Ask Mission

My answer:

As with any video poker hold, the optimal holding strategy just comes down to every possible outcome for the hand and the combined expected value related to the probabilities of each of those outcomes. In the case of the Double Double Bonus hold, which was A-A-K-K-7, holding the two pair yields only two possible results:

Two Pair: 43/47

Full House: 4/47 (Pays 8 FOR 1)

(43/47 * 1) + (4/47 * 8 ) = 1.59574468085

Therefore, the expected return of that result was 1.596x (rounded) the amount of credits bet.

The best paying hands possible holding just the Aces (which itself guarantees a minimum of 1x the bet) are four aces (160-FOR-1) and four aces with a kicker (400-FOR-1 with a kicker). Let’s determine the probability of getting just the four aces:

We draw three cards and we want two of them to be the other two aces. Let’s pretend that they are the ace of hearts and the ace of spades, here are the possible ways that could happen:

nCr(2,2)*nCr(45,1)/nCr(47,3) = 0.0027752081406105

Okay, so here is what that problem expresses in words:

(Two Aces, We Want Both)*(45 Non Aces, We Want One)/(47 Cards, Drawing Three)

Now that we know the probability of four aces is 0.0027752081406105 based on our draw, we now have to figure out the probability of getting four aces with a kicker. This is extremely easy because the probability of having a kicker card (2’s, 3’s, 4’s) is simply 12/45 assuming we already have the aces, so we do this:

0.0027752081406105 * 12/45 = 0.0007400555

The probability of the aces with no kicker is therefore:

0.0027752081406105 * 33/45 = 0.00203515263

Okay, so we multiply these results by the pays of 160 (no kicker) and 400 (with kicker) and add them together to get:

(0.0007400555*400) + (0.00203515263*160) = 0.6216466208

If we add that to the guaranteed 1-FOR-1 we get just from holding the aces, then we get a total of 1.6216466208, which is already better than the 1.59574468085 from holding the two pair, and we didn’t even need to include the possibility of getting a three-of-a-kind or still getting a full house even though we discarded a pair!

In other words, holding the pair of aces is a mathematical no-brainer.

You can also go the combinatorics route to determine the probabilities of the aces both with and without the kicker if you really want to have fun.

nCr(2,2)*nCr(33,0)*nCr(12,1)/nCr(47,3) = 0.0007400555041628 (With Kicker)

nCr(2,2)*nCr(33,1)*nCr(12,0)/nCr(47,3) = 0.0020351526364477 (Without Kicker)

How about we get a three of a kind with aces and one of the kicker cards? There are no full houses in this because the non-ace cards are automatically different.

nCr(2,1)*nCr(33,1)*nCr(12,1)/nCr(47,3) = 0.0488436632747456

How about a three of a kind (and some full houses) with no kicker cards?

nCr(2,1)*nCr(33,2)*nCr(12,0)/nCr(47,3) = 0.0651248843663275

We can also do three aces with two kicker cards that would include some full houses:

nCr(2,1)*nCr(33,0)*nCr(12,2)/nCr(47,3) = 0.0081406105457909

How many full houses? That’s easy. All you have to do is figure out the probability of getting three of a kind aces with two particular kicker cards and multiply that result by three:

(nCr(2,1)*nCr(41,0)*nCr(4,2)/nCr(47,3)) * 3 = 0.0022201665124884

The reason we multiply by three at the end is because there are three particular ranks (2’s, 3’s and 4’s) and the above formula only solves for one specific rank.

Speaking of full houses, what is the probability of getting a full house without getting any more aces? The answer to this is pretty simple and easy to discover if you know how. Here are the things about the remaining deck that are relevant:

A.) There are two aces left.

B.) There are two kings left, so no full house is possible that has three kings.

C.) There are three sevens left.

D.) There are four of every other rank left, and there are ten ranks that are not kings, aces or sevens.

