Fun with math -- TPiR stat question

[MikeK]

[quote name=\'tvrandywest\' post=\'177829\' date=\'Feb 13 2008, 05:55 PM\']1 in 20 PLUS 1 in 20? 1 in 40???[/quote]
Randy,

See me after class for help about adding fractions.

Signed,

Mr. Klauss

[mcsittel]

[quote name=\'Fedya\' post=\'177801\' date=\'Feb 13 2008, 02:32 PM\']
Probability question for the calculus students: New York used to have a lotto that was pick 6 out of 54 numbers, meaning there were 54C6, or 25,827,165 possibile combinations.  They also had a "quick pick" that would randomly pick the combination for your ticket.  Assuming that the "quick picks" are fully random (and not pseudorandom) in that on any given quick pick, each and every combination is exactly as likely to come up as any other combination, if you play 25,827,165 quick picks, what's the probability that you won't have the winning combination?
[/quote]

This is a binomial problem, as there are only two outcomes... win or lose.
P(win) = p = 1/x,
P(lose) = q = 1-(1/x)
x=54C6 ("54 choose 6", or 54!/(6!*48!).
You will purchase n tickets, where n=54C6 also.

X is the random variable representing the number of winning tickets bought.

P(X=0) = nCx * p^n * q^(n-x)

Since n=x, nCx is simply 1.  Similarly, (n-x) = (n-n) = 0, and q^0 = 1.  So this simplifies to just:

P(X=0) = p^n, or (1/54C6)^(54C6).  That comes to 0.3679.

Now let's talk about the P of being the only holder of a winning ticket...

[tpirfan28]

When the hell did we start playing "Play the Percentages"!?

/head's about ready to explode

[Fedya]

mcsittel wrote:

This is a binomial problem, as there are only two outcomes... win or lose.

{much accurate math snipped}

Generalize it for any arbitrarily large n.  And what if you buy 2n quick picks?  :-)

[Joe Mello]

[quote name=\'tpirfan28\' post=\'177837\' date=\'Feb 13 2008, 06:23 PM\']
When the hell did we start playing "Play the Percentages"!?

/head's about ready to explode[/quote]
Wait till the part where we have to asses the risk neutral probability and bullish vertical spreads.

/Yay jargon?

[Robert Hutchinson]

[quote name=\'tvrandywest\' post=\'177829\' date=\'Feb 13 2008, 05:55 PM\']Simply, round 2 of the ACES HIGH game has 2 players head-to-head trying to create a better 4 CARD poker hand using only Aces, Kings and Queens. Player 1 has already drawn cards and ends up with 4 Kings. Player 2 has 4 Queens and has to draw cards. The only winning hand would be 4 Aces.

Player 2 discards what looked like a great hand, 4 Queens, hoping that the 4 cards he draws will be all Aces. And he makes it!

Chance of each of the 4 cards being an Ace is 1 in 3. So  all four cards being Aces is 3 x 3 x 3 x 3 = 1 in 81.[/quote]
Are the cards used from a physical deck, or computer-generated in some fashion? If it's a physical deck with a whole bunch of Aces, Kings, and Queens, evenly distributed, the odds are going to be a little different based on what's already been drawn. But if it's always a 1 in 3 chance on every pick, then the above holds.

[tvrandywest]

[quote name=\'mcsittel\' post=\'177832\' date=\'Feb 13 2008, 03:13 PM\']
[quote name=\'tvrandywest\' post=\'177829\' date=\'Feb 13 2008, 04:55 PM\']
Back to TPiR: What are the odds of getting exactly a dollar in one or a combination of two spins on the big wheel? My uneducated guess: 1 in 20 PLUS 1 in 20? 1 in 40???
[/quote]

Spin 1:
1/20 chance of $1
19/20 of something else

Spin 2:
If not $1 on the first spin, 1/20 chance of getting the number that, added to your non-dollar first spin, will equal $1 total.

So...
1/20 + ((19/20)*(1/20)) = 1/20 + 19/400 = 20/400 + 19/400 = 39/400 = 0.0975.

And 1/20 + 1/20 is 2/20, not 1/40.  I've had many a student tell me otherwise!
[/quote]
Shame on me - my mother was a math teacher and a member of Mensa.

So I guess I can tell the people who repeatedly ask this question:
"It's just a little less than 1 in 10"
Hmmmm, if I had added my fractions correctly I would have been close with that 2 in 20.

Thanks for the answer. And for all you kids, finish skool!!!

Randy
tvrandywest.com