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Thanks for any help!

you do know that the past will not affect the future in this?

Thanks for responding. (Except for the wiseacre about the weighted two-ball. Heh.)

Quote:faxtorXYes, I read the posting guidelines and I pretty much get it. However, I would still like to get the odds of this run. Obviously, I am no math wiz. If I understand your comment, then I just raise .9 to the draw power to get the odds, correct? So, in this case, the answer is .9 to the 70th power, which is 0.001797. So about 18 times in 10,000, if I understand correctly?

Thanks for responding. (Except for the wiseacre about the weighted two-ball. Heh.)

Very good. I can't see why it would be wrong, but I have embarrassed myself in the past [g].

You'll get your answer in the form of a decimal. One divided by that decimal will get you your odds as usually expressed

example: 1/2 = 0.5 that's familiar, yes?

and

1/0.5= 2

so

a chance of 0.5 is 1 in 2

I'm not sure where you got that calculation but .9 ^ 70 ~ 0.0006266, which is about 1 in 160, or 62.5 times in 10,000.

Edit: Thanks og, I misplaced a decimal when I did the 1/x calculation. It should be 1 in 1595, or 6.27 times in 10,000.

noQuote:faxtorXYes, I read the posting guidelines and I pretty much get it. However, I would still like to get the odds of this run. Obviously, I am no math wiz. If I understand your comment, then I just raise .9 to the draw power to get the odds, correct? So, in this case, the answer is .9 to the 70th power, which is 0.001797.

So about 18 times in 10,000, if I understand correctly?

unless

your conclusion of 18 times in 10000 is correct only when you count a run of 70 this way

draws 1-70 no 2s = 1 count

draws 1-71 with no 2s = 2 counts

draws 1-72 with no 2s = 3 counts

this is not the way to count the number of independent streaks, imo

(these are overlapping streaks)

==============================================

over 10,000 draws (I think the VA shows over 16k have happened)

we would expect .62 on average

the number of streaks of 70+ with p=0.90

a simulation quickly shows this to be true and NOT 18 times on average, wrong results you gots, or even 6 times on average,

that others got

unless you are just counting the number of overlapping events and that method just confuses everyone

the prob of seeing at least 1 such streak of 70+ over 10k draws = abouts 46.5%

Number of Trials | 10,000 | . |
---|---|---|

Run Length (X or more) | 70 | . |

Trial Probability | 90.000000000% | . |

Event | Run Probability | 1 in |

0 runs of length 70 or more | 0.534951503 | 1.87 |

at least 1 run of length 70 or more | 0.465048497325755 | 2.15 |

at least 2 runs of length 70 or more | 0.128907708962761 | 7.76 |

at least 3 runs of length 70 or more | 0.024795602681865 | 40.33 |

at least 4 runs of length 70 or more | 0.003605819889864 | 277.33 |

at least 5 runs of length 70 or more | 0.000418089591661 | 2,391.83 |

simulation data

N=10,000 and p=0.90

group middle freq freq/100

-------------------------------------------

-0.5 <= x < 0.50 0.00 53663 53.66%

0.50 <= x < 1.50 1.00 33664 33.66%

1.50 <= x < 2.50 2.00 10218 10.22%

2.50 <= x < 3.50 3.00 2117 2.12%

3.50 <= x < 4.50 4.00 286 0.29%

4.50 <= x < 5.50 5.00 45 0.04%

5.50 <= x < 6.50 6.00 7 0.01%

-------------------------------------------

grouped data

items: 100,000

minimum value: 0.00

first quartile: 0.00

median: 0.00

third quartile: 1.00

maximum value: 6.00

mean value: 0.62

midrange: 3.00

range: 6.00

interquartile range: 1.00

mean abs deviation: 0.66

sample variance (n): 0.61

sample variance (n-1): 0.61

sample std dev (n): 0.78

sample std dev (n-1): 0.78

-------------------------------------------

cumulative

-------------------------------------------

-0.5 <= x < 0.50 0.00 53663 53.66%

0.50 <= x < 1.50 1.00 87327 87.33%

1.50 <= x < 2.50 2.00 97545 97.54%

2.50 <= x < 3.50 3.00 99662 99.66%

3.50 <= x < 4.50 4.00 99948 99.95%

4.50 <= x < 5.50 5.00 99993 99.99%

5.50 <= x < 6.50 6.00 100000 100.00%

0.90^70 = 0.000626579 or

1 in 1595.968588 ATTEMPTS,

NOT draws

each attempt is not equal to 1 draw

the length of each attempt = 9.993734213

so 1595.968588 * 9.993734213 = 15,949.6859 average number of DRAWS to see a streak of 70+

this can be done easily from

https://wizardofvegas.com/forum/questions-and-answers/math/8141-on-average-how-many-trials-will-it-take-to-see-a-streak-of-8-qs-for-fun/

10,000/15,950 = abouts .627

and this is a close estimation BTW

there are easier formulas for

1) the average number of draws to see a streak of length X (Eye linked to that)

2) expected # of runs over X draws

=(p^run) * (1+ ((trials-run)*(1-p)) )

p=0.90

trials=10000

run=70

try it!

0.622819276

3) the probability of at least 1 such run over X draws

(this has been done many times many places)

learning is fun

learning a correct way (there could be more than one correct way)

to get a correct answer = more fun

ask and you shall receive

(but you do not know what you will receive)

so the proof is in your hands

have fun winning your next Pick3

Sally

Quote:faxtorXwhich is 0.001797. So about 18 times in 10,000, if I understand correctly?

I didn't catch these errors first time through, just got to this and quit reading,

Quote:faxtorXSo, in this case, the answer is .9 to the 70th power...

sally might explain her comment,

Quote:each attempt is not equal to 1 draw

btw,Quote:odiousgambitfaxtorX says this "So, in this case, the answer is .9 to the 70th power..."

this is correct for

only

the

very

next

70 draws

so 1 in 1,596 sets of 70 draws on average will be all without a 2 in position 1

way different from the expected number of runs over 10k draws

this should be easy to visualize

Sally

are you trying to say that if we count a streak as suchQuote:wudgedEdit: Thanks og, I misplaced a decimal when I did the 1/x calculation. It should be 1 in 1595, or 6.27 times in 10,000.

SSS F as length 3 or more

SSSSSSS F as another length 3 or more (not 2 of them)

there will be on average 6.27 such streaks of 70+ (not overlapping)

in 10,000 draws

I hope nots

Sally

Quote:wudgedMaybe this will make up for my wisecrack.

I'm not sure where you got that calculation but .9 ^ 70 ~ 0.0006266, which is about 1 in 160, or 62.5 times in 10,000.

Edit: Thanks og, I misplaced a decimal when I did the 1/x calculation. It should be 1 in 1595, or 6.27 times in 10,000.

I got *my* number by the time-honored technique of making a keying error. Lol.

Thanks to everybody. I just wanted to have a sense of what the probability curve looks like and where we were on said curve and now I do.

And no, I'm not completely against a little ironic optimism. ;->

Thanks again!