# Video Poker - Probability

Tony from Chicago, Illinois, USA

According to my page on sequential royal video poker, the odds are just about one in four million.

Jim K. from Laguna Niguel, California

The number of ways to draw 3 cards out of the 47 remaining in the deck is combin(47,3)=16,215. One of those will be the three needed for a royal so the odds are 1 in 16,215.

Peter from Ottawa, Canada

Your expected return are the same on a triple play machine as a single hand machine, assuming the same pay table.

Gordon Maska from Lewiston, New York

Assuming you played conventional 8/5 strategy the return in your example would be 99.68%. However, if you played optimal strategy for this jackpot the return would be 100.08%. So, Wong was not wrong.

George W. Bordner from Clarence, USA

It is my understanding that the remaining 47 cards are continuously being shuffled until the player makes a decision what cards to draw. So, the draw cards are not predestined at all. Mathematically speaking, it doesn't make any difference.

St Flush-250

4 Aces-750

4/2,3,4-450

FourKind-250

Fullhouse-40

Flush-25

Straight-20

3Kind-10

2Pair-10

JacksBetter-5

"Anonymous" .

The return is 99.9367%.

Jeff from Granger, Indiana

You're welcome, thanks for the kind words!

The general formula is combin(X,Z) × p^{Z} × (1-p)^{X-Z}, where p = 1/combin(47,5-Y).

Combin is an Excel formula, which equals X!/[Z! × (X-Z)!].

Let's look at an example of 10-play video poker where the player holds four to a royal.

### 10-Play with Four to a Royal

Royals | Probability |
---|---|

10 | 0.0000000 |

9 | 0.0000000 |

8 | 0.0000000 |

7 | 0.0000000 |

6 | 0.0000000 |

5 | 0.0000010 |

4 | 0.0000378 |

3 | 0.0009943 |

2 | 0.0171513 |

1 | 0.1753242 |

0 | 0.8064914 |

Total | 1.0000000 |

Secondly, I am wondering which, if any, online casinos currently advise the player of a shuffle in blackjack (multi-deck, of course). Also, do you know, among the majority who do not, which shuffle after each hand and which just do not advise of a shuffle (although it actually occurs after many hands)? It would be great to have this knowledge. A follow up question would be, if they do indeed shuffle at regular casino intervals, can a player assume that if he enters a private table that he beings with a full shoe? Thanks again for your great web site, and I look forward to your response to my questions.

Tony from Columbus, Ohio

Thanks for the kind words. Yes, the probability of a three of a kind depends on the pay table, which affects player strategy. My video poker program always makes the optimal play for every hand by looping through all the possible cards on the draw. However, creating a strategy in writing is very time consuming.

David B. from El Cajon, California

The odds of being **dealt** a natural royal flush are 1 in 649,740 in any 52-card video poker game.

Dave from Mulvane, USA

I hope you're happy, I spent all day on this question. Please visit my new video poker appendix 1 for the answer. There is no easy way to get a risk of ruin figure with just the variance. It depends on exactly what the returns are for each hand and their probability.

Derek G from Vegas, baby!

No, it does not mean that. Contrary to popular myth, there is no cycle. Every hand is independent. It would take an infinite number of hands, played perfectly, to guarantee reaching the theoretical 99.54% return.

Here are some figures for you. Royals contribute 1.98% to the return in 9-6 jacks or better. That means you can expect the game to return 97.56% between royals. The standard deviation of one hand is 4.42. The standard deviation of the return of 40,391 hands, the average number between royals, is 2.20%. So, even after a complete royal cycle you can still be a long ways from a 99.54% return. Thee is a 95% chance you'll be somewhere in the range of 95.24% and 103.85%.

About the pop-ups, I hate them too. However, something has to put rice on the table. Consider them the cost of the information you're getting.

Meudon from Moisan, France

There are combin(52,5)=2598960 possible combinations of the first five cards. You don’t have to analyze all of them. Personally I break them down into 191659 different kinds and weight each one with the number of similar hands. For example the odds are the same with four aces and a king singleton regardless of the suit of the king. You don’t have to analyze four hands for each possible suit of the king, just one of them and multiply by four. Once you have a hand there are 2^{5}=32 ways to play the hand. I analyze each way and take the play with the greatest expected value. To determine the expected value of a play you have to analyze all the ways the replacement cards can fall and score each hand. In the case of throwing all five cards away there are combin(47,5)= 1533939 possible replacement hands. The total number of hands that must be analyzed to determine the best play of a specific hand is combin(47,5)+5*combin(47,4)+10*combin(47,3)+10*combin(47,2)+5*47+1, which coincidentally also equals 2598960. So if we took no short cuts at all we would have to analyze 2598960^{2}= 6,754,593,081,600 hands. Just reducing the initial hands to 191659 we still have 498,114,074,640 hands to analyze. Clearly more short cuts are in order. It would take a desktop computer several hours at least to work through this many hands. Personally I don’t actually score any hand but use carefully chosen formulas to determine the probability of improving a hand. For example with any pair and 3 singletons the probability of improving the hand to a two pair is always the same. Things get more complicated with straights and flushes but still manageable. My program can calculate the expected return for a game of jacks or better in about one minute. Considering it used to take me over a day I’m rather proud of it. I hope this answers your question.

Gary from Milwaukee, USA

Let’s assume you hold the ace of spades and toss away four non-ace singletons. There are 44 ways you can get a four of a kind in aces. The 44 is the number of possible singletons you could get on the draw along with the other three aces (52 cards less 4 aces and the 4 singletons you discarded). You might also get a four of a kind in one of the other 8 ranks besides aces and the four you discarded. So the total ways to get a four of a kind on the deal is 44+8=52. The total number of combinations on the deal is combin(47,4)=178365. So the probability of a four of a kind is 52/178365 = 1 in 3430.

1. one card

2. two cards

3. three card

4. four cards

5. Dealt a royal flush

I am asking the question because I recently hit a royal flush after holding 2 cards the ace and jack of diamonds and then drew the ten, queen and king of diamonds. I know the odds of drawing 3 cards to the royal must be very high. Then last week , I was sitting next to a man who held the Ace of diamonds and drew 4 cards to complete his royal. I was amazed. Thanks for your answer.

