Wikipedia:Reference desk/Archives/Mathematics/2016 January 10

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January 10

What are the odds of winning the $1 billion Powerball lottery?

Today, the Powerball lottery in the USA has a jackpot approaching $1 billion. My first question is "What are the odds of winning the $1 billion Powerball lottery?". And my second question is slightly modified: "What are the odds of solely winning the $1 billion Powerball lottery?". In other words, in the second (modified) question, you do not have to share the jackpot with anyone else; you win it fully by yourself. Thanks. Joseph A. Spadaro (talk) 00:36, 10 January 2016 (UTC)

I think that the exact odds on the second question, being the single winner, depend on the exact number of tickets sold. While that number of tickets sold is no longer changing, since the sale of tickets closed at 0300 UTC, it won't be known until after the drawing at 0400 UTC. Interesting question. Robert McClenon (talk) 03:54, 10 January 2016 (UTC)

No, it depends on how many of those tickets have number combinations that nobody else duplicated. I have never seen this information published for lotteries here in Canada, and I don't imagine it is for Powerball either. --76.69.45.64 (talk) 04:29, 10 January 2016 (UTC)

Correct. The second question is two-fold. (A) I have the exact matching number. What are the odds of that? And (B) No one else also selected that same number combination that I did. So, we need the odds of both "A" and "B" happening, simultaneously. From a statistics (probability) perspective, where would one even start? Joseph A. Spadaro (talk) 04:54, 10 January 2016 (UTC)

First question, what is the probability of jackpot. The answer is one chance in Comb[69,5] * 26 = 11238513 * 26 = 292201338 175.45.116.66 (talk) 23:50, 10 January 2016 (UTC)

If the mean number of winners is M then the actual number of winners i has a poisson distribution

$p(i)=e^{-M}M^{i}(i!)^{-1}$

and the probability that there is exactly one winner is

$p(1)=e^{-M}M$

If the lottery is arranged such that the expected number of winners is M=1 then

$p(1)=e^{-1}=0.367879$

Bo Jacoby (talk) 07:49, 10 January 2016 (UTC).

So, I have a 37% chance of winning the lotto, without having to share the jackpot with another winner? Impossible. What does the 0.367879 mean? Thanks. Joseph A. Spadaro (talk) 07:59, 10 January 2016 (UTC)

Not you personally, but someone who bought a winning ticket has a nearly 37% probability of being a sole winner (assuming that M = 1). As you explained above, you need to multiply this by the small probability that you choose the exact numbers. Dbfirs 08:33, 10 January 2016 (UTC)

This assumes that ticket numbers are chosen uniformly at random. Since people have the option of choosing their own numbers, I strongly suspect that's not the case. For example, I imagine tickets that start with a pattern that can be interpreted as a date are more commonly selected, as people choose to play dates which are significant to them.--66.162.254.21 (talk) 09:09, 10 January 2016 (UTC)

Yes, people do. Here are some statistics from four Canadian lotteries on the number-combinations that (stupid) people most frequently choose to play. --76.69.45.64 (talk) 10:42, 10 January 2016 (UTC)

Why are you calling them "stupid"? I am just curious. Do you mean "stupid" for selecting such number combinations? Or "stupid" in general just for playing the lottery? Thanks. Joseph A. Spadaro (talk) 22:17, 10 January 2016 (UTC)

For selecting such combinations, of course. As the article says, "lottery purchases can be explained by decision models based on expected utility maximization", or as one lottery's slogan puts it, "you can't win if you don't buy a ticket". --76.69.45.64 (talk) 10:02, 11 January 2016 (UTC)

Yes, but how is that a "stupid" strategy? Because it is more likely that they will have to share the jackpot? Is that your point? Joseph A. Spadaro (talk) 21:18, 11 January 2016 (UTC)

The current projections suggest about 400 million tickets will be sold. Assuming random choices, the odds of any one ticket winning is about 1 in 292 million. Hence M may be around 1.35. Dragons flight (talk) 17:01, 10 January 2016 (UTC)

The poisson distribution assumes that each of a big number n of tickets wins with a small probability p. The mean number of winners is M=np. If a significant number of players choose the same combination, then the probability, that there will be exactly one winner, drops. (Popular Number Combinations December 8, 2014 to December 8, 2015 Top 10 LOTTO MAX Number Sequences Played 36674+21565+7020+6890+6648+6183+5278+5154+5045+4597 = 105054). 105054 tickets per year may be small compared to the (unknown) total number of tickets per year. Bo Jacoby (talk) 16:07, 10 January 2016 (UTC).

According to the data provided by Dragons flight: the number of tickets is n=400000000, the winning probability is p=1/292000000=0.000000003424658, the expected number of winning tickets is M=np=1.36986, the probability that there is exactly one winning ticket is P(1)=M¹ e^−M (1!)⁻¹=0.348139412478960, the probability that your ticket is the only winner is p P(1)=0.000000001192258. Bo Jacoby (talk) 20:26, 10 January 2016 (UTC).

Er? What is the probability of ZERO winning tickets? 175.45.116.66 (talk) 23:44, 10 January 2016 (UTC)

The probability of zero winning tickets is P(0)=M⁰ e^−M (0!)⁻¹=0.254141771109641 . Bo Jacoby (talk) 05:26, 11 January 2016 (UTC).

Has anyone worked out the expected return for this powerball ticket? Is it more than $2?? 175.45.116.66 (talk) 00:07, 11 January 2016 (UTC)

expected return = P(1) * 950E6 * 1/292201338 + P(2) * (950E6 / 2) * 1/292201338 + P(3) * (950E6 / 3) * 1/292201338 + ....

