Unusual Run in Virginia State Pick 3 Lottery - Odds?

In the Virginia State Lottery Pick 3 game, no number starting with a 2 has been drawn for many draws, 70 draws as of today. I would like to know how to compute the odds of this ongoing run. I've been told that runs of 80 or so draws like this are not unusual. I searched the entire history of the VA Pick 3 numbers carefully and could not find an example involving the digit 2 being out more than 60 draws in the first position. That's what made me wonder if perhaps this is an unusual event. Anyway, I'd like to know the odds so I can understand whether this event is as unusual (or not) as it seems.

Thanks for any help!

9 chances out of 10 that the number won't start with a 2 [per trial], can you take it from there?

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

lol i hope he does, because I sense that poster is plannning on playing all of the 200's as a play

The 2 ball is weighted more so it's not going to be drawn!

Yes, 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.)

Quote: faxtorX

Yes, 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

Yea playing a due number in a game that holds 50% always works out.

Maybe 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.

1 in 1595 I get

Quote: faxtorX

Yes, 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?

no
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%


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: faxtorX

which 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: faxtorX

So, 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

Quote: odiousgambit

faxtorX says this "So, in this case, the answer is .9 to the 70th power..."

btw,
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

Quote: wudged

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.

are you trying to say that if we count a streak as such

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: wudged

Maybe 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!