What are the odds/probability of each possible five-card hand that could be dealt for from a standard 52-card deck where the suit of a single, unduplicated card in a hand is considered generic? For example, consider the hand A♠ A♣ A A 2. In this hand the suit of the 2 is disregarded and would represent any of the four 2's in the deck. Another example is A-J-8-6-5. In this hand the suit of all 5 cards is disregarded so that only one such combination can occur. Another example: 3♠ 3 7♣ 7♠ Q. In this hand the suit of the two 3's and two 4's is not disregarded because there is more than one in the hand but the suit of the Q is generic. In other words the suit of any card that is not duplicated in a hand is disregarded and that hand is considered to be one of the possible hands even though there would be many possibilities of the hand re-occuring if the suit of each card was not disregarded. Thus a straight or flush made up of five specific cards; say Q J♣ 10 9 8♠ or A♠ J♠ 8♠ 7♠ 3♠ could only occur once since any other combination of those same cards even though in a different suit would be a duplication. Therefore, using this criteria, what is the O/P of any single hand being dealt? In other words how many such five-card hands exist in a std 52 card deck? Thanks for your input. - Mike from Lavallette, NJ

I'm very familiar with this concept. There are 134,459 unique five-card hands. The reason I know this is my first video poker program analyzed all 2,598,960 possible hands and took days to run through a pay table. However by running just one of each of the 134,459 classes of hands, weighting by the number of all hands in its class, you can cut down the run time by 95%. Sometimes when a game is suit specific like my recent analysis of Black Jack Bonus Poker, I have to dust off my old program and do it the slow way. Feb. 11, 2006

A colleague of mine was recently playing at [an online] casino playing 10-line Jacks or Better Video Poker. Money was deposited and 10 hands were played. All 10 hands (and thus all 100 lines) failed to bring up a single win. Please can you calculate the probability of drawing a blank on 10 hands of 10 line JoB. Also, would the probabilty you calculate be evidence of a rigged game? Thanks in advance and keep up the (very) good work.

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 0.025914¹⁰ = 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. April 3, 2005

How likely is it to play 17.76 royal cycles of video poker and hit only three royals?

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*mⁿ/n!. So in this situation the probability is e^-17.76*17.76³/3! = 0.00001808, or 1 in 55321. Jan. 9, 2005

My friend and I went gambling and she got a Royal Flush on Bonus video poker in the morning. Later on in the same day she got another Royal Flush on a different machine but in the same row of machines. I was wondering what the odds were to get two Royal Flushes in the same day?

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. Dec. 5, 2004

What is the probability over the course of 1 million hands that there is a royal flush drought extending for 200,000 hands? I'm more interested in the solution than the answer itself.

It isn't often I say this but I tried for hours but the math on this one was simply over my head. So I turned to my friend and math professor Gábor Megyesi. Here is his formula for any "drought" problem.

Let p be the probability of winning any given hand.

Let d be the length of the drought.

Let n be the number of hands played.

Set k=dp and x=np.

If k=1 then let a=-1, otherwise find a such that k=-ln(-a)/(1+a). (a is a negative number, if k>1 then -1 < a < 0, if k < 1 then a < -1, and a needs to be calculated to high accuracy.) [Wizard's note: This kind of solution can be easily found in Excel using the Goal Seek feature under the tools menu.]

if k=1 then let A=2, otherwise let A=(1+a)/(1+ak).

The probability of no drought of length d in n hands is approximately Ae^a^x.

In this particular problem p=1/40391, d=200000, n=1000000, k=4.9516, x=24.758, a=-0.0073337, A=1.03007. So the probability of no drought is 1.03007*e^{-0.0073337*24.758} = 0.859042. Thus the probability of at least one drought is 1-0.859042 = 0.140958.

Here is Gábor Megyesi's full 5-page solution (PDF). Thanks Gábor for your help.

I did a random simulation of 32,095 sets of one million hands. The number with at least one drought was 4558, for a probability of 14.20%.

After I posted Gábor's solution Joshua Green supplied a shorter one by another method. Sept. 30, 2004

What is the probability of being dealt four to a royal?

There are four possible suits for the royal. There are five possible gaps for the missing card. The fifth card can be one of 47 other cards. So, there are 4 × 5 × 47 = 940 ways to get four to a royal. There are (52 × 51 × 50 × 49 × 48)/(1 × 2 × 3 × 4 × 5) = 2,598,960 total combinations. So the probability is 940/2,598,960 = 1 in 2,765. Jan. 20, 2004

What would be the probability of hitting a royal flush in video poker if you always played the best strategy to do so, which would consist of always keeping 1 or more to a royal and throwing away all cards that don't compose a royal flush? What would be the housed advantage in this situation? Just curious. Thanks.

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

Dec. 17, 2003

Could you tell me the odds when holding 3 and drawing 2 to a Royal Flush ? My wife and I will often throw away a high pair to draw 2 to a royal. THANKS, 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. Aug. 2003

I have something rather interesting to calculate. I was playing Deuces Wile Video Poker when I was dealt "Garbage". When I threw all five cards away, I was given 4 deuces on the draw, 1,000 coins! What is the probability for 4-Deuces to appear on the draw after throwing all 5 cards away on the deal? Thanks for your time and keep up the good work on your website! - Saucedo, Anthony

P.S. On the same machine, I switched to Deuces Wild Bonus Poker and was dealt 3 wilds w/4&5 of diamonds (straight flush), I threw away the 4&5 and hit 4 deuces w/Ace for another 2,000! What a lucky machine! This was at Soboba Casino in Southern California.

