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Tell Me! What Are The Odds?

Discussion in 'Casino Gaming' started by DOUBLE B, Feb 6, 2013.

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  1. DOUBLE B

    DOUBLE B Tourist

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    I have been "DEALT" two Diamond Royal Flushes within the past 30 days. On $.25 and played about a total of 20 hours on single line machines. The two royals were not in the same sequence but they were both dealt.
     
  2. mike_m235

    mike_m235 Tourist

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    The odds of being dealt a royal with no wildcards in a 52 card deck is about 1 in 650,000 (that's rounded). To figure out the odds of being dealt two in a given period of time, we'd need to know how many hands you played. To be dealt two back to back, you'd be talking about one in 421 billion.

    The odds of both of them being the same suit are obviously four times less likely.

    To give you some idea of the magnitude, you have a 2400 times better chance of winning powerball than hitting back to back dealt royal flushes.

    Since you played 20 hours, if we give you 600 hands an hour you're probably talking around 12,000 hands. The math isn't simple from there and I'm not doing it, but to give a reasonable estimate of the odds (since the idea of repeat events is pretty minimal) you'd be looking at something on the order of one in 35 million to hit two dealt royals in 12,000 hands.

    But that's really flawed, because it doesn't take into account how many hands you played before you hit the first one, and is just for any 12,000 hand sequence.

    Before any math guys flame me, I took a LOT of shortcuts on this because I was doing it quickly. It's by no means exact...but it's close.
     
  3. Auggie

    Auggie Dovahkiin

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    The odds are 1 in 38, or about 2.6%

    That might sound really low, but here is the key:

    We are talking about two events here: royal flush 1 and royal flush 2

    As mike_m235 said above: we aren't counting how many hands you played before the first, we are simply going off of you hit royal flush 1 and then 20 hours worth of play later you hit royal flush 2

    I am going off of 850 hands per hour, that is more in line with the speed a regular video poker player will play at (a slow beginner will usually be around 500 hands per hour, a veteran pro will be around 1,200+ hands per hour).

    If you played 20 hours at 850 hands per hour that would be a total of about 17,000 hands of video poker.

    The odds of being dealt a royal flush is 1 in 649,740.

    So 17,000 / 649,740 = 0.0261 = 2.61%

    Which is about 1 in 38


    Of course thats the "odds" and doesn't take in to account things like variance, which in this case would be sky high because a dealt royal flush is such a rare event.
     
  4. WrongWayWade

    WrongWayWade VIP Whale

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    I'm not sure of how you really calculate this, but you can't just divide the number of hands you play by the odds of the event. Otherwise, you'd deduce if you played 649,740 hands it would be a 100% chance of getting the royal, and obviously, that's not true. It's not like picking lottery numbers.
     
  5. Auggie

    Auggie Dovahkiin

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    Thats why I included the: "Of course thats the "odds" and doesn't take in to account things like variance"

    If you played 650,000 hands your expectation is that you would have been dealt one royal flush in that time... that doesn't mean its going to happen, just that its your expectation.
     
  6. mike_m235

    mike_m235 Tourist

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    The problem is that calculating the actual probability is really hard, and isn't something that more than a handful of people can do off the top of their heads. I can do it, but not without looking it up and taking a lot of time. Which is why I did it via estimates.

    The smaller your number of events, the closer you're going to get to your estimate being accurate. For example: 10,000 hands is going to be much closer to the true odds when you divide it into 649,740 hands than 100,000 hands. At 1,000 hands, you're approaching complete accuracy with the simplified equation.
     
  7. Nevyn

    Nevyn VIP Whale

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    Ignoring Wade's point that you aren't meant to simply divide, what you have given him is the odds of hitting the second royal given that he hit the first.

    And you also aren't factoring in royal of the same suit.
     
  8. 4Eyes

    4Eyes Low-Roller

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    Very simple calculation:

    Probability of getting dealt a royal (of any suit) = (20/52 x 4/51 x 3/50 x 2/49 x 1/48) = 480 / 311,875,200 = 1 in 649,740 = 0.000154%.

    Once the first royal is received, the probability of getting dealt at least one additional royal (of any suit) in the next 17,000 hands = 1 - (probability of NOT getting dealt a second royal (of any suit) in the next 17,000 hands) = 1 - ((649,739/649,740)^17,000) = 1 - .97417 = .02582 = 2.582%

    P. S.: "Getting dealt" means first five cards, no draw. I am using Auggie's assumption of 850 hands per hour, or 17,000 hands over 20 hours.
     
  9. mike_m235

    mike_m235 Tourist

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    And when you look at the more accurate calculation and see that it's within a couple hundredths of a percent of auggie's simple division, you get what I told you...which is that it approaches accurate.

    But both you and Auggie basically calculated for the odds of a single event based on the first one already happening. Calculating the exact odds of exactly two of those events in a given number of hands is more complicated.

    For example, if you left tomorrow, going to vegas, and said you were going to play 12,000 hands, your odds of being dealt two royal flushes in those hands is about one in 35 million...which is what I showed above.
     
  10. Nevyn

    Nevyn VIP Whale

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    That's a really simple calculation, but still not exactly the original question.

    Given 17,000 hands (we'll stick with that assumption), what are the odds of getting dealt two royal flushes of the same suit (for fun we'll make it 2 or more, not exactly 2) within those hands. So you have to get the first one within those hands and then get the next to hit within however many you have left.
     
