1. By continuing to use the site, you agree to the use of cookies .This website or its third-party tools use cookies, which are necessary to its functioning and required to achieve the purposes illustrated in the cookie policy.Find out more.
    Dismiss Notice
  2. Follow Casinomeister on Twitter | Facebook | YouTube | Casinomeister.us US Residents Click here! |  Svenska Svenska | 
Dismiss Notice
REGISTER NOW!! Why? Because you can't do diddly squat without having been registered!

At the moment you have limited access to view most discussions: you can't make contact with thousands of fellow players, affiliates, casino reps, and all sorts of other riff-raff.

Registration is fast, simple and absolutely free so please, join Casinomeister here!

Maths Geeks - assistance please :)

Discussion in 'Online Casinos' started by Nifty29, Feb 17, 2011.

    Feb 17, 2011
  1. Nifty29

    Nifty29 Dormant account

    Occupation:
    PAID CASINO SHILL
    Location:
    Turn right, then right. then right again
    Hey gang

    Question for the maths heads out there....

    In video poker, what are the odds of the following:

    1. You hold 2 to a RF, and hit only 2 of the required cards (e.g. no RF)

    1. You hold 3 to a RF, and hit only 1 of the required cards

    1. You hold 1 to a RF and hit only 3 of the required cards

    I've been curious for a while, but I also have a sneaking suspicion about something and would like to possess some facts :)

    Thankyou in advance!!
     
  2. Feb 17, 2011
  3. 3Dice

    3Dice I-Gaming Industry Representative

    Occupation:
    -
    Location:
    -
    Hi Nifty,

    holding 3 cards to royal flush, drawing 2 out of 47 remaining cards :

    odds RF : 1 / ( 47*46/ 2 ) = 1 / 1081 = 0.09 %

    odds one of : 2 * 45 / ( 47*46 / 2 ) = 90 / 1081 = 8.3 %

    magnitude .. 1 in 10

    holding 2 cards to royal flush, drawing 3 out of 47 remaining cards :

    odds RF : 1 / ( 47*46*45/ (3*2) ) = 1 / 16215 = 0.006 %

    odds one of : 3 * 44 / ( 47*46*45 / (3*2 ) ) = 132 / 16215 = 0.81 %

    magnitude .. 1 in 100

    holding 1 card to RF, drawing 4 out of 47

    odds RF : 1 / ( 47*46*45*44/(4*3*2)) = 1 / 178365 = 0.00056 %

    odds one of : 4 * 43 / ( 47*46*45*44/(4*3*2)) = 172 / 178365 = 0.09 %

    magnitude .. 1 in 1000

    Cheers,

    Enzo
     
    6 people like this.
  4. Feb 17, 2011
  5. 3Dice

    3Dice I-Gaming Industry Representative

    Occupation:
    -
    Location:
    -
    Was just rereading this .. and realized I'm a geek :D

    let me try and put it in a more understandable way.

    If you hold 3 cards of a RF in first deal, then you should expect to see 90 'near RF's' for every RF (which you should expect 1 in 1081 after holding 3 RF cards).

    If you hold 2 cards of a RF in first deal then you should expect to see 132 'near RF's' for every RF (which you should expect 1 in 16215 after holding 2 RF cards)

    if you hold 1 card of a RF in first deal then you should expect to see 172 'near RF's' for every RF (which you should expect 1 in 178365 after holding 1 RF card)

    So in general expect to be seeing the 'near RF' 100 or even 200 times before you get the full RF.

    Cheers,

    Enzo
     
  6. Feb 17, 2011
  7. rainmaker

    rainmaker I'm not a penguin CAG webmeister

    Occupation:
    -
    Location:
    -
    I bet that someone here did well in school :rolleyes:

    Good replies Mister Enzo.
     
    1 person likes this.

Share This Page