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Table Games OK you math experts, help me out...

Discussion in 'Table Games' started by Reverb, Dec 22, 2014.

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  1. Reverb

    Reverb Tourist

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    I like to bet $50 a hand at blackjack. Every time I win I press my bet by $25. Whenever I lose I revert back to the $50 base bet. So...what's my average bet? I could have figured this out once upon a time but my math isn't what it used to be. Thanks!
     
  2. Reverb

    Reverb Tourist

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    And let me add that I'm assuming for simplicity sake that I win 50% of hands and lose 50%.
     
  3. stackinchips

    stackinchips High-Roller

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    It's impossible to answer. Even if we assume it's a 50/50 proposition, the distribution of your winning hands and losing hands isn't even. In other words, you don't win a hand, lose a hand, rinse lather repeat. If you did, your average would be 62.50. However, you could win 20 hands in row (which your average would be 287.50). You could also lose every hand you play and your average would be $50.

    Reality is that your average is going to be more than $50, but how much more is entirely dependent on the distribution of possible outcomes.
     
  4. tringlomane

    tringlomane STP Addicted Beer Snob

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    Your average answer is dependent on where you stop loss/stop win points are. Also how long you're playing before quitting.
     
  5. Kickin

    Kickin Flea

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    It's definitely possible to answer if we assume you mean over time what does you average converge to. When I was in school I know I would have been able to do this easily, but that education is wasted now for practical purposes.

    I think there are a few ways to look at it. You can look at it as a Markov chain for probabilities of different sequences of wins in a row, or you can look at it as a set of negative binomial probabilities and you take the average across them. But to work with those I'd need to refresh myself on them.

    Since you're using 50/50 I think it converges quite simply to $100. I'm taking a simplistic approach and my head is tired so I may be overlooking something obvious, but since every loss resets to $50 I am taking the average as $50 + $25*prob(n wins).

    Its a summation that would go:

    50 + Σ[n=0 to ] (25n)*(0.5^n) = 100

    I don't have complete confidence in this answer though and won't take offence if you punch holes in it.
     
  6. Labeeb

    Labeeb Low-Roller

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    Kudos to you for not only the answer (or attempt thereof), but also the ability for get sigma and infinity into a board post. I haven't a clue about your equation, but I was impressed by those symbols. Well done!
     
  7. Snidely

    Snidely VIP Whale

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    It's not n wins that count but the streaks of n wins. Don't you have to calculate the probability of streaks and integrate them 0 to infinity?
     
  8. mrem3200

    mrem3200 High-Roller

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    I thought that at first too (but you would also need to consider the streaks of losses), but I don't think it matters since each hand is independent of the one before (so you don't consider the probability of streaks) and assuming a 50-50 chance, the losing streaks of 4 in a row (or whatever) and keeping the bet flat would offset the winning streaks that would increase the bet. So as simple as it seems assuming 50-50, the average bet would be $62.50. It has been a looooong time since my last STAT class so feel free to tell me I don't know what I am talking about

    Edit: It would be higher than $62.50 since you do have to factor in the probability of upside streaks but the losing streaks keep the bet at $50. But I don't think it would be much higher since the probability goes down the longer the winning streak is.
     
    Last edited: Dec 22, 2014
  9. Kickin

    Kickin Flea

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    The summation only adds sequences, since each loss just resets it to $50. So its like saying half your wins will be single wins, then a quarter will be two in a row, an eighth will be 3 in a row, etc... and adding those up.

    But like I said I'm not sure. I can see (theoretically) at least two precise ways to do it through a series summation of a Markov chain or a negative binomial distribution. And integrating some distribution probably makes sense too. But I'd need to school myself again to start doing any of this stuff and years of dealing with spreadsheets has pretty much fried my brain. :faint:
     
  10. Snidely

    Snidely VIP Whale

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    I'm in the same boat as you. I majored in stats but now I'm known as the Excel guru at work. I full agree with the bolded part you wrote above. If that is what your integral does then I think it's correct.
     
  11. TIMSPEED

    TIMSPEED !địt mẹ!

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    In the real-world..you'd hope you get $100 average...but more likely a $75 average...although if you have a dick for a floor supervisor, then you may only get a $50 average...
     
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  12. Aces and Eights

    Aces and Eights VIP Whale

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    I may be incorrect, but I think that the equation takes into account win streaks at 0.5^n probability without taking into account losing streaks which would also occur at 0.5^n probability. This may lower the average bet closer to $75.
     
  13. Snidely

    Snidely VIP Whale

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    I ran a simulation and got $70 so I think you're right; the answer is about $75.
     
  14. Kickin

    Kickin Flea

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    You're right. I was treating it as a $50 average bet plus the weighted average of the increase in his bet for different win streaks, which ends up not taking into account losing streaks pulling the average down. Instead of adding $50 to the summation you should add $25, since that's the probability weighted average of wagers when you lose.

    I just did a quick excel simulation as well with 10,000 trials and it also comes to around $75. Should have just done that from the beginning. :eek:
     
  15. Snidely

    Snidely VIP Whale

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    I ran some simulations with 10,000 hands and also get $75. I'm confident that $75 is the answer.
     
  16. tringlomane

    tringlomane STP Addicted Beer Snob

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    For the 50/50 scenario, $75 makes sense. It will obviously be slightly lower for an actual blackjack game though.
     
  17. nostresshere

    nostresshere Mr. Anti Debit Card

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    All of the answers only are valid if your get the normal "probability" on win or loss streaks.

    Since nobody can calculate your actual win/loss streaks, there is no answer.
     
  18. Snidely

    Snidely VIP Whale

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    Of course it's all presumed to be based on probability. But there is a >95% probability that if >10000 hands were played using this betting method that the average bet would work out to between $70 and $80. As the number of hands approaches infinity, the average bet approaches $75.
     
  19. Malibugolfer

    Malibugolfer High-Roller

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    Why do I have the funny feeling that there are not a whole lot of pit critters who will use Kickin's formula? Just a wild guess on my part.
     
  20. nostresshere

    nostresshere Mr. Anti Debit Card

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    Sure, I would buy that. How many hands do we think the poster can play? Limited by ability to sit there, and funds of course.
     
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