4 to the Royal 3 times?????

[GaryP]

There is a gentleman who sends out e-mails about video poker, this gentlemen sent out an e-mail describing a trip to Vegas. He was dealt 4 to a Royal 3 times in a row, on the 3rd time he got the Royal. Does anyone on this board know what the odds of this would be, it seems unlikely to me.

[Eduardo]

Very very very rare. But I guess it could happen. But unless this person has seen this frequently, it's no reason to question the randomness of video poker. You see weird things happen all the time with random events.
 
Can you ever flip a coin tails 100 times in a row? Try right now, not going to happen. But if you get a bunch of people to go out and o it, it will happen eventually. That doesn't mean the person it happens to has a trick coin. You just see weird things like that sometimes.

[Eduardo]

Also I doubt they are reporting it 100% acurately though it is possible. They probably saw 4 to the royal a few times in a very short period and are embellishing or even misremembering. People have a tendency to do that. If there were even a 1 deal gap in between or this happened over just 5 or 10 deals instead of three, then the odds of it happening go up significantly.


 
So yes... possible but very unliekly. Which is probably why none of us has ever seen it before either.
 
I also have never hit the monster wins like we see in the contests here but I know people do. unlikely doesn't mean impossible.
 
If they got it dealt three times, I hope they hit at least one of them.  That would make for a very good day!

[shadowman]

4RF occurs on average every 2765 hands. Doing it 3 times in a row would be 2765^3 = about one in 21 billion. Which, if you think of all the VP hands played, is not that unusual.

[Eduardo]

1 in 21 billion sure sounds hard until you put it in perspective.
 
Winning the california lotto jackpot odds are 1 in 175,711,536. But someone does it almost every week.

[New2vp]

Shadowman is more succinct than I am, but here is a bit more detail...likely more than you want to know.Playing off Eduardo's initial response, getting 34 tails in a row from flipping a coin (probability of 0.0000000000582) is slightly more likely than getting 3 4-to-the-Royals in a row (probability of 0.0000000000473).  If you did get this hand 3 times in a row, you could draw at least one Royal Flush about 6.25% of the time.Could it happen?  You can expect the 3 dealt hands to happen about once every 21 billion times you try it as reported by S-man.  Throw in the fact that you want to also hit the royal at least once, and this will occur once about every 338 billion times that you try it.  Since this is a 3-hand parlay, that means, you would be trying 3 times as many hands or a little more than a trillion video poker hands.It is 1925 times more likely that you could win the next Mega Millions jackpot on a single $1 ticket than see this result the next time you look at 3 hands.  And here is a sobering statistic that I saw on Wikipedia's article on Mega Millions:"To put [the Mega Millions] odds in perspective, in the US in 2008 there were 1.03
deaths per 100 million vehicle miles traveled.
A person living one mile from a retailer selling Mega Millions tickets
is 3.6 times as likely to die in an accident traveling to and from that
store (2 miles) than winning the Mega Millions jackpot on a $1 play."  (So, does this mean you should only buy tickets from stores 1/4 mile from your house??  Or you need to buy 4 or more tickets when you go to the store to make it worth the risk??)Did it happen?  Who knows?  There's lots of time and lots of repetitions in the world.  Certainly more than 338 billion things happen in the world every day, so it's possible.  Will it ever happen for you?  I'd bet against it.Calculation of 4-royal probability:  There are 940 different 4-Royals that can be dealt:  4 suits x 5 sets of ranks (KQJT, AQJT, AKJT, AKQT, AKQJ) x 47 (remaining cards that are not in the deck that are not the other royal flush card) = 940.  Since there are 2,598,960 possible deals, the odds of getting a 4-royal dealt 3 straight times are (940/2598960)^3.Some may consider only 936 legitimate 4-royals since the 940 number contains 4 instances of KQJT9 suited which is usually considered a straight flush with all 5 cards held instead of playing it as a 4-royal in most non-deuces-wild games.

[faygo]

new2vp, here is one for you from an '07' post of mine.
 
http://www.videopoker.com/forum/forum_p ... iffin&PN=6

[spxChrome]

Yea I'm currently 94 times getting Trip aces and not getting the 4 one.  Ranted in that time span I have got quad aces twice but they came from a 4 draw and a 3 draw.

[New2vp]

my story;
 
A number of years ago, I was on a business trip that took me to
Metropolis,IL. A lovely little town in the South of Il. The self
proclaimed home of Superman. They even have a big statue of him
downtown. Ah, but I digress.
 
At that time there was a little riverboat casino owned by Merv Griffin.
The wife and I decided to give it a try. She was off playing her
slots and I sat down at an old slant top coin chunker $1 machine. I must
have played about 5 hands when I was dealt Ks,Js,10s,Qs and a card I
don't remember, tossed the odd card and got nothing. Loaded the
machine(coin chunker remember) and it dealt me Ks,Js,10s,Qs and a non-
descript card. tossed the odd card and caught the As. (always thought
the first one was an omen)
 
Now my question is what are the odds of not only being dealth 4 to a
RF back to back, but to have it be in the same suit, same cards, in the
same order???At the time, Webman and Shadowman took shots at this for you; I'll try to fill in some blanks after making some assumptions.  If we choose to consider KQJT9 suited as a straight flush and not a 4-royal, we have to consider the KQJT case separate from the other 4-royals.If we want to know the odds on the next two hands of hitting 4-royals in which the 4 cards are in the same order and are of the same rank and suit allowing the odd card to be in any position and to be different cards in the 2 hands, the calculation for the probability is:[KQJT]  (184 / 2598960) x (46 / 2598960) x (1/24) +[AQJT, AKJT, AKQT, AKQJ]  (4 x 188 / 2598960) x (47 / 2598960) x (1/24) = 43,808 / 162,110,233,958,400  or about 1 chance in 3.7 billion.The first factor is based on getting any 4-royal in any order and any suit on the first hand; the second factor is the number of odd cards that are possible to be matched with the 4-royal (with KQJT, I don't allow the suited 9 to be a possibility, which is why there is one less card there); and the third factor is there because there are 24 possible orderings of the 4 cards in the royal hold.  If you also wanted the odd card to be in the same position as the first hand, you can change this last factor from 1/24 to 1/120 (or multiply the odds that I state here by 5).As Webman pointed out at the time, the odds are quite a bit less if you simply consider the chances of the same draw on the next hand only occurring AFTER you have already been dealt one hand with a 4-royal.  But they are still mighty slim.  If the 4-royal were KQJT, the odds would be 1,355,978 to 1; otherwise, the odds would be 1,327,128 to 1.  You get this by eliminating the first factor in the previous calculation.All in all, pretty odd, my friend--another once in a lifetime experience!More importantly though, good luck to marie's and your Spartans.  My Buckeyes came up a bit short tonight.  Congratulations to Tennessee fans out there.

[Eduardo]

That Tenessee win won my bracket contest for me today. Not bad, locking it up before the Elite Eight, huh?