Paul

1. 1/47

2. 1/combin(47,2) = 1/1081

3. 1/combin(47,3) = 1/16215

4. 1/combin(47,4) = 1/178365

5. 4/combin(52,5) = 1/2598960

Bradford from Houston, USA

There are combin(47,4) = 178365 ways to choose 4 cards out of the 47 remaining. Only one way will result in the three cards you need. So the probability is 1 in 178365.

^{3}= 0.0625, correct? But how do you determine the odds of hitting exactly 1, 2 or all 3 Royals?

John from Milwaukee, USA

The probability of hitting x royals in an n-play machine when drawing to a 4-card royal is combin(n,x) * (1/47)^{x} * (46/47)^{n-x}. For an explanation of the combin(n,x) function visit my section on probabilities in poker. In the case of 3-play the probabilities are as follows:

0 royals: 0.937519

1 royal: 0.061143

2 royals: 0.001329

3 royals: 0.000010

"Anonymous" .

To answer your first question, there are 2598960 ways to choose 5 cards out of 52 for the initial hand. On the draw there are 1, 47, 1081, 16215, 178365, or 1533939 ways to draw the replacement cards, depending on how many card the player holds. The least common denominator for these numbers is 7669695. The actual combinations are weighted to get a total of 7669695. So the total number of combinations is 2,596,960*7,669,695=19,933,230,517,200. To answer your second question video poker machines simply pick numbers at random from 1 to 52 and assign them to a card. The random number generators themselves are very complicated but the object is simple.

D.S.M.

Nevada Gaming Control Board regulation 14.040.1(a) states that gaming devices must return at least 75% assuming optimal player strategy. To answer your second question I modified my video poker program to always make the worst possible play. For example, keeping all five cards on a non-paying hand , and tossing part or all of pat hands. Based on 9/6 Jacks or Better this strategy results in a return of 2.72%, or house edge of 97.28%. Following is the complete return table. Such a player would not be able to sue the casino because it was his fault for playing so badly.

### Jacks or Better - Worst Possible Player

Hand | Payoff | Number | Probability | Return |
---|---|---|---|---|

Royal flush | 800 | 48564 | 0.000000 | 0.000002 |

Straight flush | 50 | 2058000 | 0.000000 | 0.000005 |

4 of a kind | 25 | 38040380 | 0.000002 | 0.000048 |

Full house | 9 | 292922028 | 0.000015 | 0.000132 |

Flush | 6 | 336550092 | 0.000017 | 0.000101 |

Straight | 4 | 6239759724 | 0.000313 | 0.001252 |

3 of a kind | 3 | 12510891616 | 0.000628 | 0.001883 |

Two pair | 2 | 34968642984 | 0.001754 | 0.003509 |

Jacks or better | 1 | 334574728656 | 0.016785 | 0.016785 |

Nothing | 0 | 19544266875156 | 0.980487 | 0.000000 |

Total | 19933230517200 | 1.000000 | 0.023717 |

Ron

The probability is 1/combin(47,2) = 1 in 1081. In every game I have studied a high pair is a stronger hand than 3 to a royal, except in the game Chase the Royal.

"Anonymous" .

If you hold three deuces there are 46 ways you can get the other deuce and another card. There are combin(47,2)=1081 to choose two cards out of 47 left in the deck. So the probability of getting four deuces on the draw with three held is 46/1081 = 4.26% = 1 in 23.5. If you hold two deuces there are 45 ways to get two more deuces plus another card. There are combin(47,3)=16215 ways to choose 3 cards out of 47. So the probability of getting four deuces on the draw after holding two is 45/16215 = 0.28% = 1 in 360.33.

"Anonymous" .

If your strategy were to maximize the number of royals at all costs then you would hit a royal once every 23081 hands. I assumed that given two plays of equal royal probability the player will choose the play which maximizes the return on the other hands. The house edge of this strategy on a 9/6 jacks or better game is 51.98%. Below is a table showing the probability and return of each hand.

### Royal Seeker Return Table

Hand | Payoff | Probability | Return |

Royal Flush | 800 | 0.000043 | 0.034661 |

Straight Flush | 50 | 0.000029 | 0.001472 |

4 Of A Kind | 25 | 0.000222 | 0.005561 |

Full House | 9 | 0.001363 | 0.012268 |

Flush | 6 | 0.00428 | 0.025681 |

Straight | 4 | 0.004548 | 0.018191 |

3 Of A Kind | 3 | 0.020353 | 0.061058 |

Two Pair | 2 | 0.046374 | 0.092749 |

Jacks Or Better | 1 | 0.228543 | 0.228543 |

Nothing | 0 | 0.694243 | 0 |

Total | 0 | 1 | 0.480184 |

"Anonymous" .

Thanks! Yes, I said before that if I had a betting system that had just a 1% advantage I could turn $1000 into $1,000,000 by simply grinding out that edge. This would also be possible in video poker but it would take much longer because the 0.77% advantage game (full pay deuces wild) can only be found in the quarter level. Assuming you can play 1000 hands per hour (a speed few can attain) and played perfectly that would result in an average income of $9.63 per hour. To reach $1,000,000 would require working 11.86 years non-stop. $1000 would also be very undercapitalized to play quarter video poker, so the risk of ruin would be quite high. It would be faster to reach the $1,000,000 with the same edge in a table game because the player can bet more.

"Anonymous" .

There is a simple formula for this answer. It the initial investment divided by the house edge. In this case the answer is $100/0.02 = $5000. However due to the volatility of video poker, most of the time the $100 won’t last this long.

"Anonymous" .

The probability of any three of a kind or full house, based on "9/6" jacks or better is 0.085961. To make things easy I’ll divide by 13 to get the probability that the rank of the three of a kind is threes. This is obviously overstating the probability because you will see more in jacks through aces because correct strategy is to hold those cards more often. 0.085961/13 = 0.006612. Tripling the wins for 9 games is like getting 18 free games. 18* 0.006612= 0.119023. To this I would apply some kind of fudge factor to account for the disproportionately fewer three of a kinds in threes, perhaps 75%. 0.119023*0.75 = 0.089267. So whatever your normal return is multiply it by 1.089.