((nCr(4,3)*nCr(43,0)/nCr(47,3)) * 10 = 0.0024668516805427

This can also be expressed this way:

(4/47 * 3/46 * 2/45) * 10 = 0.00246685168

Okay, so the only thing left to figure out is the probability of a full house with two aces and three sevens, which is easy:

((nCr(3,3)*nCr(44,0)/nCr(47,3)) = 0.0000616712920136

Or, if you prefer:

(3/47 * 2/46 * 1/45) = 0.00006167129

Okay, so we add the two together and the overall probability of getting a full house without getting a third ace is 0.00252852297 or about 1 in 395.49.

Anyway, I just included those silly examples as a way to do combinatorics if you might be interested in that. I think that combinatorics are fun because it offers an easy opportunity to come up with all sorts of highly ridiculous specific things in an easy way. For instance, let’s say that I wanted a seven, a nine and a ten for some bizarre reason:

nCr(3,1)*nCr(4,1)*nCr(4,1)*nCr(36,0)/nCr(47,3) = 0.0029602220166512

Anyway, when it comes to any card drawing probability, it’s an easy way to figure out the probability of any specific result as long as you ask the question correctly. Also, make sure the front numbers on the left side of the equation (before the division symbol) add up to the front number on the right side, and same with the second numbers in each parenthesis, or you will end up with a nonsensical answer.

My Answer:

First of all, dear readers, Ultimate X is a game in which a player plays video poker and winning hands are awarded not only the pay on the hand, but a multiplier that applies to any wins that may be gained on the following hand. Playing ONLY when there are already multipliers is known as, “Vulturing,” Ultimate X.

To answer the question, the first thing that we have to understand is how Ultimate X works:

A max bet in Ultimate X is a ten credit per hand bet, but the game only pays as if the bet were five credits per hand, with multipliers being applied to future hands.

For one example, let’s look at 8/5 Jacks or Better with three-play on Ultimate X that has a return of 0.978730:

Normally, if you had to bet ten credits on an 8/5 paytable that paid as if the bet were five credits, the game would have a return to player of 0.486492, or 48.6492%. What that means is that 49.2238% of the Ultimate X machine return comes from the multipliers and strategy changes that are made accordingly.

Imagine that someone is going to get a 2x multiplier on the following hand. There are two ways to look at the return, which are percentage basis and actual expected monetary profit basis. Most who vulture Ultimate X (myself included) would only care about the latter. However, imagine that the multiplier is on the bottom of the three hands.

With a 250-FOR-1 Royal, 8/5 Jacks would return 96.0635%, or .960635 credits without a multiplier. If we double that amount, then the return is 192.1270% or 1.921270. Since we would be betting one credit, which we will make dollars to keep it simple, that is also the expected return in dollars.

With our five-credit per hand bet, we are getting the full 800-FOR-1 (per credit) value on the Royal Flush, which means that the game has the full 97.2984% or .972984 return per the normal. What this means is that this is happening:

((.972984 + .972984) + (.972984 * 2))/3 = 1.297312 or 129.7312% total return on the total play.

In the case of the five credits per hand bet, we would be betting a total of $15, so our expected cash return is 15 * 1.297312 = $19.45968, which represents an expected profit of $4.45968 as opposed to the expected cash profit of $0.921270 if we just played the one credit on the first line. In other words, our PERCENTAGE expected profit is lower, but our cash expected profit is greater.

Using our Ultimate X example of 8/5 Jacks or Better, this is what happens when we bet ten credits per hand:

((.486492 + .486492) + (.486492 *2) + (.492238 * 3))/3 = 1.140894 or 114.0894%

When it comes to the expected return of the $30 total bet, here is what happens: 30 * 1.140894 = $34.22682.

So, here are our expected cash profits and percentage profits for each proposition assuming the 2x multiplier is on the bottom hand only:

One Line, One Credit: $0.921270---192.1270%

Three Lines, Five Credits: $4.45968---129.7312%

Three Lines, Ten Credits: $4.22682---114.0894%

What betting max does in this scenario is it dilutes the profitability of playing off the multipliers because we are making the extra five credits bet per hand for future multipliers, but those future multipliers are not only NOT adding value, the potential value of getting additional multipliers is less than 100%. In other words, that part of the bet is actually at a NEGATIVE expectation. That’s the reason why the percentage advantage is trimmed by a little more than half in this case.

