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Thread: Wolverine: probability question

  1. #1
    Wolverine
    Guest

    Wolverine: probability question

    Okay, I think I know how to do this, but I'm going to ask the professionals to be sure I get the CORRECT answer rather than my INCORRECT answer (and I think I can only do the first part of it, besides).

    Here goes: recreational player, 2 years of playing experience. Absolutely no posibility of more than 100 hours of playing time in those 2 years, and probably closer to 50 hours. Plays at moderately full shoe games, so "100 hands an hour" would be a definite no-no. I believe 60 hands an hour would be fair. Given that range, my best guess is that this person has played anywhere from 3000 to 6000 hands of blackjack (60 x 50 to 60 x 100). Feel free to provide your own more accurate figures if you think you can.

    This same person has been the (un)lucky witness to the dealer pulling 3 blackjacks in a row and now more recently, FOUR naturals in a row. The 3 BJs came in a 6D shoe game (no recollection of how many other players), the 4 BJs came during a double deck game (with one other player).

    I think the math goes like this---help me here---blackjack occurs about 1 in every 21 hands. 4.8% I seem to recall from some text, but I'll let the experts present and then use their number for their calculations. So, to see 3 BJs come at you in a row, the chances of a 1 in 21 event occuring 3 times in a row is 21 x 21 x 21 or 1 in 9,261. I was told earlier, and obviously this would be true, if you play enough blackjack, you're going to see a dealer or player receive 3 blackjacks in a row. In order to get 4 in a row, the number above gets multiplied again by 21 to get to the highly unlikely odds of 1 in 194,481. Please confirm or correct these calculations as needed.

    Now, given the original premise of the player being a "newbie" and playing so "few" hands relative to most people, what is the likelihood of seeing BOTH of these events in such a short span by the same player? Furthermore, for extra credit, since the 3 BJs in a row happened twice (the first time, and then the more recent time, in which the string continued on to a fourth sighting--what are the odds on seeing those same events (3 BJs in a row, twice)?

    Last piece to the puzzle: they were both at the same casino. Luckiest dealers ever?

    I can't wait for the "voodoo" howls to come in from the usual suspects regarding this, but c'mon...this is beyond the "pretty rare." Don't you agree?

    Thanks in advance for your thoughts....I can't wait to read them.

  2. #2
    Don Schlesinger
    Guest

    Don Schlesinger: Amended answer

    I posted something earlier that was wrong. I must have hit some wrong keys on my calculator. Below is the corrected version.

    Sorry.

    > Given that range, my best guess is that this
    > person has played anywhere from 3000 to 6000 hands of
    > blackjack (60 x 50 to 60 x 100). Feel free to provide
    > your own more accurate figures if you think you can.

    Let's settle on one number, say 5,000 hands.

    > This same person has been the (un)lucky witness to the
    > dealer pulling 3 blackjacks in a row and now more
    > recently, FOUR naturals in a row. The 3 BJs came in a
    > 6D shoe game (no recollection of how many other
    > players), the 4 BJs came during a double deck game
    > (with one other player).

    Probabilities for a natural vary by number of decks shuffled, as you know. For simplicity, let's just use 1 in 21 for all the calculations.

    > I think the math goes like this---help me
    > here---blackjack occurs about 1 in every 21 hands.

    Close enough. See above.

    > 4.8% I seem to recall from some text, but I'll let the
    > experts present and then use their number for their
    > calculations.

    4.78% (1 in 20.92, for DD); 4.75% (1 in 21.06, for 6-deck).

    > So, to see 3 BJs come at you in a row,
    > the chances of a 1 in 21 event occuring 3 times in a
    > row is 21 x 21 x 21 or 1 in 9,261. I was told earlier,
    > and obviously this would be true, if you play enough
    > blackjack, you're going to see a dealer or player
    > receive 3 blackjacks in a row.

    All right so far.

    > In order to get 4 in a
    > row, the number above gets multiplied again by 21 to
    > get to the highly unlikely odds of 1 in 194,481.
    > Please confirm or correct these calculations as
    > needed.

    Right so far.

    > Now, given the original premise of the player being a
    > "newbie" and playing so "few"
    > hands relative to most people, what is the likelihood
    > of seeing BOTH of these events in such a short span by
    > the same player?

    Highly unlikely! :-) But, it happened. Go figure.

    > Furthermore, for extra credit, since
    > the 3 BJs in a row happened twice (the first time, and
    > then the more recent time, in which the string
    > continued on to a fourth sighting--what are the odds
    > on seeing those same events (3 BJs in a row, twice)?

    You're not being clear here. Did you see two separate streaks of three naturals in a row AND another of four, or did you see three in a row once and four in a row once? If the latter, then why call it three in a row twice?

    > Last piece to the puzzle: they were both at the same
    > casino. Luckiest dealers ever?

    A bit strange.