"Anonymous" .

We can see from my deuces wild section that the probability of four deuces in any one hand is 0.000204. So the probability of not getting four deuces in any one hand is 1-0.000204 = 0.999796. The probability of going 14000 hands without four deuces is 0.999796^{14000} = 5.75%.

"Anonymous" .

It is not that unusual. Sometimes Vegas casinos have a promotion in which the second royal hit in a 24-hour period pays double. Let’s assume you play for 8 hour at a speed of 400 hands per hour, or 3200 hands total. The probability that one hand is a royal flush is 0.00002476. The probability of getting zero royals in 3200 hands is (1-0.00002476)^{3200} = 0.923825. The probability of getting one royal is 3200*0.923825*(1-0.923825)^{3199} = 0.073198. So the probability of getting two or more is 1- 0.923825 - 0.073198 = 0.002977, or about 1 in 336.

"Anonymous" .

This is a good question for the Poisson distribution. If an event is equally likely any given moment and independent of other events, and the mean number you can expect is m, then the probability of n events is e^{-m}*m^{n}/n!. So in this situation the probability is e^{-17.76}*17.76^{3}/3! = 0.00001808, or 1 in 55321.

"Anonymous" .

Here is the probability of winning zero per game according to the number of plays.

### Probability of Winning Zero in n-play Video Poker

Plays |
Probability |

3 |
0.26260274 |

5 |
0.1301204 |

10 |
0.02591377 |

15 |
0.00649444 |

25 |
0.0007854 |

50 |
0.00002178 |

75 |
0.00000076 |

100 |
0 |

The table is based on a random simulation. I know it is theoretically possible to get a win of zero in 100-play, but in 15,820,000 games it just never happened. So please don’t write about that. The table shows the probability of getting zero in 10-play is 0.025914, or 2.59%. The probability of this happening ten times in a row is 0.025914^{10} = 1 in 7,323,073,295,177,980.

I tried the software in question in free-play mode and my results seemed fine. In particular in 10 games I won something every time. However as far as I know no casino offers this software and takes real money players from the U.S. I’ll plan to do some further investigating but don’t want to explain how in this forum.

"Anonymous" .

A strategy of going for a royal at all costs, as if all the other hands paid zero, would result in a return of 47.85% on a 9/6 Jacks or Better game. The expected frequency of a royal would increase from once every 40388 hands to once every 23081.

TS from Santa Barbara

Almost. If more than one royal per deal in 5-play counts as only one sighting then you will have sightings slightly less than 5 times as often. This is because the total number of royals will be five times as much, but sometimes they will be clumped together in the same play, usually when you get a royal on the deal, and thus 5 on the draw.

The following table shows the probability of making a royal in 1-play according to the number of cards to the royal held, assuming full pay optimal strategy.

### Royal Flush Probability in 1-Play Video Poker

Card Held | Probability on deal | Probability on draw | Total probability |

0 | 0.19066396 | 0.0000014 | 0.00000027 |

1 | 0 | 0.00000561 | 0 |

2 | 0.01969711 | 0.00006167 | 0.00000121 |

3 | 0.01299751 | 0.00092507 | 0.00001202 |

4 | 0.0003309 | 0.0212766 | 0.00000704 |

5 | 0.00000154 | 1 | 0.00000154 |

Total | 0.22369101 | 0 | 0.00002208 |

What this table shows is that 22.37% of the time you will have a possible royal draw. The rest of the time a royal will be impossible, for such reasons as you held a wild card or a pair. The lower right cell shows the overall royal probability is 0.00002208, or 1 in 45282.

The next table shows the same thing but for 5-play, and the probability of at least one royal.

### Royal Flush Probability in 5-Play Video Poker

Card Held | Probability on deal | Probability on draw | Total probability |

0 | 0.19066396 | 0.00000698 | 0.00000133 |

1 | 0 | 0.00002803 | 0 |

2 | 0.01969711 | 0.00030832 | 0.00000607 |

3 | 0.01299751 | 0.0046168 | 0.00006001 |

4 | 0.0003309 | 0.10195134 | 0.00003374 |

5 | 0.00000154 | 1 | 0.00000154 |

Total | 0.22369101 | 0 | 0.00010268 |

Note the probability of at least one royal is 0.00010268. This is 4.65 as high as the probability for one-play. The reason is the probability of making at least one royal is always less than five times that of 1-play. For example the probaiblity of hitting a royal holding for to the royal is 1/47 in 1-play. However in 5-play the probability of making at least one royal is 1-(1-(1/47))^{5} = 0.101951341, which is about 4.79 times as high.

Gerald from Coal Valley, IL

In games like Bonus Poker and Double Bonus I assume they pay more for certain four of a kinds to give the player a better chance at a big win, at the cost of smaller small wins of course. It makes sense to have four aces as the premium four of a kind, because aces are the highest card in regular poker. The reason I think that four twos pays more than four kings is because players don’t hold low cards as often, and thus four twos comes up less often than four kings. So although the probability of each card is the same, player behavior causes less of the low four of a kinds, thus it makes it easier for the game maker to pay more for the low four of a kinds.

Jon from Lafayette, CO

Let’s examine the general case first.

Define p as the probability that the next four of a kind will be one that you need for the promotion.

Define q as 1 - p.

Define m as the expected number of four of a kinds to get one that you need.

The sum of probabilities is 1. Thus,

(1) p + p×q^{1} + p×q^{2} + p×q^{3} + p×q^{4} + ... = 1

The following is the formula for m in terms of p and q.

(2) m = 1×p + 2×q×p^{1} + 3×q^{2}×p + 4×q^{3}×p + 5×q^{4}×p + ...

Multiply both sides of (2) by q.

(3) mq = 1×pq + 2×p×q^{2} + 3×p×q^{3} + 4×p×q^{4} + 5×p×q^{5}

Subtract (3) from (2)

(4) m - mq = p + pq + pq^{2} + pq^{3} + pq^{4} + ...

