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Probability of streaks

  1. Sep 17, 2006 #1
    hi everybody, I have no idea how to figure this out and would really appreciate some help. Ok I'll try and explain this as best i can, and if I miss anything or you need more information let me know and I'll try and explain more. Suppose you have 37 balls, 24 blue and 13 red. after each pick the chosen ball gets put back in the bag so it can be picked again. So the probability of drawing 3 red balls in a row is roughly 4%. now suppose you pick 3 red balls,then a blue ball, 3 more red balls,then a blue ball (it doesn't matter how many blue balls, i just picked 1 to make it simple, it could be 10 or 40, it doesn't matter), and then 1 red ball. If the next ball picked is blue, i win 1 point. But if the next ball picked is red, i lose 2 points. I know that the chance of getting another red ball is less than 1 in 1,000. What i'm trying to figure out though is if this situation comes up more than once in a thousand ball picks, because it would have to come up at least 3 times in order for me to win consistently (if it comes up once i lose a point, twice i break even, 3 times win). Anyone know how to figure this out?
    Thanks,
    Tim
     
  2. jcsd
  3. Sep 17, 2006 #2

    CRGreathouse

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    I don't understand the situation at all. What is it, exactly, that makes you win or lose? What function does the picking of 3 red balls, then a blue ball, 3 more red balls,then X blue balls, then 1 red ball serve?

    If the question is, "How likely am I to pick the sequence RRRBRRRB(B*)RB" that's one answer. If it is "How likely am I to pick the sequence B having already picked RRRBRRRB(B*)R" then it's the same as picking B alone, since you're drawing with replacement.
     
  4. Sep 17, 2006 #3
    My question is how likely is it to pick B after this
    streak: RRR(b*)rrr(b*)r. It has to be this streak, the
    next blue ball isn't treated as an independent event.
    So after this streak comes up, if the next ball is red,
    I win 1 point, if it's blue i lose 2. In other words,
    when this situation comes up I need to win twice for every
    1 loss just to have a score of zero, 3 times to have won.(3 wins minus loss of 2=1)
    What I don't know is how important the number of ball picks
    are, and how to find out. Ex. suppose this streak RRR(b*)rrr(b*)r
    comes up only twice every 1,000 picks. If that were the case
    then I would ultimately lose I think, because the chance
    at this point for another blue ball is almost 99%? Does it make
    any sense now CR?
     
  5. Sep 19, 2006 #4

    CRGreathouse

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    You're replacing the ball, so it is an independant event. The balls don't 'remember' if they were pulled out before.

    Let x = 24/37. Then the chance of

    RRR(B+)RRR(B+)R : [itex]x^5(1-x)^2\approx1.42\%[/itex]
    RRR(B+)RRR(B+)RR: [itex]x^6(1-x)^2\approx0.92\%[/itex]
    RRR(B+)RRR(B+)RB: [itex]x^5(1-x)^3\approx0.50\%[/itex]
    RRR(B*)RRR(B*)R : [itex]x^5\approx11.5\%[/itex]
    RRR(B*)RRR(B*)RR: [itex]x^6\approx7.45\%[/itex]
    RRR(B*)RRR(B*)RB: [itex]x^5(1-x)\approx4.03\%[/itex]

    where (B+) means 1 or more blue balls and (B*) means 0 or more blue balls. Even though I don't seem to understand you, this should answer your question.
     
    Last edited: Sep 19, 2006
  6. Sep 19, 2006 #5
    That would answer my question, except I don't understand why it would be an independant event. Because I'm not just trying to find the next ball, it has to be all the balls I said in that order, so why would that be independant of the past results? I know the number of balls doesn't change, but isn't each pick in this case dependant on the last one? I mean if I get 2 red balls and then a blue,it doesnt count for anything, i have to wait until 3 red 1 blue....
    Correct me if I'm wrong on this next example. But say you flip a coin 10 times and it comes up heads every times. Now for the next flip, you have a 50/50 chance to get heads. But when you're asking after every flip what my chance is to get heads on each flip for the next 10 flips,first flip is 50%, next flip is 25%...it goes down. That's the same thing here I think.
     
    Last edited: Sep 19, 2006
  7. Sep 19, 2006 #6
    Let x = 24/37. Then the chance of


    RRR(B+)RRR(B+)RR: [itex]x^6(1-x)^2\approx0.92\%[/itex]
    RRR(B+)RRR(B+)RB: [itex]x^5(1-x)^3\approx0.50\%[/itex]



    Also, why is the chance of getting RRR(b+)RRR(B+)RR
    greater than the chance of getting RRR(B+)RRR(B+)RB, since there are more blue balls than red by almost double?

    Thanks for your help by the way, sorry I don't make much sense.
     
    Last edited: Sep 19, 2006
  8. Sep 19, 2006 #7

    CRGreathouse

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    With a fair coin, the chance to get heads is 1/2. The chance to get 10 heads in a row is 1/1024. The chance to get heads, given that your last 9 flips are heads, is 1/2.

    Sorry, I flipped the probabilities. Make x = 13/37 and adjust the percentages accordingly.
     
  9. Sep 19, 2006 #8
    So if you agree that the chance of getting 10 heads in a row is 1 in 1024, wouldn't it stand to reason the same logic can somwhow be applied to my question?
     
