Need a Little Help With a Statistics Problem...

Discussion in 'General Discussion' started by UncleMorgan, Mar 12, 2017.


  1. UncleMorgan

    UncleMorgan I like peeling bananas and (occasionally) people.

    I'm fairly good with practical math, but never got much into Statistics.

    Now I have a dinky little problem that's driving me nuts.

    My grandson popped this one on me. It's part of his high school homework.

    I not only couldn't figure it out. I couldn't googleate it.

    It would have been a snap with coins, because everybody has stuff on the Net about coin probabilities.

    I'm hoping there's somebody out there in Monkeyland that can make the solution to this problem as simple as I wish it was. Here's the problem:

    To Sit With the Captain: A Ship’s Pool Question

    On a ship, a daily pool will be held each day for six days.

    Tickets will be sold for each pool.

    Each ticket buys one number in one daily pool.

    A passenger can pick the number they want in each pool.

    A passenger can buy as many tickets as they wish for each pool—but only before the ship sails.

    Whoever wins all six pools gets to sit at the Captain’s table for the rest of the voyage.

    The first pool has 23 numbers.

    The second pool has 27 numbers.

    The third pool has 33 numbers.

    On fourth pool has 34 numbers.

    The fifth pool has 29 numbers.

    The sixth pool has 20 numbers.

    Q1: What are the chances of a passenger winning all six pools with a total of six tickets?

    Q2: How many tickets would a passenger have to buy to be certain of winning all six pools?

    I am seriously monkey-stumped here. Any help would be much appreciated.
     
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  2. GrayGhost

    GrayGhost Monkey++

    It's way too early for this. Sorry...no help here.
     
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  3. chelloveck

    chelloveck Diabolus Causidicus

    It's not so much a statistical problem but a problem of probability....beyond that, I'm not so much of a help....

    but the following reference might be helpful to you and your grandson....

    All I can say, is that the Captain's table scam would be a good little earner for the Captain's retirement fund, and the Captain could be assured of a very small and quiet Captain's table. ;)

    I'm not much of a mathematician...or a gambler come to think of it, but the odds are going to be long of getting a seat at the captain's table.
     
  4. Airtime

    Airtime Monkey+++

    Chell is right, this is a probability question.

    However, you have two problems here actually. How the solve the math and how to teach the kid. Lets focus on the second.

    The easy way to think of probabilities is as a ratio or rather a fraction. The numerator is the number of desired outcomes possible and the denominator is the total number of outcomes possible.

    So, flipping a coin once and calling heads: n=1 and d=2 so probability of heads is 1/2.

    Roll a 5 on one die: n=1 and d=6 so answer is 1/6

    Roll a pair of dice (one red and one green) looking for a 11: n=2 [5 on red and 6 on green or roll a 6 on red and 5 on green] and d=36 (6 possibilities on red and for each of those there are 6 possibilities for the green for a total of 6x6 or 36). So probability of rolling an 11 is 2/36 or 1/18 or 5.55%.

    What are odds of rolling a 7 on a pair of dice? d=36. For n, how many of the possible outcomes are =7? [1&6,2&5,3&4,4&3,5&2,6&1] So, 6 possible desired outcomes from 36 total possible outcomes: 6/36 or 1/6.

    Look at your problem. For first pool question n=1 and d=23 but what happens when there are multiple pools. Lets just consider the first two pools. How many possible outcomes and how many desired outcomes?
    For possible outcomes, think of a rectangle drawn on a piece of graph paper that is 23 boxes wide and 27 boxes tall. Each box across is one possible outcome for first pool and each box down is possible outcome for second pool. How many boxes total are inside the rectangle? That is your denominator d. How many correct ( desired) boxes in total are inside the rectangle? That is your numerator n.

    So, what is the probability of getting just the first two pools correct?

    Now what if you consider the first three pools? Think of a cube. How many possible outcomes? See a pattern in calculating the denominator? What is the total number of possibilities for all 6 pools? (d=?) And is there only one desired answer? (i.e. is n=1?)

    AT
     
    Last edited: Mar 12, 2017
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  5. Airtime

    Airtime Monkey+++

    Did you get 1/404,121,960?
    I think they call this Powerball or something like that down at the convenience store.

    Edit: its much worse than powerball, just looked up those odds: 1 in 292,201,338.
     
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  6. Lone Gunman

    Lone Gunman Draw Varmint!

    The total number of events is 6.

    The odds of winning vary from event to event. There are six events. Each must be individually analyzed.

    In the first event the odds are 22:1. (23)

    In the second event the odds are 26:1. (27)

    In the third event the odds are 32:1. (33)

    In the fourth event the odds are 33:1. (34)

    In the fifth event the odds are 28:1. (29)

    In the sixth event the odds are 19:1. (20)

    The total number of possible wins is 6. The total number of possible losses is 160. The total number of chances is 166. The combined odds for a single one time event are 160:6. However this problem is complicated by the fact that there are also 6 individual events. Each of which consists of a different probability ratio. What you’re looking for is not an average, but a mean average for all 6 events. Then:

    Event one has a 0.0454 chance of success. (1/22)

    Event two has a 0.0384 chance of success. (1/26)

    Event three has a 0.0313 chance of success. (1/32)

    Event four has a 0.0303 chance of success. (1/33)

    Event five has a 0.0357 chance of success. (1/28)

    Event six has a 0.0526 chance of success. (1/19)

    (.0454 + .0384 + 0.0313 + .0303 + .0357 + .0526 = .2337 chance to win.)