The right side of (4) equals 1 from (1).

(5) m - mq = 1

(6) m×(1-q) = 1

(7) m = 1/(1-q) = 1/p.

So, if the probability of an event is p, then on average it will take 1/p trials to occur.

To get back to the problem at hand, it will obviously only take one four of a kind to cross the first one off the list. The probability the next four of a kind will be one that you need is 12/13. So, on average, it will take 13/12=1.0833 trials to get it. Once you have two crossed off the list, the probability the next one will be one that you need is 11/13, so that will take 13/11=1.1818 more trials to get the third one.

Following this pattern the total expected number of four of a kinds to get at least one of each kind is

1 + (13/12) + (13/11) + (13/10) + ... + (13/1) = 41.34173882.

John L. from Bouldter

Yes. Let’s assume you are playing 9/6 Jacks or Better. The variance per final hand is n*1.966391 + 17.548285, where n is the number of plays. So the variance per hand in 10 play is 10*1.966391 + 17.548285 = 37.2122, and in 1-play is 1*1.966391 + 17.548285 = 19.51468. The variance of 1,000 initial or 10,000 total hands of 10-play is 10,000*37.2122 = 372,122. The variance of 10,000 hands of 1-play is 10,000*19.51468 = 195,149. However, standard deviation is what I think we should be talking about, which is the square root of the variance. The standard deviation of 10,000 hands of 10-play is 372,122^{0.5} = 610.02. The standard deviation of 10,000 hands of 1-play is 195.149^{0.5} = 441.75. As long as the total final hands are the same, 10-play will always be 38.1% more volatile, in 9/6 Jacks or Better. For more information visit my section on the standard deviation in n-play video poker.

Steve from Oxnard

In 9/6 Jacks or Better with perfect strategy you will see a royal on the draw once every 40,601 hands, but four to a royal once every 460 hands. For every royal you see, you will be one card away 88.33 times. Of the four to a royal hands, 50.37% will pay nothing, 24.89% will pay as a pair, 7.89% as a straight, 16.16% as a flush, and 0.69% as a straight flush. Here are the exact numbers.

### Possible Outcomes in 9/6 Jacks or Better

Hand | Combinations | Probability |

Four to royal + straight flush | 299529168 | 0.000015 |

Four to royal + flush | 7005972000 | 0.000351 |

Four to royal + straight | 3420857076 | 0.000172 |

Four to royal + pair | 10793270244 | 0.000541 |

Four to royal (non-paying) | 21844510692 | 0.001096 |

Royal flush | 490952388 | 0.000025 |

All other | 19889375425632 | 0.9978 |

Total | 19933230517200 | 1 |

The expected number of royals for 170 four to a royals is 170/88.33 = 1.92. The probability of seeing zero with a mean of 1.92 is e^{-1.92} = 14.59%.

Kevin from Long Island, New York

The Poisson distribution can be used to answer this kind of question. The general formula is e^{-m}*m^{x}/x!, where x is the number of the event you observed, and m is the expected number. In this case x is 3. The probability of a royal flush in "Not so Ugly Ducks deuces wild" is 0.000023. So the expected number in 10,000 hands would be 0.23. Thus the probability of hitting exactly three royals in 10,000 hands is e^{-0.23}*0.23^{3}/3! = 0.161%. The formula in Excel for this is poisson(3,0.23,0).

Dave S. from New Haven

I assume that you assume the probability of a royal is 1 in 40,000. Playing 4,000 hands per session the expected number of royals per session is 0.1. A very close appoximation for the probbility of zero royals per session is e^{-0.1} = 90.48%. The reason it is not 90% is because sometimes you will get more than one royal per session. The expected number of royals in 50 sessions is 0.1 × 50 = 5. The probability of zero royals in 50 sessions can be closely approximated at e^{-5} = 0.67%. The exact probability is (39,999/40,000)^(200,000) = 0.67%, as well.

Dave from Las Vegas

If you were playing single line it would be easy. $800,000 is 160,000 $5 hands. That is 3.9616 royal cycles. The probability of no royals can be closely approximated as e^{-3.9616} = 1.9%.

The math gets messier with mutli-line games. I think the easiest way to answer the question is by random simulation. My video poker appendix 6 shows the probability of getting at least one royal per hand in 50-play 9/6 Jacks or Better is 0.00099893. Each hand of $1 50-play costs $250. So you would have played 3,200 initial hands. The expected number of hands with a royal in 3,200 hands is 3.1966. By the same method of approximation, the probability of getting zero royals is e^{-3.1966} = 4.09%. The exact answer, based on the simulation results, is (1-0.00099893)^3200 = 0.04083732, or 4.08%.

J.B. from Las Vegas

You’re welcome. I ran some random simulations in 9/6 Jacks or Better, to get at the answer to your question. The following table shows the covariance for 2 to 9 lines played, in 9/6 Jacks or Better. The variance would be the same as the base game played.

### Covariance in 9/6 Jacks or Better Spin Poker

Lines | Covariance |

2 | 1.99 |

3 | 3.70 |

4 | 9.62 |

5 | 15.27 |

6 | 19.53 |

7 | 23.37 |

8 | 27.94 |

9 | 33.46 |

Let’s look at an example of 9-line 9/6 Jacks or Better. The variance of the base game is 19.52. The covariance is 33.46. So the total variance is 19.52 + 33.46 = 52.98. The standard deviation is 52.98^{1/2} = 7.28.

"Anonymous" .

As long as she is flat betting at a steady rate, yes, by all means take the bet. Either she is using some kind of worthless progression, or this is second-hand exaggeration. This got me to thinking, what would be the optimal number of hands for your friend’s side. Assuming 9/6 Jacks or Better, and optimal strategy, the probability of being ahead is maximized at 136 hands, with a probability of 39.2782%.

Nathan from Edina, MN

The return of 6/5 double double bonus is 0.946569, to be exact. My table says the probability of a royal is 0.000025. However, I like to use more significant digits that that, so let’s take the return, divided by the win, which is 0.020297/800 = 0.00002537. The return of all the wins besides the royal is 0.926273. Let’s call j the breakeven jackpot amount. Solving for j:

1 = 0.926273 + 0.00002537*j

j = (1-0.926273)/ 0.00002537 = 2,906.