    Last edited: Feb 7, 2013
  11. mike_m235

    mike_m235 Tourist

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    Being dealt two royals (at least) in 17,000 hands is .000666667, or .067%

    For having it happen in the same suit the math is easy...just divide by four.
    Because although you have to count on the odds to hit the first royal, you don't have to account for suit on the first one, because regardless of the suit of the first, you just have to match it.

    So in the same suit, it's .0001667, or .017%
     
  12. Nevyn

    Nevyn VIP Whale

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    Weird. I got something slightly different.

    Trying to figure out why. Its been awhile since I did this type of thing.

    What I come out with is .008538013% chance of it happening (about half as likely as you got).
     
  13. mike_m235

    mike_m235 Tourist

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    To get the odds of it happening at least twice, you just square the results of it happening once. To get the results of it happening exactly twice, you have to square the results of it happening once then subtract the odds of it happening more than twice (which in this case, wouldn't change it much since the odds of it happening more than twice are so small.)
     
  14. Nevyn

    Nevyn VIP Whale

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    Are you sure about that?

    Squaring the odds would give you the chances that you hit at least one royal in 17000 hands, and then the odds that you would again hit one royal in another (different) 17000 hands, wouldn't it? And that's a different thing, right? Maybe I'm just going nuts.


    Example with a coin flip:

    The chance that I flip a coin at least one in 3 tries? 87.5%

    But the chance that I flip it at least twice in 3 flips isn't .875^2=76.5625%. Its actually 50%.

    You take the odds of heads on the first flip (.5) multiplied by the odds of heads on at least one of the next two (.75), and then add the odds of tails on the first flip multiplied by the odds of heads twice in a row (for when you get there by getting it on the last 2 flips).

    .5*.75 = .375
    .5*.25 = .125

    add them for .500

    I came out with my solution using a spreadsheet, basically applying the above method, but with 16999 rows (seperate odds) added up. But I screwed it up, too and will need to redo it. I forgot one important element (discounting the chances that you've already hit the first, because otherwise you double count). I'll update tomorrow with the new number assuming no one has corrected me since.
     
    Last edited: Feb 7, 2013
  15. 4Eyes

    4Eyes Low-Roller

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    Nevyn and Mike,

    Yes, I misread the question and assumed the first royal was a given. It's been a while since I've used the Binomial Theorem, and Excel can't calculate 17,000! so, for now at least, I will punt on the precise calculation and let you guys figure it out. :)
     
  16. mike_m235

    mike_m235 Tourist

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    Well I suppose I should subtract one hand from the second number...because you're talking about the same 17K hands, but if you subtract out the one that you hit, you have 16,999 possibilities in which to hit the next one.

    Your way is more accurate, treating everything as a separate event. But we've got to be close i think.
     
  17. Nevyn

    Nevyn VIP Whale

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    What I now get is 0.008464002% from the spreadsheet. I'll doublecheck my math in the morning.

    If people want to duplicate or check my methodology, here's what I have in Excel

    16999 rows

    column a is a countdown. row 1 is 16999, row 16999 is 1
    column b is a count up, matches row #
    column c is your formulas. row 1 formula is =(1/649740*(1-((2598959
    /2598960)^A1))). row 2 formula is =(1/649740*(1-((2598959
    /2598960)^A2))*((649739/649740)^B2)). Copy and paste that formula for remaining rows.

    Then sum column C. Multiply by 100 for %.

    Explanation

    1/649740 is odds of a royal in one hand. So first formula is odds your first hand is a royal, and you hit it again in the next 16999. Next formula takes odds second row is a royal times odds you get it again in next 16998 times odds you did NOT hit it on previous spins (to avoid double counting).

    This is usually the point that a true math expert joins and says "you didnt need to do that, he's your formula", but I think this is right.
     
  18. Kickin

    Kickin Flea

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    Nevyn you really went for the brute force method eh? You should be a password cracker. Its a binomial probability question with simplifies to the following:

    Using the assumption of 17k hands and 1/649740 prob of dealing a royal -

    17000C2*(1/649740)^2*(649739/649740)^16998 = 0.0333427%

    That means roughly 1/3000 sessions of 17,000 hands...or somewhere in about once in 51,000,000 hands for any royal.

    Divided by 4 for just a diamond is 0.00833% along the lines Nevyn got or once in about 204 million hands.
     
  19. Nevyn

    Nevyn VIP Whale

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    Hey, when you can't remember your college math, and 2 minutes of googling doesnt get it done, go nuts in Excel. With copy/paste it didnt take long

    edit: did some quick googling ... isn't that the formula for exactly 2? I went for 2 or more ... is there a simpler formula for that? I'll assume in the meantime thats why my # was slightly higher
     
  20. Kickin

    Kickin Flea

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    Yeah the formula I listed is for exactly 2 royals being dealt out of 17,000 hands. If you want a simplified method for calculating the probability of it happening at least twice. You can do it two ways off the top of my head, both using the binomial probability function:

    1) Calculate prob of happening 0 times and 1 time:
    P(0) = 17000C0*(1/649740)^0*(649739/649740)^17000 = 97.417%
    P(1) = 17000C1*(1/649740)^1*(649739/649740)^16999 = 2.548%

    Take 1 - [P(0) + P(1)] = 0.0336354%. Divide that by 4 for one suit and you get 0.008409%.

    Which is along the lines you got. The probability of it happening exactly twice and at least twice are not much different because it becomes so incredibly improbable for each subsequent one.

    2) The other way is to treat it as a simple summation of the binomial probability function from i = 2 to 16998.

    By the way, excel has the formula built in. To calculate the original question of exactly 2 royals dealt out of 17,000 hands type in the following:

    =BINOMDIST(2,17000,1/649740,0)
     
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