    .2337/6 = .0400

    There is, then, a generalized 4% probability of a win; and the question becomes: How probable is it that a single winning event might be repeated by the same ticket holder as many as 6 times?

    (.0400/6 = .00666)

    Conclusion? The probability of winning this particular drawing is universally too close to zero; and the Captain is going to be dining alone!
     
    Last edited: Mar 12, 2017
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  7. Airtime

    Airtime Monkey+++

    Hmmm.... here is a challenge for you LG. Change the pool sizes from 23, 27, 33, 34, 29 and 20 to just 2 for all six pools. Re-run the calculations exactly as performed before and tell us what you get for the probability of winning all six pools with one set of six guesses. Does the answer seem plausible?
     
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  8. Pax Mentis

    Pax Mentis Philosopher King |RIP 11-4-2017

    However, for Q2, one must only buy 166 tickets in order to assure winning all 6...since each is a 1 number draw lottery.
     
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  9. Airtime

    Airtime Monkey+++

    Yes! Buy every number for every pool.
    Another way to look at this is odds of winning first pool times odds of winning second times odd of winning third... etc. Buy just one ticket for each pool and its 1/23 x 1/27 x 1/33 x 1/34 x 1/29 x 1/20 or buy all the tickets and its 23/23 x 27/27 x 33/33 x 34/34 x 29/29 x 20/20.
     
    Last edited: Mar 12, 2017
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  10. Lone Gunman

    Lone Gunman Draw Varmint!

    Assuming a fair and open lottery then the answer is no, but then again, your question doesn't actually apply to the problem that's being considered. Perhaps you should go into more detail.

    (The 'buy all the tickets reply' is amusing; but, here, the only available outcome is for there to be only one winner: The person who's holding all of the tickets! The question is no longer a matter of probability; it's a certainty!)

    The numbers in your above reply are also not correctly stated. It is NOT: 1/23, 1/27, 1/33, etc., etc. A correct numerical statement for the first 1/23 example is actually 1:22::X:100. (Because 1 is NOT competing against 23; it is competing against 22, instead!)
     
    Last edited: Mar 12, 2017
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  11. techsar

    techsar Monkey+++

    Q1- 1:2.474500519595619e-9
    Q2- 166
    How dare you ask a question like that before coffee??[loco]
     
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  12. Airtime

    Airtime Monkey+++

    Happy to.
    The hypothetical was instead of you having to the buy a ticket for the right answer in 6 pools with the options in each being 23, 27, 33, 34,29 and 20, what if the captain said there were just 2 options for each pool, a one or a two (or he could have called them heads or tails)?

    I copied the logic you had earlier and changed numbers to red to represent the replacements for when there are just two options in each pool,

    ------=-

    The total number of events is 6.

    The odds of winning vary from event to event. There are six events. Each must be individually analyzed.

    In the first event the odds are 22:1. (23) 1:1 (2)

    In the second event the odds are 26:1. (27) 1:1 (2)

    In the third event the odds are 32:1. (33) 1:1 (2)

    In the fourth event the odds are 33:1. (34) 1:1 (2)

    In the fifth event the odds are 28:1. (29) 1:1 (2)

    In the sixth event the odds are 19:1. (20) 1:1 (2)

    The total number of possible wins is 6. The total number of chances is 166 12. The total number of possible wins is 6 6. The total number of possible losses is 160 6.

    The combined odds for a single one time event are 160:6. However this problem is complicated by the fact that there are also six individual events. Each of which consists of a different probability ratio. What you’re looking for is not an average, but a mean average for all six events. Then:

    Event one has a 0.0454 chance of success. (1/22) 1/1 is 1.0

    Event two has a 0.0384 chance of success. (1/26) 1/1 is 1.0

    Event three has a 0.0313 chance of success. (1/32) 1/1 is 1.0

    Event four has a 0.0303 chance of success. (1/33) 1/1 is 1.0

    Event five has a 0.0357 chance of success. (1/28) 1/1 is 1.0

    Event six has a 0.0526 chance of success. (1/19) 1/1 is 1.0

    (.0454 + .0384 + 0.0313 + .0303 + .0357 + .0526 = .2337 chance to win.)
    (1.0 + 1.0 + 1.0 + 1.0 + 1.0 + 1.0 = 6.0 chance to win

    .2337/6 = .0400
    6.0 / 6 = 1.0

    There is, then, a generalized 4% 100% probability of a win; and the question becomes: How probable is it that a single winning event might be repeated by the same ticket holder as many as 6 times?

