by multiplier not by hand type, with the probabilities given on the WOO page. The fact it comes to 149% EV led me to surmise that the formula is simplified for strategy generation purposes.

It still left me with this question for single line: what is the exact formula? Is it a slightly simplified version of the Markov decision problem shown for multiline by Koehler?

Since the number of hands is 1 and multipliers is 7 so there are only 7 multiplier combinations times 10 hand outcomes = 70 states, it is not a combination of hand outcomes like it is for multiline.

I suspect this is the case and is how the Wizard calculates the EV for ultimate X games, although it appears that the WOO analyzer uses an approximation also, based on an average optimal strategy, and doesn’t match his exact EV calculation shown for 8/6 jacks or better.

Now that you clarified that the probability column includes the 4ok, sf and rf, it leaves me confused about how you are calculating the EV column when there are hands of different EVs mixed in.

But I think looking at the EV column again led me towards to an even bigger issue. Should the EV be the expected value of the current hand given the current multiplier? Or should the EV include the expected value of the multiplier generated by the hand? I believe you are including the value of the future multiplier but don’t think you should.

To the best of my knowledge, optimal play decisions consider the potential multipliers of future hands.
When betting 5 coins per line, there is no future multiplier, and optimal strategy does not differ from the standard play.
At 10 coins per line, optimal play does adjust for potential future multipliers.

Quote: Dieter

To the best of my knowledge, optimal play decisions consider the potential multipliers of future hands.
When betting 5 coins per line, there is no future multiplier, and optimal strategy does not differ from the standard play.
At 10 coins per line, optimal play does adjust for potential future multipliers.
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If this was a response to me, I thought this thread was about calculating the EV of the game for a fixed strategy, not making the strategy.

I did not calculate the EVs, they are from the WOO page on single line Ultimate X. There are strategy links for each multiplier. That is where I got them.

Later I did verify them myself with my own program using the formula provided, win + multiplier - 1, although as I noted above, the win is multiplied by the current multiplier, then you add the multiplier that would be awarded for the next hand and subtract 1.

For example, the value for the royal for 12x multiplier is not 800 + 2 - 1, but rather 12x800 + 2 - 1, or 9601.

So yes the EV of each strategy includes possible results for the next hand, since in the formula you’re adding the multiplier (with a zero offset - nothing is added for 1x).

And the EV of the 2nd hand includes the 3rd, and so on. So I do see this as a series and suspect it still requires the Markov solution like multiline does, albeit simplified. So I’m going to read the Koehler paper again and try to figure it out for single line lol

Quote: randomperson

Quote: Dieter

To the best of my knowledge, optimal play decisions consider the potential multipliers of future hands.
When betting 5 coins per line, there is no future multiplier, and optimal strategy does not differ from the standard play.
At 10 coins per line, optimal play does adjust for potential future multipliers.
link to original post

If this was a response to me, I thought this thread was about calculating the EV of the game for a fixed strategy, not making the strategy.
link to original post

Fair enough, I probably misinterpreted something.

Quote: randomperson

But I think looking at the EV column again led me towards to an even bigger issue. Should the EV be the expected value of the current hand given the current multiplier? Or should the EV include the expected value of the multiplier generated by the hand? I believe you are including the value of the future multiplier but don’t think you should.
link to original post

Just make sure you include future multiplier EV somewhere, or your calculations will absolutely be off.

I thought of a way to demonstrate the point since nobody seems to believe me. Imagine that you have a game where you just get exactly a pair of jacks every hand, nothing more nothing less. That is clearly a 100% return game. But what would EV be given your method of calculation?

Am I not correct that it would be 3 and then you divide it by 2 to make 1.5?

RP's instincts are correct. Even though the Wizard's strategy pages say EV, that software is designed to show correct EV only if the values for each hand are the actual payouts (and for 5-coin-in bets). So, for the latter, you do need to divide by 2. And for the former, you need to adjust out the future multipliers that you need in order to determine strategy.

The Wizard's method is in fact an estimate; however, for strategy purposes for this game, it is an excellent estimate. If you use a transition matrix to iteratively solve for both the strategy and EV, you will get some improvement over the Wizard's estimates; however, the improvement is on the order of 0.000001%. You will have a few different penalty situations; but when I calculated this several years ago, I got identical strategies for 1x, 3x, and 12x with minor changes for the others.

Because the strategies are so close here, you can IN FACT use the Wizard's estimates to get an incredibly close EV for the game. You simply have to use the pages and substitute in the values that are only the CURRENT multiplier times the Base Payout, not the CURRENT multiplier times the Base Payout PLUS the NEXT HAND multiplier minus one. And then divide by two to account for the 10-coin per hand bet instead of a 5-coin per hand bet.

With this single line game, there are only 7 states, not 7 x 10 = 70. You only need a 7 x 7 transition matrix, not a 70 x 70 transition matrix, so EXCEL can do a fine job of calculating this. You do not need special software other than the Wizard's strategy calculation plus Excel. For each of the 7 current multiplier states, you can combine the probabilities of hands that share common NEXT HAND multipliers in constructing the transition matrix.

The Wizard's page for "Ultimate X (single line)" shows an EV for Jacks 8-6 of 0.98573037. Using an exact method, I calculated an EV of 0.985730379. I don't know if the Wizard's own number is rounded or truncated, but as I originally stated, his method is an excellent estimate. In fact, for canned strategy software that requires integer entries, you will have some difficulty to get any better.

