What does Perfect Play mean for Ultimate X Games?

Perfect Play for Ultimate X games is a bit harder to explain than is Perfect Play for regular Video Poker games. The reason is that what one does with the current hand not only affects what will be the immediate return, but also what the next round of play might yield since the outcomes of the current hands establish multipliers for the next round of hands.

Suppose you are playing 10 Line Jacks or Better with the following pay table and multipliers:

Perfect Play for Ultimate X games is a bit harder to explain than is Perfect Play for regular Video Poker games. The reason is that what one does with the current hand not only affects what will be the immediate return, but also what the next round of play might yield since the outcomes of the current hands establish multipliers for the next round of hands.

**Naïve Expected Value:**The Naïve Expected value of a hand is the optimal expected outcome of all the hands times their current multipliers without regard for any future impact from the resulting new multipliers. This is Perfect Play ONLY for your last hand of play, since you are indifferent to any future impact. It is NOT Perfect Play in general. In fact, on your last hand of play you should only bet 5 units, not the max bet of 10 units since you are not interested in establishing new multipliers.**Perfect Play Expected Value:**The expected value of Perfect Play takes future returns into account. The academic paper linked below gives the math behind a complete explanation (see Equation 1). In short, one must add a factor to the Naïve Expected return of a hold hand that captures both the probabilities of going to various different multiplier sets and the optimal expected returns of possible future hands.Suppose you are playing 10 Line Jacks or Better with the following pay table and multipliers:

Now suppose you are dealt the following hand:

Suppose the current multipliers were all one, meaning the sum of the multipliers is 10. The WHY? button gives us the following information. We see that Perfect Play would have us hold the Ace, Seven and Eight of Spades. The Expected Value is 1.0471. This includes the current outcomes plus future potential outcomes. If this was our last hand of play, we see that the Naïve Expected value for holding just the Ace of Spades is higher than holding the three Spades (0.4642 versus 0.4468).

However, if the sum of current multipliers is 83, the WHY? button shows the following. Holding just the Ace of Spades is a Perfect Play (actually even if this is the last hand of play since its naïve expected value is also higher).

This example illustrates several things. Perfect Play is dependent not only on the dealt hand, but also on the sum of the current multipliers and on whether or not this is the last hand of play.

## Multiple Best Perfect Play Holds

As for typical Video Poker games such as Jacks or Better, it may be the case that multiple hold hands can lead to the same expected value of outcomes. Indeed, in the example above when the multipliers summed to 83, the expected returns of the top two hold hands are very similar (33.0163 versus 33.0082), although not identical. Instead of repeating the arguments used for Video Poker games, we just refer the reader to that material. However, breaking ties using the variance is problematic for Ultimate X because computing the variance not only depends on the exact multipliers (not just the sum of the multipliers) but is also very complex and time-consuming.

Thus, we break expected value ties using the variance or standard deviation of the naïve return. The standard deviation is just the square root of the variance and is often easier to work with because it has units in common with the expected value. The user can use the standard deviation of the naïve return as a judge of possible volatility of outcomes. Even this is complicated by the fact that the standard deviation of the naïve return depends on the actual multipliers, not on the sum of the multipliers. When we know the actual multipliers (for example, in the Game Mode), we can compute the exact standard deviation of the naïve return. Just take the square root of the sum of each multiplier squared and multiply the naïve standard deviation by this value.

When all we know is the sum of the multipliers, we compute a range of standard deviations. For example, in the 10 line Jacks or Better game above, suppose the sum of the current multipliers is 22. This might have come from multipliers (4,4,4,4,1,1,1,1,1,1) or from (4,2,2,2,2,2,2,2,2,2,2) or from (7,7,1,1,1,1,1,1,1,1), and possibly others. The base standard deviation for just one outcome would be multiplied by the square roots of 70, 52 and 106, respectively, for these three cases. Clearly, the standard deviation (the possible spread) of outcomes is greatest for (7,7,1,1,1,1,1,1,1,1), and least for (4,2,2,2,2,2,2,2,2,2,2). We show the range of these possible values in the Why? listing.

If the expected values and naïve standard deviations are tied, we break ties as in Video Poker games. See here for details.

Thus, we break expected value ties using the variance or standard deviation of the naïve return. The standard deviation is just the square root of the variance and is often easier to work with because it has units in common with the expected value. The user can use the standard deviation of the naïve return as a judge of possible volatility of outcomes. Even this is complicated by the fact that the standard deviation of the naïve return depends on the actual multipliers, not on the sum of the multipliers. When we know the actual multipliers (for example, in the Game Mode), we can compute the exact standard deviation of the naïve return. Just take the square root of the sum of each multiplier squared and multiply the naïve standard deviation by this value.

When all we know is the sum of the multipliers, we compute a range of standard deviations. For example, in the 10 line Jacks or Better game above, suppose the sum of the current multipliers is 22. This might have come from multipliers (4,4,4,4,1,1,1,1,1,1) or from (4,2,2,2,2,2,2,2,2,2,2) or from (7,7,1,1,1,1,1,1,1,1), and possibly others. The base standard deviation for just one outcome would be multiplied by the square roots of 70, 52 and 106, respectively, for these three cases. Clearly, the standard deviation (the possible spread) of outcomes is greatest for (7,7,1,1,1,1,1,1,1,1), and least for (4,2,2,2,2,2,2,2,2,2,2). We show the range of these possible values in the Why? listing.

If the expected values and naïve standard deviations are tied, we break ties as in Video Poker games. See here for details.

## Math of Ultimate X Games

Below is an academic analysis of Ultimate-X Video Poker by one of our principals, updated to correct typos and minor points. The original version is posted at: http://wizardofodds.com/videopoker/images/UltimateX.pdf.

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