CONCLUSION: That doesn’t mean there’s anything fundamentally wrong with betting max credits when faced with a situation in which you start with multipliers. Maybe you want to play until you run out of multipliers, which is fine, since any hand you play WITH multipliers puts you at some sort of advantage.*** However, the best expected monetary return on the hand and best percentage return of the play (of the two) is to bet five credits per hand.

***(Traditional Ultimate X and Ultimate X Spin Poker only, not Ultimate X Bonus Streak)

Answer: You definitely guess wrong when it comes to the overall return of the game. The 4OaK is the only area that hurts, otherwise that would be a decent Joker Poker (Kings) paytable. 96.3848% return using this.

As it stands above.

Now, I don't know how to get you the exact answer that you want, but what I do have some idea how to do is get close by making an extrapolation. According to the calculator for the Kings or Better game, the probability of getting Kings or Better is 0.140901, which is also the return that comes from that result. I would say that we could take this to mean, without making any other strategy changes than you almost always hold a pair of jacks or queens, that you double the frequency of Kings or Better and that is your approximate frequency of Jacks or Better. That would increase the return of the game by about 14% to over 110%.

Again, there would be some strategy changes (a few obvious) and a few other assumptions we could make. Straight Flushes and Flushes now become less likely because we are holding JJ or QQ over a four-flush or various straight flush draws. Royals become less likely (and wild royals) because we now hold JJ or QQ over certain draws of that type. 3OaK, 4OaK and 5OaK all become more likely because we are (naturally) holding more pairs.

If it sounds too good to be true...I'm not going to ask where this is at (PM me if you like) but are you sure it is a Class III machine and not a Class II?

ADDED: It turns out that this was on a variant where the max bet that had to be made reduced the overall return of the game. There was no choice but to max bet.

Variance: Playing triple-play would have less variance than playing single-handed at $3.75/hand. The same number of deals would have more variance than playing single-handed at $1.25/hand.

Triple Play does not affect the house edge. You have less variance on the draws, so I believe you actually approach the, "Long-term," on fewer dealt hands.

ADDED: I later clarified this answer to mean that playing triple-play at $3.75 per round would have less variance than single-hand at $3.75 per round, if such a thing were possible. In the case of the poster, he is referring to playing Triple-Play quarters as opposed to single-play quarters, but the latter would have less cash variance (in raw dollar terms) because the player is betting less. The house edge is unaffected, but the more money that is bet in any case, the greater the expected loss.

A more accurate example might be playing five-play nickels ($1.25 total bet) vs. single-line quarters (also $1.25 bet) the nickels game would have less variance because there is a wider range of possible results close to the expected value of the bet. One key here, of course, would be ensuring that the paytable (and therefore the house edge) is the same for both games.

First of all, using 2% of Max Value as a general rule is not a good idea. The reason why is because some games actually have a very short range, (Must Hit by $500's that start at $450, for example) which means that the base jackpot accounts for a greater percentage.

Anyway, if we look at $4,900 on this machine, here is what we get:

Meter Moves Until Guaranteed: 10,000

Coin-In Until Guaranteed: 50,000

Expected Coin-In: 25,000

Expected Hit Point: $4,950.00

Another thing when you look at the Minor is that, if the Minor hasn't already hit by then, it will be at $444.29

Meter Moves Until Guaranteed: 5571

Coin-In Until Guaranteed: 27885

Expected Coin-In: 13942.5

Expected Hit Point: 472.15

The good news there is that, if you have to push the major all the way, you'll get an extra $444.29-$500.00 in EV for your trouble. That definitely makes it a more attractive play if the Minor can be a lock when the major needs pushed all the way.

With an expected coin-in for the Major (starting at $4900) of $25,000 and an expected hit point of $4,950, that means that you will need to run at:

20050/25000 = 80.2% On the Reels and with the Minor to break even if it hits EXACTLY at the expected point.

If you push it all the way:

45000/50000 = 90% On the Reels and with the Minor if you push it all the way.

I can't make any absolute guarantees on the return, but one thing I can say is that I SERIOUSLY DOUBT if the Reels + Minor pays 90%+ from base, so just know that it is possible to lose a ton on these. You should also understand that this is one of the worst games there is from a Variance standpoint, (NOTE: The machine in question is a high-Variance title that I am familiar with) so it is nothing to run below 80% for an extended period of time.