The 2,906 is measured in bet units. For a $1 machine ($5 total bet) the breakeven point would be $5*2,906 = $14,530. So, $12,000 is still a long way away from break-even. Before some perfectionist writes me, as the progressive goes up, the optimal strategy will change, to be more aggressive towards playing for royals. My answer assumes the player follows the same 6/5 optimal strategy the entire time.

A simple approximation for any 52-card video poker game is to add 0.5% for every extra 1,000 coins in the meter. In the case of a $10,100 meter, that is $6,100 higher than a non-progressive. It is a dollar game, so that is 6,100 coins, so add 0.5% × (6,100/1,000) = 3.05% to the base return. The base return is 92.63%, so the total return could be approximated as 94.66% + 3.05% = 97.71%. The actual return for a $10,100 meter is 97.75%, so pretty close.

"Anonymous" .

There are 4 suits to choose from for the 3 to a royal. There are combin(5,3)=10 ways to choose 3 out of the 5 ranks. There are combin(47,2)=1,081 ways to choose the other two cards. There are combin(52,5)=2,598,960 ways to choose 5 cards out of 52. So the probability of getting 3 to a royal is 4×10×1081/2,598,960 = 1.66%.

David from Fort Worth, Texas

For the benefit of other readers, the coefficient of skewness (skew) for any random variable is a measure of which direction has the longer tail. A negative skew means the most likely outcomes are on the high side of the distribution, offset by the extremes tending to be on the low side. A positive skew is the opposite, where the most likely outcomes are the low side, but with the extremes tending to be on the high side. The mean is less than the median with a negative skew, and greater with a positive skew. An exact formula can be found at Wikipedia, or lots of statistics books.

Loosely stated, skewness is going to correlate with how often you get a win in a session. In Jacks or Better, for the most part, you are not going to get a winning session over a few hours if you do not hit a royal. You can sit down at Double Double Bonus and be a winner after a few hours more often because of the big quad payouts. Because most people are subject to cognitive biases, the pain from a loss is twice the pleasure from a win. People do not really play Double Double Bonus because they like the variance, they play because they have a greater shot at winning. The following table shows some key statistics for four common video poker games. It is interesting to note that skew is greatest for Jacks or Better.

### Key Video Poker Statistics

Statistic | JoB — 9/6 | BP — 8/5 | DDB — 9/6 | DW — NSUD |
---|---|---|---|---|

Return | 0.995439 | 0.99166 | 0.989808 | 0.997283 |

Variance | 19.514676 | 20.904113 | 41.985037 | 25.780267 |

Skew | 147.114643 | 134.412152 | 66.495372 | 101.23991 |

(Excess) Kurtosis | 26,498 | 23,202 | 6,679 | 14,550 |

JoB — 9/6 = Full pay Jacks or Better

BP — 8/5 = Standard pay Bonus Poker

DDB — 9/6 = Standard pay Double Double Bonus Poker

DW — NSUD = "Not so Ugly Ducks" Deuces Wild

How can knowing this actually help the video poker player? I suppose one could say that a game with a large skew has a greater chance of a loss over a session of a few hours. For example, in Jacks or Better, if you don’t hit any royals, the house edge will probably eventually grind your bankroll down. However in a game like Deuces Wild or Double Double Bonus, the second highest wins can pull you out of the hole over a session. In other words, the skew keeps you from winning when you are not hitting royals. Knowing the skew won’t increase your odds, but it is mentally helpful to know what to expect. So, the next time you take a beating in 9/6 Jacks, blame it on the skew.

My thanks to Jeff B. for his help with this question.

Robert from Biloxi, MS

Nice find! You didn’t say what denomination you are playing, which is important, so I’m going to assume dollars. For five-coin maximum bet, the number of doubles required for a win of w (where w<1200) is 1+int(log(1200)-log(w))/log(2).

The following table shows for each initial hand the pre-double win, pre-double probability, number of doubles required, post-double win, and probability achieving the post-double win, including the $250 bonus. The lower right cell shows a return of 115.5%. You will get a jackpot every 297 hands on average, with an average jackpot of $1,717.46.

### 8-5 Triple Bonus Return Table with $250 Bonus for Wins of $1,200 or More

Pre-Double Win | Pays | Pre-Double Probability | Doubles Required | Post-Double Win | Post-Double Probability | Return |

Royal flush | $4000 | 0.000026 | 0 | $4250 | 0.000026 | 0.02193 |

Straight flush | $500 | 0.000118 | 2 | $2250 | 0.00003 | 0.013322 |

4 aces | $1200 | 0.000235 | 0 | $1450 | 0.000235 | 0.068227 |

4 2-4 | $600 | 0.000542 | 1 | $1450 | 0.000271 | 0.078557 |

4 5-K | $250 | 0.001629 | 3 | $2250 | 0.000204 | 0.091637 |

Full house | $40 | 0.010546 | 5 | $1530 | 0.00033 | 0.100842 |

Flush | $25 | 0.011055 | 6 | $1850 | 0.000173 | 0.063913 |

Straight | $20 | 0.012738 | 6 | $1530 | 0.000199 | 0.060902 |

3 of a kind | $15 | 0.075542 | 7 | $2170 | 0.00059 | 0.256136 |

Two pair | $5 | 0.123065 | 8 | $1530 | 0.000481 | 0.147101 |

Jacks or better | $5 | 0.211575 | 8 | $1530 | 0.000826 | 0.252898 |

Total | 0.447071 | 0 | 0 | 0.003364 | 1.155465 |

James from Spencer, MA

The following table shows the probability of each kind of royal, according to the number of cards held, given that there was a royal. It shows that 3.4% of royals are from holding one card. The probability of a royal to begin with is 1 in 40,391, so the unconditional probability of a royal holding one card is 1 in 1,186,106.

