Odds Question

Dan808Dan808
in Off Topic Lounge
just a quick search turned up no answers...i don't think...
in a 6 player game, what are the odds of getting an Ahigh straight flush?
if u want, i wouldn't mind knowing about getting one on the turn, since only 1 hole card contributes to the straight.

ok, thanks.

Comments

  • MickeyHoldemMickeyHoldem
    Dan808 wrote:
    just a quick search turned up no answers...i don't think...
    in a 6 player game, what are the odds of getting an Ahigh straight flush?
    if u want, i wouldn't mind knowing about getting one on the turn, since only 1 hole card contributes to the straight.

    ok, thanks.

    The odds of you getting the royal flush don't change whether there are 6 opponents, 10 or just 1.

    Royal odds by RIVER using 1 hole card = (4 choose 4) * 46 / (50 choose 5) = 0.0000217108 = 1 in 46060

    Royal odds by TURN using 1 hole card = (4 choose 4) / (50 choose 4) = 0.0000043422 = 1 in 230300

    If both your hole cards can make a possible royal (if you hold KJo) then it is twice as likely... 1 in 23030 and 1 in 115150 respectively.
  • ScottyZScottyZ
    Royal odds by RIVER using 1 hole card = (4 choose 4) * 46 / (50 choose 5) = 0.0000217108 = 1 in 46060

    Royal odds by TURN using 1 hole card = (4 choose 4) / (50 choose 4) = 0.0000043422 = 1 in 230300

    Don't forget that these computations assume that you play every hand you are dealt to the river (respectively, the turn).

    Since in real poker games you fold many of your hands, the probabilities of such events happening in a real poker game are lower.
    The odds of you getting the royal flush don't change whether there are 6 opponents, 10 or just 1.

    Again, assuming that you play every hand you're dealt to the river (resp. turn).
    If both your hole cards can make a possible royal (if you hold KJo) then it is twice as likely... 1 in 23030 and 1 in 115150 respectively.

    Having two Royal hole cards greatly improves the chances of making a Royal (once again, running the hand hot & cold to the river/turn). The probabilities should be far greater than just double the previous figures.*

    For example, the probability of making a Royal by the river (playing hot & cold) beginning with two Royal cards is:

    (3 choose 3) * (47 choose 2) / (50 choose 5)

    = 1 * 1081 / 2,118,760

    = 0.00051

    or around 1960 to 1.

    ScottyZ

    *Similarly, beginning the hand with s000ted hole cards increases the probability of making a (regular) flush by the river by far greater than a factor of 2.
  • boxcardboxcard
    I would like to point out that the calculations performed to this point on this thread ASSUME THAT THE PLAYER ALREADY HAS THE APPROPRIATE HOLE CARDS TO WORK FROM.
    just a quick search turned up no answers...i don't think...
    in a 6 player game, what are the odds of getting an Ahigh straight flush?

    I believe the proper way to calculate this would be to consider there are 1326 different combinations of hole cards, 19600 different combinations of flops, 47 different turns, and 46 different rivers. This means there are 1326*19600*47*46=56,189,515,200 different possible outcomes.

    If we work with the idea of finding the probability of getting a royal flush in one particular suit, and then multiply our results by 4 to take in each of the 4 suits, the following cases can be found:

    1) Royal in 5 cards: 2 hole, 3 flop, 0 turn, and 0 river

    =(5C2/1326)*(3C3/19600)*(47C1/47)*(46C1/46)*4 suits= 86480/56189515200

    2) Royal in 6 cards: 2 hole, 2 flop, 1 turn, and 0 river

    =(5C2/1326)*(3C2*47C1/19600)*(1C1/47)*(46C1/46)*4 suits= 259440/56189515200

    3) Royal in 6 cards: 1 hole, 3 flop, 1 turn, and 0 river

    =(5C1*47C1/1326)*(4C3/19600)*(1C1/47)*(46C1/46)*4 suits= 172960/56189515200

    4) Royal in 7 cards: 2 hole, 2 flop, 0 turn, and 1 river

    =(5C2/1326)*(3C2*47C1/19600)*(46C1/47)*(1C1/46)*4 suits= 259440/56189515200

    5) Royal in 7 cards: 2 hole, 1 flop, 1 turn, and 1 river

    =(5C2/1326)*(3C1*47C2/19600)*(2C1/47)*(1C1/46)*4 suits= 259440/56189515200

    6) Royal in 7 cards: 1 hole, 3 flop, 0 turn, and 1 river

    =(5C1*47C1/1326)*(4C3/19600)*(46C1/47)*(1C1/46)*4 suits= 172960/56189515200

    7) Royal in 7 cards: 1 hole, 2 flop, 1 turn, and 1 river

    =(5C1*47C1/1326)*(4C2*46C1/19600)*(2C1/47)*(1C1/46)*4 suits= 259440/56189515200

    8) Royal in 7 cards: 0 hole, 3 flop, 1 turn, and 1 river

    =(47C2/1326)*(5C3/19600)*(2C1/47)*(1C1/46)*4 suits= 86480/56189515200

    These calculations give us the following results:

    Odds of getting a royal in exactly 5 cards = 649739:1

    Odds of getting a royal in exactly 6 cards = 129947:1

    Odds of getting a royal in at least 6 cards = 108289:1

    Odds of getting a royal in exactly 7 cards = 54144:1

    Odds of getting a royal in at least 7 cards = 36095.6...:1
    Don't forget that these computations assume that you play every hand you are dealt to the river (respectively, the turn).

    Since in real poker games you fold many of your hands, the probabilities of such events happening in a real poker game are lower.


    Quote:
    The odds of you getting the royal flush don't change whether there are 6 opponents, 10 or just 1.



    Again, assuming that you play every hand you're dealt to the river (resp. turn).

    These facts are VERY important to consider.
  • MickeyHoldemMickeyHoldem
    ScottyZ wrote:
    MickeyHoldem wrote:
    If both your hole cards can make a possible royal (if you hold KJo) then it is twice as likely... 1 in 23030 and 1 in 115150 respectively.
    Having two Royal hole cards greatly improves the chances of making a Royal (once again, running the hand hot & cold to the river/turn). The probabilities should be far greater than just double the previous figures.

    For example, the probability of making a Royal by the river (playing hot & cold) beginning with two Royal cards is:

    (3 choose 3) * (47 choose 2) / (50 choose 5)

    = 1 * 1081 / 2,118,760

    = 0.00051

    or around 1960 to 1.
    Just to clarify... I was calculating the chance that you could make a royal using only 1 hole card, but allowing for the fact that you could use either of them. Your calculation is obviously correct for suited cards.
    These calculations give us the following results:

    Odds of getting a royal in exactly 5 cards = 649739:1

    Odds of getting a royal in exactly 6 cards = 129947:1

    Odds of getting a royal in at least 6 cards = 108289:1

    Odds of getting a royal in exactly 7 cards = 54144:1

    Odds of getting a royal in at least 7 cards = 36095.6...:1
    I get some different answers here:
    Exactly 5 = (4 / 52c5) = 1 in 649740 (same)
    Exactly 6 = (4*47 / 52c6) - (4 / 52c5) = 1 in 129948 (same)
    Exactly 7 = (4*47c2 / 52c7) - (4*47 / 52c6) - (4 / 52c5) = 1 in 43316 (different)

    Atleast 6 = (4*47 / 52c6) = 1 in 108290 (same)
    Atleast 7 = (4*47c2 / 52c7) = 1 in 30940 (different)
  • boxcardboxcard
    I get some different answers here:
    Exactly 5 = (4 / 52c5) = 1 in 649740 (same)
    Exactly 6 = (4*47 / 52c6) - (4 / 52c5) = 1 in 129948 (same)
    Exactly 7 = (4*47c2 / 52c7) - (4*47 / 52c6) - (4 / 52c5) = 1 in 43316 (different)

    Atleast 6 = (4*47 / 52c6) = 1 in 108290 (same)
    Atleast 7 = (4*47c2 / 52c7) = 1 in 30940 (different)

    My bad. Thanks Mickey. My error was in calculating #7. the formula is correct, but the result is 518880/56189515200. I stand corrected.
  • ScottyZScottyZ
    Just to clarify... I was calculating the chance that you could make a royal using only 1 hole card, but allowing for the fact that you could use either of them.

    I see what you meant. That's certainly true that the chances are doubled when you have two separate one-card draws.

    ScottyZ
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