In fact, I had a must hit by $500 (Except $500 was the Major on that one) that I started playing at $493.70, hit at $498.55, and still lost $220 overall, betting fifty cents per spin. I hit the Minor once, too, so my overall return was:

(498.55-493.70) = $4.85----485*5 = $2,425 coin-in.

(2425-220-498.55)/2425 = 70.37%

As you can see, I got spanked on the reels + Minor running 70.37%, so that could theoretically happen to you, too, only you're playing a lot more coin-in.

Let's take a look at your machine assuming an 85% return on the Reels + Minor and starting at $4900: We'll assume it hits at $4950, which is $25,000 coin-in. Just for the sake of argument, we'll say you don't hit the Minor in that time.

(25000*.85) + 4950 = $26,200

Okay, so that would be an expected profit of $1,200 if you could assume that kind of return on the reels. That would represent a 4.8% gain on coin-in and would be your assumed advantage.

Now, let's say you have to push it to 4990.00, which would be $45,000 coin-in, at the same 85% return on the Reels, again, ignoring the Minor:

(45000*.85) + 4990 = $43,240

That would have you losing $1,760.

SHORT ANSWERS:

1.) Yes, I think that $4,900 is a good number because I assume that the reels return more than 80.2%, though almost definitely less than 90%.

2.) Just remember that you can be absolutely punished if you don't run well on the reels. In effect, it's quite possible that $5,000 wouldn't be enough and you'll run out of money before the thing hits. In fact, I would call that scenario quite possible just based on the fact that you would have to run at 80.2% to have it hit at the expected point. These machines can run much worse than that over an extended period of time. They are High-Variance, relying almost exclusively on hitting the special symbol in the middle of reels 2 & 4 at the same time, as well as doing the same thing during bonus games.

3.) I can't emphasize this enough: If you sit down with five grand, then don't be surprised if you lose all of it and have to go get more money to continue. I don't know what a, "Safe," amount to sit down with is, to the extent that it would get you through it even if you run abysmally, but probably more like $15,000.

Or, better yet, hope some people play it up without hitting giving you a bigger advantage. Although, there's a not unreasonable chance it hits before it even makes it to $4900.

Follow-Up Question: bring $15k? uggg...

another question: How much on the meter will it be guaranteed +EV?

within 1%? (ie: $4950?)

NOTE: The minimum allowed return in that state is 85%. I was asked not to reveal the state or casino.

Answer: The minimum return is good information, but keep in mind that return is also going to include the Base Progressives based upon the average hit point. Do you know what the Major starts at? I'm assuming the Minor is $250 to start.

If the Major starts at 4,000, then it is expected to hit at $4,500. The expected amount of coin in it would take to hit would be $250,000, which would mean the Major contributes 4500/250000 = 1.8% to the return of the game, on average. If the Major starts higher, then it contributes more to the game, on average, because the average hit point will be higher.

I think 85% on the reels is a fairly good assumption.

4950 would be the same way of figuring it out. You would have 2,500 meter moves, so $12,500 coin in.

(12500 * .85) = $10,625 --- Loss of $1,875 on the reels, gain of $4,975 on the Major, total +$3100.

(25000 * .85) = $21250--- Loss of $3,750 on the reels, gain of $5,000 on the Major, +$1,250.

Even at that number, you could still get punished. Let's say you ran at 70% on the reels:

(12500 * .7) = $8,750 --- Loss of $3,750 on the reels, gain of $4,975 on the Major, total +$1,225

(25000 * .7) = $17,500 --- Loss of $7,500 on the reels, gain of $5,000 on the Major, total -$2,500.

What I would say is, while unlikely, even starting at $4950.00, it's reasonably possible to find yourself down $5,000+ at a given point. Definitely more than 1% likely. The great irony of must-hits is that, even though they must do just that, there are no guarantees.

The lesson here is that a person, even while on a positive play, must take bankroll into consideration. A gambler must also make a practical judgment of, “Is the potential downside worth it?”, even when it comes to playing at a positive expected value.