Good luck.

Quote: randomperson

I thought of a way to demonstrate the point since nobody seems to believe me. Imagine that you have a game where you just get exactly a pair of jacks every hand, nothing more nothing less. That is clearly a 100% return game. But what would EV be given your method of calculation?

Am I not correct that it would be 3 and then you divide it by 2 to make 1.5?
link to original post

Yes if you mean an Ultimate X game that always returns 2 units for a 2 unit bet, or 10 for 10, the EV would be 100% or 1.0.

And the formula current_multiplier*win+multiplier-1 for strategy generation would yield 1.5. If you remove the future multiplier-1 from the formula, the result is 1.0.

I get your point. Thanks

Quote: drrock

RP's instincts are correct. Even though the Wizard's strategy pages say EV, that software is designed to show correct EV only if the values for each hand are the actual payouts (and for 5-coin-in bets). So, for the latter, you do need to divide by 2. And for the former, you need to adjust out the future multipliers that you need in order to determine strategy.

The Wizard's method is in fact an estimate; however, for strategy purposes for this game, it is an excellent estimate. If you use a transition matrix to iteratively solve for both the strategy and EV, you will get some improvement over the Wizard's estimates; however, the improvement is on the order of 0.000001%. You will have a few different penalty situations; but when I calculated this several years ago, I got identical strategies for 1x, 3x, and 12x with minor changes for the others.

Because the strategies are so close here, you can IN FACT use the Wizard's estimates to get an incredibly close EV for the game. You simply have to use the pages and substitute in the values that are only the CURRENT multiplier times the Base Payout, not the CURRENT multiplier times the Base Payout PLUS the NEXT HAND multiplier minus one. And then divide by two to account for the 10-coin per hand bet instead of a 5-coin per hand bet.

With this single line game, there are only 7 states, not 7 x 10 = 70. You only need a 7 x 7 transition matrix, not a 70 x 70 transition matrix, so EXCEL can do a fine job of calculating this. You do not need special software other than the Wizard's strategy calculation plus Excel. For each of the 7 current multiplier states, you can combine the probabilities of hands that share common NEXT HAND multipliers in constructing the transition matrix.

The Wizard's page for "Ultimate X (single line)" shows an EV for Jacks 8-6 of 0.98573037. Using an exact method, I calculated an EV of 0.985730379. I don't know if the Wizard's own number is rounded or truncated, but as I originally stated, his method is an excellent estimate. In fact, for canned strategy software that requires integer entries, you will have some difficulty to get any better.

Good luck.
link to original post

First, thanks for the feedback. I think I understand now to use the multiplied paytable values, but to omit the future multiplier-1.

In his “Analysis of 8-6 Jacks or Better” table on the single-line Ultimate X page, the Wizard uses a 7x10 matrix. I understand u could reduce this to 7x7 if u wish to by combining probabilities that result in the same multiplier.

I understand his matrix, the only problem I have is with reproducing the probability column, which is where I started in my original post.

For example, a royal flush with a 10x (flush) multiplier has a probability of 0.00000033. I understand using the royal flush probability from the 10x page of 0.000022. But where would i get the probability of the flush? It seems to me the flush probability would have to be the weighted average from all multiplier pages. Is that right? The value would have to be close to 0.00000033/0.000022= 0.015. If I sum the weighted flush probabilities I do get 0.015. So now I think I can complete the table, if that’s correct.

But that leaves another question: how do I calculate the multiplier probabilities provided on the Wizards page? It seems those would be dependent on the probability of the individual outcomes that produce each multiplier, and if I make the probability of the individual outcomes dependent on the probability of the multipliers, I have a circular dependency.

To get the exact solutions for EV and strategy, you do have to iterate?

Any help is appreciated. Thanks

PS - I believe both the wizard strategy maker and analyzer do allow decimals

Each of the 7 pages that you have already cited from the strategy analyzer contains the transition probabilities from one state to itself and the other 6. Each page allows the construction of one of the 7 rows of the transition matrix. After that, you can use that transition matrix to determine what the ultimate steady state matrix looks like. One way of doing this is to multiply the matrix by itself over and over again until it converges. For such a Markov chain, each row of that eventual product matrix will converge to the same unique stationary distribution vector of probabilities.

It has been a while since I looked at Dr Koehler's paper and I calculated things in a little different manner than he did, but it may contain the rest of the info you need to understand. If that doesn't get you to the answers you want, you may want to search the internet for examples of the properties of Markov chains and how to work with transition matrices.

Best of luck.

Quote: drrock

Each of the 7 pages that you have already cited from the strategy analyzer contains the transition probabilities from one state to itself and the other 6. Each page allows the construction of one of the 7 rows of the transition matrix. After that, you can use that transition matrix to determine what the ultimate steady state matrix looks like. One way of doing this is to multiply the matrix by itself over and over again until it converges. For such a Markov chain, each row of that eventual product matrix will converge to the same unique stationary distribution vector of probabilities.

It has been a while since I looked at Dr Koehler's paper and I calculated things in a little different manner than he did, but it may contain the rest of the info you need to understand. If that doesn't get you to the answers you want, you may want to search the internet for examples of the properties of Markov chains and how to work with transition matrices.

Best of luck.
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Yes I wasn’t familiar with Markov until I read his paper. And transition matrices are new to me. I’m familiar with matrices from computer programming but never took matrix math. I did math up to calculus and later statistics. I will have to look into it further. Thanks!