Need a Little Help With a Statistics Problem...

Oh, yeah. There is one pool drawing a day for the first six days after leaving the dock.
So the first six days the Captain eats alone, I guess.

Then six days later, after the last pool, the lucky jackpot winners (if any) join him.

 

Close, but no, not quite. If we draw the picture with the first pool possibilities along the vertical axis and second pool possibilities along the top axis you have the grid like below. Lets say we pick 5 for the first pool and 10 for the second pool. All the boxes represent the full set of possible selections but you see there is only one box (green) for picking 5 and then the 10.



So the probability is 1 out of 621 (.161%).
We can get this a different way. We pick the 5 in first pool and the odds that it is the winner is 1/23 or 4.35%. We pick
10 in the second pool and odds that it wins is 1/27 or 3.70%. If we multiply these for the odds of getting both correct, then .0435 x .037 = .001609 or .161% as we saw previously.

Yes, as you can see from the box above, it would cover every row for the first pool and then every column for the second.

For buying just one ticket for each pool, there would be only one scenario where you got the right number each time for the 3 pools. So it would be 1 desired outcome (three winning numbers, just one for each of the three pools) for the 20,493 possibilities.

Buying just 6 tickets, one for each pool, means only one winning combination for the entire event so buying just 6 tickets means your odds are just 1 out of 404,121,960.


There is another little trick to use to help check if you got things right; run the numbers for winning and then again for losing and the total should equal 100%. So say we buy 7 tickets in pool one, our odds of winning would be 7/23 or 30.43% and the odds of losing would be 16/23 or 69.57%. Add 30.43 plus 69.57 and we get 100%.

Hope this all helps, Uncle.
AT

 

It may be because if once sailing, you lucked out at draw number1, there would be no incentive to buy tickets in draws 2-6...same goes to failing punters in draws 2-5.....if you're still in the run for a winning ticket...at draw number six, there'll only be a few, if any who could be bothered to buy a ticket at that stage of the voyage......The Captain is a greedy ba$tard, so folks buying tickets have to buy them all up front before the voyage begins.

 

Ok--the magic smoke has mostly cleared away and I have plugged my ears with Silly Putty to keep in what small amount yet remains.

Just for giggles and (maniacal) grins, I used the formula that the Florida Lottery uses to calculate the odds on each of the six pools individually, then averaged the results.

I was looking for the chance of winning six-number lotteries for each size pool.

That formula is at
http://www.flalottery.com/exptkt/lottoodds.pdf

The results were:
Pool size (1 chance in
20 38,760
23 100,947
27 296,010
29 475,020
33 1,107,568
34 1,344,904
Average: 1 chance in 560,534.8333 (Repeating)

This is based on having to pick all six numbers on one ticket before any are drawn, lottery-style.

The number of tickets question seems fairly straightforward, now, if the Ship's Pool is run like a lottery.

One ticket for each chance to win: 560,534.8333(R) tickets.

560,535 tickets at the dock.

I think this approach is right, or right enough for high school, anyway.

 

FOR REAL !!
You remember flash cards ?? make up a few & see if the High School Student can answer them !!

For Real ??
I bet NOT .
Don't go wild , no 3 ,5s,7s or Never a 9 !!
Math , Accountants know 9's power !!
Only in AME do we have this Problem

Sloth

 

lol.. way too early for the ones that only made a c in stat.
but i'm firmly convinced that there is a 100% chance there are some real smart guys on this board!

 

The correct answer is, of course: purple.

 

* In numerical probabilities: Numbers which approach zero are considered to be indicative of events which are less likely to occur; and numbers which are closer to 1 (or 100%) are considered to be indicative of events which are more likely to occur.

Let’s look at this question in a slightly different way: An overall independent event probability (Which this example is looking for.) is determined by multiplying the total number of separate probabilities together: (0.0454 x 0.0384 x 0.0313 x 0.0303 x 0.0357 x 0.0526) = 0.0000000031

[(P)(A + B + C + D + E + F)] = [(PxA) + (PxB) + (PxC) + (PxD) + (PxE) + (PxF)]

The statistical probability for this series of independent events to produce 6 consecutive winning results is ≤ 0.0000003%. What, then, is the actual meaning of the above cumulative product?

IT IS A MEAN AVERAGE PROBABILITY; AND THE FARTHER THE FINAL PRODUCT IS AWAY FROM 1.0 — OR, CONVERSELY, THE CLOSER IT IS TO 0.0 — THE LESS LIKELY THE EVENT IS TO OCCUR!

My best guess for this ship's combined lottery to be won? Not a snowball's chance in hell!

 

Thanks, Lone Gunman.

Never did like that damn Captain, anyway.