### 9/6 Jacks Royal Combinations

Cards Held | Combinations | Probability |
---|---|---|

0 | 1,426,800 | 0.002891 |

1 | 16,805,604 | 0.034053 |

2 | 96,804,180 | 0.196154 |

3 | 195,055,740 | 0.395240 |

4 | 152,741,160 | 0.309498 |

5 | 30,678,780 | 0.062164 |

Total | 493,512,264 | 1.000000 |

Mic

There are combin(52,5)=2,598,960 possible combinations on the deal. The reason my video poker return tables have almost 20 trillion combinations is you also have to consider what could happen on the draw. Here are the number of combinations according to how many cards the player discards.

### Combinations on theDraw in Video Poker

Discards | Combinations |

0 | 1 |

1 | 47 |

2 | 1,081 |

3 | 16,215 |

4 | 178,365 |

5 | 1,533,939 |

The least common multiple of all those combinations is 5×combin(47,5)= 7,669,695. Regardless of how many cards the player discards, the return combinations should be weighted so that the total comes to 7,669,695. For example, if the player discards 3, there are 16,215 possible combinations on the draw, and each one of them should be weighted by 7,669,695/16,215 = 473.

So the total number of combinations in video poker is 2,598,960 × 7,669,695 = 19,933,230,517,200 . For more on how to program video poker returns yourself, please see my page on Methodology for Video Poker analysis.

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

rudeboyoi

My best guess is Royal Aces Bonus Poker. I’ve seen it only once in Mesquite years ago. It pays 800 for four aces, but compensates with a lowest paying hand of a pair of aces, as opposed to the usual jacks. Here is the return table.

### Royal Aces Bonus Poker

Hand | Pays | Combinations | Probability | Return |
---|---|---|---|---|

Royal flush | 800 | 490,090,668 | 0.000025 | 0.019669 |

Straight flush | 100 | 2,417,714,292 | 0.000121 | 0.012129 |

Four aces | 800 | 4,936,967,256 | 0.000248 | 0.198140 |

Four 2-4 | 80 | 10,579,511,880 | 0.000531 | 0.042460 |

Four 5-K | 50 | 31,662,193,440 | 0.001588 | 0.079421 |

Full house | 10 | 213,464,864,880 | 0.010709 | 0.107090 |

Flush | 5 | 280,594,323,000 | 0.014077 | 0.070384 |

Straight | 4 | 276,071,121,072 | 0.013850 | 0.055399 |

Three of a kind | 3 | 1,470,711,394,284 | 0.073782 | 0.221346 |

Two pair | 1 | 2,398,705,865,028 | 0.120337 | 0.120337 |

Pair of aces | 1 | 1,307,753,371,584 | 0.065607 | 0.065607 |

Nothing | 0 | 13,935,843,099,816 | 0.699126 | 0.000000 |

Total | 19,933,230,517,200 | 1.000000 | 0.991982 |

The standard deviation is 13.58! That is over three times as high as 9-6 Jacks or Better at 4.42.

However, if you limit me to games that are easy to find, my nomination is Triple Double Bonus, with a standard deviation of 9.91. Here is that pay table.

### Triple Double Bonus Poker

Hand | Pays | Combinations | Probability | Return |
---|---|---|---|---|

Royal flush | 800 | 439,463,508 | 0.000022 | 0.017637 |

Straight flush | 50 | 2,348,724,720 | 0.000118 | 0.005891 |

4 aces + 2-4 | 800 | 1,402,364,496 | 0.000070 | 0.056282 |

4 2-4 + A-4 | 400 | 3,440,009,028 | 0.000173 | 0.069031 |

4 aces + 5-K | 160 | 2,952,442,272 | 0.000148 | 0.023699 |

4 2-4 + 5-K | 80 | 6,376,626,780 | 0.000320 | 0.025592 |

4 5-K | 50 | 31,673,324,076 | 0.001589 | 0.079449 |

Full house | 9 | 206,321,656,284 | 0.010351 | 0.093156 |

Flush | 7 | 311,320,443,672 | 0.015618 | 0.109327 |

Straight | 4 | 252,218,322,636 | 0.012653 | 0.050613 |

3 of a kind | 2 | 1,468,173,074,448 | 0.073655 | 0.147309 |

Two pair | 1 | 2,390,581,734,264 | 0.119929 | 0.119929 |

Jacks or better | 1 | 3,944,045,609,748 | 0.197863 | 0.197863 |

Nothing | 0 | 11,311,936,721,268 | 0.567491 | 0.000000 |

Total | 19,933,230,517,200 | 1.000000 | 0.995778 |

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

- 6-5 Bonus Poker progressive.
- 2% meter rise on royal flush.
- 5-coin game.

Now assume the following about me.

- Minimum return to play of 100.5%.
- I’m capable of playing a progressive until it hits.
- I know perfect 6-5 Bonus Poker strategy for a 4000-coin royal.

What is the least the jackpot should be for me to play?

Mark

7,281.8 coins. It is interesting to note that if you played only once at exactly that meter then the return would be 98.5% only. The reason you should play at that point is because of the assumption you are capable of playing until you pop the jackpot. That is like having a 2% cash back slot club. 98.5% + 2% = 100.5%.

I might add that if you start playing 4000-coin jackpot strategy at exactly a 7,281.8 jackpot, you can expect to profit 201.18 bets. However, if you took the time to learn the strategy changes for a 7,281.8 coin jackpot, then your expected profit would be 234.31 coins.

On a related note, I just finished reading The Secret World of Video Poker Progressives by Frank Kneeland. This book has lots of formulas for much more complicated progressive situations, as well as practical advice and stories based on his years running a team of progressive hunters. I recommend it for advantage progressive video poker players.

Frank

For a near-exact answer to streak questions such as this we need to use matrix algebra. I answered a similar, yet easier, question in my June 4, 2010 column. If your matrix algebra is rusty I would look at that one first.