    > I can't wait for the "voodoo" howls to come
    > in from the usual suspects regarding this, but
    > c'mon...this is beyond the "pretty rare."
    > Don't you agree?

    The four in a row, given only 5,000 hands total is, obviously, pretty rare. I'm not at all sure if this is the correct way to calculate this, but, say you have 5,000 hands. You can start your streak of four naturals on any hand from 1 to 4,997. And, let's say we're trying to avoid the streak of 4, which means that, each time, we have, roughly, 191,999/192,000 of "succeeding." So, to succeed 4,997 times in a row, we raise 191,999/192,000 to the 4,997th power. Then, we subtract that answer from 1, to see what the chances are for four naturals in a row, with only 5,000 hands as our sample space. We get, roughly, 2.6%, or 1 chance in about 39.

    > Thanks in advance for your thoughts....I can't wait to
    > read them.

    As you can see, not at all impossible.

    Don

  3. #3
    Don Schlesinger
    Guest

    Don Schlesinger: By the way

    If I'm doing the math correctly, the chance of seeing three naturals in a row, with 5,000 hands and a 6-deck game, is about 41.6%, or one chance in 2.4, so not at all uncommon.

    Don

  4. #4
    Wolverine
    Guest

    Wolverine: Thanks

    sorry I was unclear. I will try to clarify.

    The first instance was a streak of 3 dealer naturals, with a 6D shoe.

    The second instance was a streak of 4 dealer naturals, this time on a DD.

    I thought seeing 3 come up would be "rare" enough that I thought figuring witnessing two such streaks would be a "challenge" to believe, let alone having seen one continue on to a streak of 4.

    So if I am reading your answer right, 1 in 39 players in a subset of 5000 hands is going to witness a streak of 4 dealer blackjacks? How often would one person see a streak of 3 AND a streak of 4 in that same subset of 5000 hands?

    Thanks for the quick answer.

  5. #5
    Don Schlesinger
    Guest

    Don Schlesinger: Re: Thanks

    > sorry I was unclear. I will try to clarify.

    > The first instance was a streak of 3 dealer naturals,
    > with a 6D shoe.

    That's how I understood it.

    > The second instance was a streak of 4 dealer naturals,
    > this time on a DD.

    Right.

    > I thought seeing 3 come up would be "rare"
    > enough that I thought figuring witnessing two such
    > streaks would be a "challenge" to believe,
    > let alone having seen one continue on to a streak of
    > 4.

    Understood. To tell you the truth, I'm not happy with the 1 in 39 answer, although I don't see anything wrong with my math. The problem is that is simply doesn't jibe with my intuition. I can't believe that, on average, for every 50 hours of play (assuming 100 hands per hour, which is normal for me), I would witness the dealer getting four naturals in a row. I don't think I've ever seen a dealer get four naturals in a row in the 30 years that I'm playing the game!

    Something seems wrong.

    > So if I am reading your answer right, 1 in 39 players
    > in a subset of 5000 hands is going to witness a streak
    > of 4 dealer blackjacks?

    That's what I wrote, but I'm not sure I believe it.

    > How often would one person see
    > a streak of 3 AND a streak of 4 in that same subset of
    > 5000 hands?

    I don't know. There is some overlapping of the two sample spaces, so I don't think you can just multiply one by the other. I think it is a very, very rare event.

    > Thanks for the quick answer.

    Sorry I can't do better. Maybe someone can comment on my math.

    Don

  6. #6
    Wolverine
    Guest

    Wolverine: Glad I got you thinking!!!!! :-)

    > Understood. To tell you the truth, I'm not happy with
    > the 1 in 39 answer, although I don't see anything
    > wrong with my math. The problem is that is simply
    > doesn't jibe with my intuition. I can't believe that,
    > on average, for every 50 hours of play (assuming 100
    > hands per hour, which is normal for me), I would
    > witness the dealer getting four naturals in a row. I
    > don't think I've ever seen a dealer get four naturals
    > in a row in the 30 years that I'm playing the game!

    And this person plays with other players at the table (often full!) so I know that 100 hands per hour is too high as an estimate. But, hey, it makes the math easy and it is still such a rare event that 100 hands or 100,000 hands, it just shouldn't be an event a beginner has to live through once, let alone twice!

    > I don't know. There is some overlapping of the two
    > sample spaces, so I don't think you can just multiply
    > one by the other. I think it is a very, very rare
    > event.

    But if they didn't, wouldn't it just be the 9,261 x 194,481 = 1,801,088,541 or almost a 2 billion to one shot!?! [And how can this person manage to miss the Powerball numbers?!?] Wouldn't a divisor of 2 approximate the odds since there are 2 parallel subsets running? That still makes it a 900,000,000-to-1 shot at seeing both events in a lifetime. I'm guessing that I am wrong about that, because that number seems ridiculous compared to the 39-to-1 odds per 5000 hands you are talking about.