  10. Sep 19, 2006 #9
    Does the day of the week affect the probability of picking a red ball out next? Does the weather affect it? Why not? Because those don't change the scenario: you're still drawing out of a bag of 37 balls. Now, ten times, you draw a ball one at a time and put it back. Is your bag now, after you've drawn and replaced the balls, any different than it was when you made the first draw? Is it the case that you taking out and putting back in marbles changed the probability that the very next ball you pick out will be red? No! The bag is identical to the way it was before, just as when we started. It's still a bag of so-many reds and so-odd blues. You still have the same chance of pulling out a red as you did the first time you made a draw.

    The other example: the chance of getting 10 heads in a row on a coin is 1 in 1024. What if you've already rolled 1,414,512 tails: does that change the probability that the next time you flip a coin, you'll get heads? No! Nothing has changed about the coin!

    Thinking further: What if you didn't put back the balls which you had drawn? After ten draws, will you still have the same bag configuration?
     
    Last edited: Sep 19, 2006
  11. Sep 19, 2006 #10
    Well obviously not, however that's irrelevant to this example.



    While it's true nothing has changed about the coin, the odds have changed for the next heads if you're thinking about it in terms of past results: after 1,414,512 flips with the coin and all heads, what now (astronomical of course) is the chance of getting another heads. Yes, thinking about it independantly, the chance is still only 50/50 to get heads, even after that many heads in a row. I'm sure you know that's not the case if you count past flips though. And if you don't believe me, just try and flip 20 heads in a row, and tell me if it felt like you had a 50% chance of doing it. Try it twice, give yourself a 100% chance lol.
     
    Last edited: Sep 19, 2006
  12. Sep 19, 2006 #11
    And while you might expect that after 10 draws, you would have a ratio of about 1 red to 2 blues (13/37 beforehand, 10/30 after), that isn't the case, because of standard deviation.
     
    Last edited: Sep 19, 2006
  13. Sep 19, 2006 #12

    CRGreathouse

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    No. If you flip a fair coin 1414512 times and get heads each time, the chance of the next flip being heads is exactly 50%.

    I don't know what you mean. Counting past flips is irrelevant.

    I also don't know what experiment you suggest. Here's the 'correct' one: flip fair coins one by one until you get HHH. Whenever this happens, record the next flip. (Don't overlap these sequences!) Do this until you have recorded 10 flips. These are distributed exactly like 10 unconditional flips would be.

    I'm sure you know the chance of at least one success wih two independant 50% chances is 75%.
     
  14. Sep 19, 2006 #13

    CRGreathouse

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    No. The chance of RRR(b+)RRR(B+)R is similar, but the chance of the conditional probability RRR(b+)RRR(B+)RR given RRR(b+)RRR(B+)R is not.

    If you are counting the number of times RRR(b+)RRR(B+)RR comes up in a fixed interval, you need to find the (small) chance of it happening and work appropriately (depending on how you handle overlap and so forth). If you are simply drawing until it happens, and then looking at the next draw, then the simple probability 13/37 is enough.

    The chance of having 3 consecutive heads in 6 flips of a fair coin is 5/16. The chance of having 3 consecutive heads if you flip a fair coin until you get exactly 2 heads in a row and then flip once more is 1/2.
     
  15. Sep 20, 2006 #14
    Maybe I didn't word that right, I know while the odds don't "actually" change, it's still 50/50, but the chance of getting 1,414,513 heads after flipping 1,414,512 in a row heads is astronomical. That's what I meant by counting past flips. I know that an independant flip after this many is still 50/50, but that's not what i was saying.
     
    Last edited: Sep 20, 2006
  16. Sep 20, 2006 #15
    I was being sarcastic, because that punk 16 year old posts rude replies to other people, and he completely missed the point.
    And I was referring to the chance of getting 21 heads in a row, not 1 head twice.
     
  17. Sep 20, 2006 #16
    Ok, it finally makes sense, thanks for being patient. One more question which I think I alread know the answer to, but I just want to make sure. Suppose you have the streak we've been using: RRR(b*)RRR(B*)R, and the next ball is R. The chance of that R coming up was 13/37. Now, to find the odds of that same streak coming up again (not independantly, after this streak) you would do what?
    ex. RRR(B*)RRR(B*)RR(B*)RRR(B*)RRR(B*)R after RRR(B*)RRR(B*)RR just came up.
     
  18. Sep 20, 2006 #17

    CRGreathouse

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    It would be purely the probability of the remaining part coming up, to wit:
    RRR(B*)RRR(B*)RR(B*)RRR(B*)RRR(B*)R -
    RRR(B*)RRR(B*)RR =
    (B*)RRR(B*)RRR(B*)R

    Using r and b for their associated probabilities:
    (B*)R: 1
    RR: r^2
    (B*)R: 1
    RR: r^2
    (B*)R: 1

    Total probabilty: r^4.

    If you instead meant "one or more B" (B+), then it would be r^4 * b^2 = r^4(1-r)^2.
     
  19. Sep 20, 2006 #18
    Ya that's what I meant. That's everything I think.
     
  20. Sep 20, 2006 #19

    CRGreathouse

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    I'm glad I could help.
     
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