No problem. There are combin(47,5)=1533939 ways to arrange 5 cards out of the 47 left in the deck. 43 of those will result in four deuces (the 5th card has 47-4=43 possibilities). So the probability of getting four deuces on the draw is 43/1533939 = 1/35673 = 0.000028032. The probability of drawing a fourth deuce after keeping 3 is (47-1)/combin(47,2) = 46/1081 = 0.0426. In my video poker appendix 5 you can see how the probability distribution in deuces wild of the number of cards drawn to any given hand. For example 2.62% of all four deuces will be obtained by getting all four on the draw. The probability of this happening on any given hand is 0.000005. For more information on the combin function visit my section on probabilities in poker. March 21, 2003

How does one calculate the probability of hitting on a *specific* number of draws in n-play video poker? Example: drawing to a four-card Royal on a triple-play machine, the odds of *at least* one hit are 1-(46/47)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   Feb. 4, 2003

If I hold just the queen of clubs what are the odds (ten million to one etc.) of drawing to a royal flush? - 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. Sept. 24, 2002

Dear Wizard, In video poker, what are the odds of drawing the following cards to a royal flush?

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) = 1/1,081
3. 1/combin(47,3) = 1/16,215
4. 1/combin(47,4) = 1/178,365
5. 4/combin(52,5) = 1/649,740

When playing video poker with a single deck, what are the odds of getting 4 of a kind when you hold just one card. This happened to me this last weekend. I wound up with 4 aces and 4 kings when just starting with one of each. I know holding a pair and drawing the other two to make your quad is 360 to one, but I have never seen odds for drawing 3 to make four of a kind. - 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+4=48. The total number of combinations on the deal is combin(47,4)=178365. So the probability of a four of a kind is about 1 in 7432. Nov. 11, 2001

First of all, thanks for your very informative, comprehensive, and overall helpful site. I have a couple of questions for you. I have noticed in your tables of probabilities and expected returns for video poker, that the probabilities (and corresponding number of hands) for each hand vary for the same type (jacks or better, for example) from one pay out chart to another. For example, on the first jacks or better chart, the probability of forming a three-of-a-kind is .074344, but on the second that same probability is listed as .074449. Why would this discrepancy exist? It seems that the only possibility is that the game is being played with a different strategy. Otherwise, the probability of forming any hand should be the same in that type of game, no matter what the pay outs are. If you have indeed devised a unique playing strategy for each pay out schedule, would you mind sharing that info with us? Secondly, I am wondering which, if any, online casinos currently advise the player of a shuffle in blackjack (multi-deck, of course). Thanks again for your great web site, and I look forward to your response to my questions. - Tony from Columbus, Ohio

Yes, the probability of a three of a kind depends on the pay table. I have software that plays out the optimal strategy for every hand under the given pay table to determine the frequency of each hand and the return. However creating a strategy in writing is very time consuming. I have only done that for full pay deuces wild and full pay jacks or better. I figure anyone who is willing to bother learning the strategy will also take the bother to find a good machine. April 15, 2001

Q: What are the odds of getting a Royal Flush on a video poker game by holding two cards and drawing three to make the royal flush? - Jim of Laguna Niguel, California

A: 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. July 9, 2000

Q: In video poker what are the odds of drawing a sequential royal flush? Not being dealt it, but getting it including a draw. -Tony of Chicago, US

A: I'm not sure exactly but I can tell you that playing optimal strategy on a full pay jacks of better game you will average one royal flush every 40387 plays. There are 5!, or 120 ways to arrange the five cards and only two of them are sequential so the odds of a royal being sequential are only 1 in 60. So assuming no strategy adjustments you would get a sequential royal once every 2,423,263 plays. However it would be wise to be more aggressive going for the royal if you already had some cards in sequential place, which would increase the probability slightly. By how much exactly I don't know. circa Jan. 2000

Video Poker Q&A

Frequently-asked questions
General, inc. Jacks or Better
Deuces Wild
Other Video Poker variations (e.g., Joker Poker)

Dealing
Doubling up
Multiplay (Multihand)
Paytables
Probability

The Wizard has answered questions about...

Specific Games
General Topics

Baccarat
Bingo
Blackjack
Caribbean Stud Poker
Craps
Horse Racing
Keno

Let It Ride
Lottery
Pai Gow Poker
Poker
Roulette
Slot Machines

Spanish 21
Sports Betting
Texas Hold 'Em
Three Card Poker
Video Poker
Other Casino Games
Non-Casino Games
TV Game Shows

Betting Systems
Boyfriends
Casinos
Dealers
General
Online Gaming
Probability
Table Games in general
The Wizard himself

Please save new questions for my bulletin board, coming circa October....

The Wizard's other sites: Wizard of Macau, Sweat the Money, Math Problems