    I was not at the same table at the same time, but I have also seen 3 naturals in a row from a dealer within the last 100 hours of play (and I'm at about 100 hands per hour). Again, all these have happened in the same casino.

    Thanks for the insights.


  7. #7
    Don Schlesinger
    Guest

    Don Schlesinger: Not so rare

    I wrote, above:

    > Understood. To tell you the truth, I'm not happy with
    > the 1 in 39 answer, although I don't see anything
    > wrong with my math. The problem is that is simply
    > doesn't jibe with my intuition. I can't believe that,
    > on average, for every 50 hours of play (assuming 100
    > hands per hour, which is normal for me), I would
    > witness the dealer getting four naturals in a row. I
    > don't think I've ever seen a dealer get four naturals
    > in a row in the 30 years that I'm playing the game!

    > Something seems wrong.

    But, when you do the math, it turns out that, playing 10,000 hands of BJ a year, the chance that I would never see such a streak is still a hefty 21% -- not so rare at all. And, maybe my memory is faulty, and I actually have seen such a streak.

    Don

  8. #8
    Don Schlesinger
    Guest

    Don Schlesinger: Correction from Stewart Ethier

    Still fearing that I had done something wrong (I have learned never to doubt my intuition in blackjack matters!), I dashed off an e-mail describing the problem to our Masters and to Stewart Ethier, one of our foremost probabilists.

    His answer to me (below) confirmed my suspicions that I had done something wrong:

    "Hi Don,

    "There is a flaw in your reasoning that may be responsible for the unexpectedly high probability you calculated. What you are seeking is the probability of a run of 4 consecutive successes in 5000 independent trials, each with success probability p=0.0478. There is a (complicated) formula for just this scenario due to de Moivre. It is in Uspensky's 1937 textbook but is not found much in modern textbooks.
    I don't have it handy at the moment, but will look it up tonight or tomorrow and see if it is computable with r=4, n=5000, and p=0.0478.

    "The flaw in your argument is easily explained. Let E_n be the event consisting of 4 consecutive successes starting with the nth trial. Clearly, P(E_n)=p^4 as you observed. But then you argued that P(E_n for some n=1,2,...,4997)=1-P(E_n for no
    n=1,2,...,4997)=1-(1-P(E_n))^4997=0.0257. The next-to-last step assumes that E_1,E_2,... are independent, which is not the case. For example, to see that E_1 and E_2 are not independent, we observe that
    P(E_1 and E_2)=p^5, while P(E_1)P(E_2)=p^8.

    "I'll write again once I look up the correct formula.

    "Stewart"

    I'd like to add, in layman's terms, where my mistake was, for those of you who may not have understood everything above. I assumed that you could treat, say, hands 1-4 as one separate event, and then, say, hands 2-5 as yet another separate (independent) event. But that makes no sense, for, if hands 1-4 do not produce four consecutive naturals, then it is clearly impossible that hands 2-5 produce the desired streak. In short, the results of hands 2-5 depend (are not independent of) on the results of hands 1-4, and so my methodology breaks down.

    My sole consolation is that the formula for finding the correct answer is, apparently, a very complicated one. Now that I think of it, I seem to remember that it could once be found on Richard Reid's bjmath.com site, but I just looked there and no longer can find it. I assume that it was taken down.

    And so, we shall await Stewart's further communication.

    Don

  9. #9
    Sun Runner
    Guest

    Sun Runner: Re: Correction from Stewart Ethier

    > "The flaw in your argument is easily explained.

    Of course it is; come on Mr. S.

    > Let E_n be the event consisting of 4 consecutive
    > successes starting with the nth trial. Clearly,
    > P(E_n)=p^4 as you observed. But then you argued that
    > P(E_n for some n=1,2,...,4997)=1-P(E_n for no
    > n=1,2,...,4997)=1-(1-P(E_n))^4997=0.0257. The
    > next-to-last step assumes that E_1,E_2,... are
    > independent, which is not the case. For example, to
    > see that E_1 and E_2 are not independent, we observe that
    > P(E_1 and E_2)=p^5, while P(E_1)P(E_2)=p^8.

    See; a piece of cake. I observed that right off the bat. You thought E_1, E_2 were independent? Oh brother. Have you considered night school maybe?



    > "I'll write again once I look up the correct formula.

    Truthfully some days I just feel more stupid than others. I am able to reclaim some fractional self-worth just knowing even this guy had to do some homework!

  10. #10
    Wolverine
    Guest

    Wolverine: Thank you

    Thanks for taking this further. We are very interested in the answer. For a moment, let us consider that perhaps this event (4 naturals) is very, very rare (like we said, 1 in 194,481 rare) and 3 naturals don't happen all that often (roughly 1 in 10,000). Should we be considering the thought that the casino is cheating somehow?

    Looking forward to the new answer.

    Also, has anyone else following this thread every seen, received, or been the victim of 4 naturals in a row (or even 3)?

    Thanks

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