**Step 1**: Determine the probability of 0 to 6+ royals in the first 5,000 hands. Let's assume the probability of a royal is 1 in 40,000. The expected number in 5,000 hands is 5,000/40,000 = 0.125. Using the Poisson estimate, the probability of exactly r royals is e^{-0.125} × 0.125^{r}/r!. Here are those probabilities:

### Royals in 5,000 Hands

Royals | Probability |
---|---|

0 | 0.8824969026 |

1 | 0.1103121128 |

2 | 0.0068945071 |

3 | 0.0002872711 |

4 | 0.0000089772 |

5 | 0.0000002244 |

6+ | 0.0000000048 |

**Step 2**: Consider there to be seven states for the remaining 24,995,000 hands. For each one, the previous 5,000 hands can have 0, 1, 2, 3, 4, or 5 royals, or the player could have already achieved getting six royals in 5,000 hands, in which case success is achieved, and it can't be taken away. With each new hand, one of three things can happen to the player's state:

- Move down a level. This happens if the hand that was played 5,000 games ago was a royal, and is now dropping off, and the new hand was not a royal.
- Remain at the same level. This will usually happen if the hand played 5,000 games ago was not a royal, and the new hand is also not a royal. It can also happen if a hand 5,000 games ago was a royal, but the new hand is also a royal.
- Move up a level. This will happen if the hand played 5,000 games ago was not a royal, and the new hand is.

**Step 3**: Develop the transition matrix for the odds of each change of state for an additional game played.

The first row will correspond to level 0 before the new hand is played. The odds of advancing to level 1 in the next hand are simply 1 in 40,000. The probability of staying at level 0 is 39,999/40,000.

The second row will correspond to level 1 before the new hand is played. The odds of advancing to level 2 in the next hand are the product of the odds of not losing a royal on the hand dropping off and getting a royal on the new hand = (4999/5000)×(1/40000) = 0.0000250. The odds of going back to level 0 are the product of a royal dropping off and not getting a royal on the current game = (1/5000)×(39999/40000) = 0.0002000. The odds of staying the same is pr(no royal dropping off) × pr(no new royal) + pr(royal dropping off) × pr(new royal) = (4999/5000)×(39999/40000) + (1/5000)×(1/40000) = 0.9997750.

The probabilities for rows 2 to 6 will depend on how many royals are present in the history of the last 5,000 hands. The more there are, the greater the probability of one dropping off as a new hand is played. Let r be the number of royals in the last 5,000 hands and p be the probability of getting a new royal.

Pr(promote a level) = Pr(no royal dropping off) × Pr(new royal) = (1-(r/5000))× p.

Pr(remain at same level) = Pr(no royal dropping off) × Pr(no new royal) + Pr(royal dropping off) × Pr(new royal) = (1-(r/5000))× (1-p) + (r/5000)×p.

Pr(demote a level) = Pr(royal dropping off) × Pr(no new royal) = (r/5000)× (1-p).

Row 7 corresponds to having achieved the state of success for getting six royals in 5,000 hands. Once you achieve that accomplishment it can never be taken away, so the odds of staying in that state of success are 100%.

The rows in the transition matrix will correspond to the levels before the new hand, starting with level 0 in the top row. The columns will correspond to the levels after the new hand, starting with level 0 in the left column. The body of numbers in the matrix will correspond to the probabilities of moving from each old state to each new state in one game. Let's call this T1 =

0.999975 | 0.000025 | 0.000000 | 0.000000 | 0.000000 | 0.000000 | 0.000000 |

0.000200 | 0.999775 | 0.000025 | 0.000000 | 0.000000 | 0.000000 | 0.000000 |

0.000000 | 0.000400 | 0.999575 | 0.000025 | 0.000000 | 0.000000 | 0.000000 |

0.000000 | 0.000000 | 0.000600 | 0.999375 | 0.000025 | 0.000000 | 0.000000 |

0.000000 | 0.000000 | 0.000000 | 0.000800 | 0.999175 | 0.000025 | 0.000000 |

0.000000 | 0.000000 | 0.000000 | 0.000000 | 0.001000 | 0.998975 | 0.000025 |

0.000000 | 0.000000 | 0.000000 | 0.000000 | 0.000000 | 0.000000 | 1.000000 |

If we multiply this transition matrix by itself we get the probabilities of each change of state in two consecutive games. Let's call this T2, for the transition matrix over two games:

0.999950 | 0.000050 | 0.000000 | 0.000000 | 0.000000 | 0.000000 | 0.000000 |

0.000400 | 0.999550 | 0.000050 | 0.000000 | 0.000000 | 0.000000 | 0.000000 |

0.000000 | 0.000800 | 0.999150 | 0.000050 | 0.000000 | 0.000000 | 0.000000 |

0.000000 | 0.000000 | 0.001199 | 0.998750 | 0.000050 | 0.000000 | 0.000000 |

0.000000 | 0.000000 | 0.000000 | 0.001599 | 0.998351 | 0.000050 | 0.000000 |

0.000000 | 0.000000 | 0.000000 | 0.000001 | 0.001998 | 0.997951 | 0.000050 |

0.000000 | 0.000000 | 0.000000 | 0.000000 | 0.000000 | 0.000000 | 1.000000 |

By the way, in Excel to multiply two matrices of equal size first select the region where you want the new matrix to go. Then use this formula =MMULT(range of matrix 1, range of matrix 2). Then do ctrl-shift-enter.

If we multiply T2 by itself we get the probabilities of each change of state in four consecutive games, or T4:

0.999900 | 0.000100 | 0.000000 | 0.000000 | 0.000000 | 0.000000 | 0.000000 |

0.000800 | 0.999100 | 0.000100 | 0.000000 | 0.000000 | 0.000000 | 0.000000 |

0.000000 | 0.001598 | 0.998301 | 0.000100 | 0.000000 | 0.000000 | 0.000000 |

0.000000 | 0.000001 | 0.002396 | 0.997503 | 0.000100 | 0.000000 | 0.000000 |

0.000000 | 0.000000 | 0.000003 | 0.003193 | 0.996705 | 0.000100 | 0.000000 |

0.000000 | 0.000000 | 0.000000 | 0.000005 | 0.003989 | 0.995907 | 0.000100 |

0.000000 | 0.000000 | 0.000000 | 0.000000 | 0.000000 | 0.000000 | 1.000000 |

So keep repeating this doubling process 24 times until we get to T-16,777,216:

0.882415 | 0.110305 | 0.006893 | 0.000287 | 0.000009 | 0.000000 | 0.000091 |

0.882415 | 0.110305 | 0.006893 | 0.000287 | 0.000009 | 0.000000 | 0.000092 |

0.882413 | 0.110304 | 0.006893 | 0.000287 | 0.000009 | 0.000000 | 0.000094 |

0.882385 | 0.110301 | 0.006893 | 0.000287 | 0.000009 | 0.000000 | 0.000125 |

0.881714 | 0.110217 | 0.006887 | 0.000287 | 0.000009 | 0.000000 | 0.000885 |

0.860229 | 0.107531 | 0.006720 | 0.000280 | 0.000009 | 0.000000 | 0.025231 |

0.000000 | 0.000000 | 0.000000 | 0.000000 | 0.000000 | 0.000000 | 1.000000 |

If we doubled again we would overshoot our goal of T-24,995,500. So now we need to carefully multiply by smaller transition matrices, which we would have already calculated. You can arrive at any number using powers of two (the joys of binary arithmetic!). In this case T-24,995,500 = T-16,777,216 × T-2^{22} × T-2^{21} × T-2^{20} × T-2^{19} × T-2^{18} × T-2^{16} × T-2^{14} × T-2^{13} × T-2^{10} × T-2^{7} × T-2^{5} × T-2^{4} × T-2^{3} =

0.882375 | 0.110300 | 0.006893 | 0.000287 | 0.000009 | 0.000000 | 0.000136 |

0.882375 | 0.110300 | 0.006893 | 0.000287 | 0.000009 | 0.000000 | 0.000136 |

0.882373 | 0.110299 | 0.006892 | 0.000287 | 0.000009 | 0.000000 | 0.000138 |

0.882345 | 0.110296 | 0.006892 | 0.000287 | 0.000009 | 0.000000 | 0.000170 |

0.881675 | 0.110212 | 0.006887 | 0.000287 | 0.000009 | 0.000000 | 0.000930 |

0.860191 | 0.107527 | 0.006719 | 0.000280 | 0.000009 | 0.000000 | 0.025275 |

0.000000 | 0.000000 | 0.000000 | 0.000000 | 0.000000 | 0.000000 | 1.000000 |

To be honest, in the interest of simplicity and saving time, you don't really need to bother with those last four multiplications. These correspond to the last 56 hands only, and the odds that those 56 will make a difference in the final outcome are negligible. I'm sure my many perfectionist readers would take me to the woodshed for saying that, if they could.

**Step 4:** Multiply the initial state after 5,000 hands by T-24,995,500. Let S-0, from step 1, be as follows:

0.8824969026 | 0.1103121128 | 0.0068945071 | 0.0002872711 | 0.0000089772 | 0.0000002244 | 0.0000000048 |

So S-0 × T-24,995,500 =

0.88237528 |

0.11029964 |

0.00689251 |

0.00028707 |

0.00000896 |

0.00000022 |

0.00013632 |

The number in the bottom cell is the probability of having achieved six royals within 5,000 hands at least once during the 25,000,000 hands. So a 1 in 7,336 chance.

My thanks to CrystalMath for his help with this question.

"Anonymous" .

Before I answer, I'd like to remind everybody that the number of ways to choose k out of n items, with replacement, is combin(n+k-1,k) = (n+k-1)!/((n-1)!×k!).

That said, here are the following types of seven-card hands and the number of distinct ways to make each:

- 7 cards of one suit: combin(13,7)=1,176.
- 6 cards of one suit and 1 of another: COMBIN(13,6)×13 = 22,308.
- 5 cards of one suit and 2 of another: COMBIN(13,5)×combin(13,2) = 100,386.
- 5 cards of one suit and 1 each of another two: COMBIN(13,5)×combin(13+2-1,2) = 117,117.
- 4 cards of one suit and 3 of another: COMBIN(13,4)×combin(13,3) = 204,490.
- 4 cards of one suit, 2 of a second, and 1 of third: COMBIN(13,4)×combin(13,2)×13 = 725,010.
- 4 cards of one suit and one each of another 3: COMBIN(13,4)×combin(13+3-1,3)×13 = 325,325.
- 3 cards of two different suits and one card of a third suit: 13×((COMBIN(13,3)×(COMBIN(13,3)-1)/2+COMBIN(13,3))) = 533,533.
- 3 cards of one suit and two cards each of two other suits: COMBIN(13,3)×(COMBIN(13,2)×(COMBIN(13,2)+1)/2) = 881,166.
- 3 cards of one suit, 2 cards of a second, and one card each of the two other suits: COMBIN(13,3)×COMBIN(13,2)×COMBIN(13+2-1,2) = 2,030,028.
- 2 cards each of three suits and 1 from the fourth: ((COMBIN(13,2)×(COMBIN(13,2)+1)×(COMBIN(13,2)+2)/6) = 1,068,080.

The sum of those combinations is 6,009,159. Compared to the combin(52,7)= 133,784,560 ways to pick 7 cards out of 52, that is a 95.5% reduction in hands analyzed.

For more discussion on this question, please see my forum at Wizard of Vegas.

I was playing 10-play video poker and held a pair after the deal. All ten hands then improved to a four of a kind on the draw. What are the odds?

"Anonymous" .

The probability of a pair improving to a four of a kind is 45/COMBIN(47,3) = apx. 0.002775208.

The probability of that happening in ten out of ten hands is (0.002775208)^{10} = apx. 1 in 36,901,531,632,979,700,000,000,000.

That probability is like buying three independent and random Powerball tickets and winning on ALL three of them.

The explanation is that this is NOT a normal video poker game, with natural probabilities, in which each card had an equal chance to be drawn out of the remaining cards in the deck. No, this is what is called a "VLT," or Video Lottery Terminal. In such games, the outcome is predestined, regardless of how the player pays his hand. It is like a scratch card lottery ticket, but the outcome is displayed to the player like a game of video poker. You might ask what would happen if the player held all five cards. Then a genie would have come along to change some of the cards or the player would have won a bonus to get the final win to 2,500 credits.

This question is raised and discussed in my forum at Wizard of Vegas.