    (.0400/6 = .00666)
    (1.0/6 = .16667)
    ----------------------------

    Now, does that all seem plausible?
    The point of the hypothetical was to illustrate that the math that had been presented wasn't correct as could be seen in an extreme example where the pool size was small and the errors not obscured by the larger numbers of 23, 27, 33, etc. I also presented it as a hypothetical or challenge to prompt one to work out the math and come to the conclusion there was something amiss with the methodology and not launch an argument.
    Hope that helped.
    And with that I will probably conclude my time with this thread.
    AT

    Oh BTW UncleMorgan.... check out the Khan Academy Youtube videos and website for some very good info on all this stuff.
     
    Last edited: Mar 12, 2017
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  13. arleigh

    arleigh Goophy monkey

    The captains table is not worth it unless he's footing the bill.
     
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  14. techsar

    techsar Monkey+++

    Something isn't right here...if there are 2 numbers in each pool, there is a 50% probability on each pool...you might care to revisit this part ;)
     
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  15. 3M-TA3

    3M-TA3 Cold Wet Monkey Site Supporter++

    Real world math solution:
    A banknote of the appropriate denomination applied to the correct person increases the probability to 100%
     
  16. chelloveck

    chelloveck Diabolus Causidicus

    Oh not fair....you're applying pursers rules to the equation. :lol:
     
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  17. UncleMorgan

    UncleMorgan I like peeling bananas and (occasionally) people.

    @AT:

    Hey, AT. Thanks for the help.

    You offer a very good way to learn the process, rather than just the answer to one problem.

    For the first two pools, the box has 621 possible outcomes (d), and 2 possible desired outcomes (n).

    The probability of winning the first two pools would be 2/(23X27) = 2/621 = 1/310.5 = 0.0032 (Rounded off.)

    And you would need to buy (23+27) or 50 tickets to cover every possible ticket combination in the first two pools.

    For the third pool, you’d need a cube 23 X 27 X 33, with 20,493 possible outcomes (d) and 3 desired outcomes (n).

    Covering that, you’d need to buy (23+27+33+34+29+20) tickets

    And, following the pattern, you’d need a six-dimensional hypergon with 404,121,960 possible outcomes (d), and 6 possible desired outcomes (n). So the probability would be 6/404,121,960, or 1.4847003117573714628128597614443e-8 or 0.000000014847003117573714628128597614443. Which rounds up to a probability of 0.0000000149, or 1 chance in 67,114,094.

    (Did I get that right?)

    I think the magic smoke just came out of my ears.

    Here’s what just frazzles me: The lottery also has six “contests”. They have a choice of 53, 52, 51, 50, 49, and 48 numbers in them, respectively, since all previously picked numbers drop out.

    According to the Lottery, the chances of a jackpot are 1 in 22,957,480.

    How can this dinky little Ship’s Pool, with so much smaller fields, be almost three times harder to win?

    This isn’t just a silly little high school problem, it’s an assault on all Humanity by a legion of Demon Bookies spawned in the foulest pits of Hell.

    If I can solve this, I’ll turn it in myself. But I don’t want the extra credit—I want Eternal Life.



    @ Lone Gunman:

    Thanks for the help, also.

    I followed your math pretty well, I think.

    I see where you use the mean chance of winning.

    That’s a direction I never considered.

    (I hit this one web site, trying to find the process to solve the problem, and the first thing it said was “There are five ways to solve a problem of this kind.” Then it started out at Ph.D. level, and went up from there. A great place to find the solution for a problem--if you already know it.)

    But—when you get down to the bottom, the odds of a 6-time win are 0.00666.

    That’s, like, 1 in 150.150, isn’t it? Not really small at all.

    I thought cumulative probabilities multiplied.

    I was expecting it to be more like (1/25) ^6, or 1/244,140,624. AKA 0.000000004096.

    Which, as above, is WAY harder than a lottery.

    Whoops, have to evacuate! The magic smoke alarm just went off. Again.
     
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  18. chelloveck

    chelloveck Diabolus Causidicus

    I actually worked for a while in the NSW State Lotteries Office, for a while. It cured me of gambling for life. These days I might occasionally buy tickets in a social club meat tray raffle....more as a charitable donation than with the actual prospect of winning a tray of stewing steak....gambling odds should always be viewed from the prospect of the probability of losing the wager, rather than winning it....it puts the transaction in a more realistic light when putting up hard earned cash. Then there is the probability of legs being broken (or worse) to be considered, if the bet isn't covered should it lose...:eek:
     
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  19. Mindgrinder

    Mindgrinder Karma Pirate Ninja|RIP 12-25-2017

    "A passenger can buy as many tickets as they wish for each pool—but only before the ship sails."

    Did you give us this question EXACTLY how it was worded to the student?
    We're missing something here with the "6 days", "1 draw each day".
     
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  20. ghrit

    ghrit Bad company Administrator Founding Member

    Thunk the same, I did, and still think so. Just in case there's an explanation otherwise, I'll keep reading. So far as why the ship's odd are suckier than the lotto, they constructed it that way.
     
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