The ironic thing about must-hits is that, given how poorly they often return on the reels, an individual is expected to find himself down before the thing hits. Further, the longer the play goes on, the more they can find themselves down.

Another thing to realize is that the expected hit point (assuming halfway) is also variable. The longer it goes without hitting, the greater the number that it is expected to hit at becomes. If you start playing one of these machines at $4,900.00, for example, then it is expected to hit at $4,950.00. If you have been playing it, though, and now the meter is at $4,920.00, it is no longer expected to hit at $4,950.00; it is expected to hit at $4,960.00.

When deciding whether or not to play one of these at a small (or even large) advantage, it is absolutely essential to realize that your EV comes from it hitting at or below the expected number. You can even assume a return percentage on the reels and determine what your expected break-even point is. Let’s assume that the machine returns 85% on the reels and that you start playing at $4,920.00 with a $5/penny meter move:

The formula to determine the exact point is a little more complex, but it is easy to get a general idea with trial and error, and the best part is, anybody with an eighth grade education in math can do it. For instance, I might ask, “Am I still expected to win if the meter goes to $4,970.00?”

First, determine how much coin-in that would be:

(497000 - 492000) * 5 = 25000

Now, just take that $25,000 coin-in, assume 85%, and then add the hit amount back into it:

(25000 * .85) + 4970 = 26220

That means that you would expect a return of $26,220 if the reels ran based on your assumed 85% and you hit it at $4,970. The return is on $25,000 coin-in, so your expected win (relative to coin-in) is 26220/25000 = 1.0488, or an expected gain of 4.88% on your total coin-in.

Let’s look at if it ran up to 4980 before hitting:

(498000-492000) * 5 = 30000

(30000 * .85) + 4980 = 30,480

30480/30000 = 1.016 or an expected win of 1.6% of your total coin-in.

Now, you could look at what happens if you had to push it all the way:

(500000 - 492000) * 5 = 40000

(40000 * .85) + 5000 = 39000

Therefore, if you had to push it all the way, you would expect to lose $5,000 overall and be down $6,000 right before it hits.

Anyway, you can determine different hit points of a must-hit, assume whatever return percentage you want to, and that will give you an idea of how much you are expected to win (or possibly stand to lose) before it hits. I would say some 90% of the time, whether you profit or not largely depends on whether it hits at, or sooner, than expected. Of course, being pretty conservative in what you’re willing to play also helps.

Just make sure you have the bankroll before you start!!! If you run out of money without hitting the thing, that is the worst thing that could possibly happen. Not only are you out of money, but the only thing your money accomplished is making a better play for someone else.

Be cautious. Make sure you have the bankroll and also make sure to always understand that this is one AP play where your maximum downside will usually be a greater dollar amount than your maximum upside.

I would say that you are incorrect, with all due respect. There are a few reasons:

1.) If you can still qualify for earned multipliers by betting five coins, then ANY multiplier is good up to at least ten-play.

2.) If you have to bet eight coins, then the bottom hand can get a multiplier and every hand will have a multiplier thereafter at least once. Meantime, bottom hand could still get one.

Okay, so assuming the worst, Ten-Play with the multiplier feature might return 95% overall. (It's not that bad) Better yet, let's use an actual example from an actual paytable:

The worst Bonus Poker has a base return of .9578 and a return of .9598 with the feature. With a five credit bet, the player expects, from base, to get:

.9578 * 5 = 4.789

If you extrapolate that to ten hands, the player bets fifty credits and expects to get 47.89 back.

Now, with the feature, the player would bet eighty credits and expects to get, without a multiplier to begin with:

80 * .9598 = 76.784

What we determine is that any addition to the base pay (in the form of a multiplier) is worth 4.789 per multiplier value because it is based upon getting paid on five coins for that hand based on the base expectation of 95.78%.

If you add 4.789 to 76.784 you would get 81.573, so 81.573/80 = 101.96625%

The worst case scenario would be one 2x multiplier for the tenth hand on ten-play, but much like regular and spin Ultimate X (which is a ten-credit bet) even if you have to bet all ten credits per hand, (usually not the case) you still have an advantage.

Lesson: When it comes to vulturing, ask your friendly, neighborhood, Mission146!