Thanks, all. Joseph A. Spadaro (talk) 17:05, 12 January 2016 (UTC)

Resolved

Dot product

Is there a real proof for the Dot product definition? This question bugs me a lot because I just can't see any Rigour by just learning it out of nowhere.

I'm very scared something's fishy and circular here. Can anyone clear this anomaly? יהודה שמחה ולדמן (talk) 23:35, 10 January 2016 (UTC)

What do you mean a proof for the definition? Since when do definitions require proof? Do you mean a proof that two definitions are equivalent? Contact Basemetal here 01:04, 11 January 2016 (UTC)

The article gives two definitions so it requires proof that they are equivalent, but this is also done in the article. I think what the OP is getting at is there is little motivation for the definition. Probably the original motivation was from physics, but sometimes you just have to accept that in math there are concepts that seem to come out of nowhere but turn out to be incredibly useful. Often they are the result of a long process of evolution and students only see the end product. --RDBury (talk) 05:06, 11 January 2016 (UTC)

If you know how to add and subtract and halve and square, then you can use the expression

$\left({a+b \over 2}\right)^{2}-\left({a-b \over 2}\right)^{2}$

as the definition of multiplication. The square of a vector is the square of its length. According to the Pythagorean theorem a²=(a₁,a₂)²=a₁²+a₂², and so

$a\cdot b=\left({a+b \over 2}\right)^{2}-\left({a-b \over 2}\right)^{2}$

$=\left({(a_{1},a_{2})+(b_{1},b_{2}) \over 2}\right)^{2}-\left({(a_{1},a_{2})-(b_{1},b_{2}) \over 2}\right)^{2}$

$=\left({(a_{1}+b_{1},a_{2}+b_{2}) \over 2}\right)^{2}-\left({(a_{1}-b_{1},a_{2}-b_{2}) \over 2}\right)^{2}$

$=\left({{a_{1}+b_{1} \over 2},{a_{2}+b_{2} \over 2}}\right)^{2}-\left({{a_{1}-b_{1} \over 2},{a_{2}-b_{2} \over 2}}\right)^{2}$

$=\left({a_{1}+b_{1} \over 2}\right)^{2}+\left({a_{2}+b_{2} \over 2}\right)^{2}-\left({a_{1}-b_{1} \over 2}\right)^{2}-\left({a_{2}-b_{2} \over 2}\right)^{2}$

$={a_{1}^{2}+b_{1}^{2}+2a_{1}b_{1} \over 4}+{a_{2}^{2}+b_{2}^{2}+2a_{2}b_{2} \over 4}-{a_{1}^{2}+b_{1}^{2}-2a_{1}b_{1} \over 4}-{a_{2}^{2}+b_{2}^{2}-2a_{2}b_{2} \over 4}$

$=a_{1}b_{1}+a_{2}b_{2}$

Q.E.D. Bo Jacoby (talk) 06:06, 11 January 2016 (UTC).

Yeah, I just now found the right definition "proof" - it's so simple, seems like people just don't remember to teach it:

Given non-zero vectors:

$\color {blue}{\vec {a}}=(a_{1},a_{2})\quad ,\quad {\vec {b}}=(b_{1},b_{2})$

$\color {green}{\bigg |}{\vec {b}}-{\vec {a}}{\bigg |}^{2}=|{\vec {a}}|^{2}+|{\vec {b}}|^{2}-2|{\vec {a}}||{\vec {b}}|\cos(\theta )$

$\Downarrow$

$\color {green}(b_{1}-a_{1})^{2}+(b_{2}-a_{2})^{2}={a_{1}}^{2}+{a_{2}}^{2}+{b_{1}}^{2}+{b_{2}}^{2}-2|{\vec {a}}||{\vec {b}}|\cos(\theta )$

$\color {green}={a_{1}}^{2}+2a_{1}b_{1}+{b_{1}}^{2}+{a_{2}}^{2}+2a_{2}b_{2}+{b_{2}}^{2}={a_{1}}^{2}+{a_{2}}^{2}+{b_{1}}^{2}+{b_{2}}^{2}-2|{\vec {a}}||{\vec {b}}|\cos(\theta )$

$\color {green}=-2a_{1}b_{1}-2a_{2}b_{2}=-2|{\vec {a}}||{\vec {b}}|\cos(\theta )\quad {\color {black}/}:-2$

$\color {red}|{\vec {a}}||{\vec {b}}|\cos(\theta )=a_{1}b_{1}+a_{2}b_{2}$.

But now Another question:

Is this Cauchy–Schwarz inequality proof acceptable?

Given non-zero vectors and from the dot product

$\color {blue}\cos(\theta )={\frac {{\vec {a}}\cdot {\vec {b}}}{|{\vec {a}}||{\vec {b}}|}}\quad ,\quad -1\leq \cos(\theta )\leq 1$

$\Downarrow$

$\color {green}-1\leq {\frac {{\vec {a}}\cdot {\vec {b}}}{|{\vec {a}}||{\vec {b}}|}}\leq 1\,{\color {black}\implies }\,{\bigg |}{\frac {{\vec {a}}\cdot {\vec {b}}}{|{\vec {a}}||{\vec {b}}|}}{\bigg |}\leq 1\,{\color {black}\implies }\,{\frac {|{\vec {a}}\cdot {\vec {b}}|}{|{\vec {a}}||{\vec {b}}|}}\leq 1\quad {\color {black}/}\cdot |{\vec {a}}||{\vec {b}}|$

$\color {red}|{\vec {a}}\cdot {\vec {b}}|\leq |{\vec {a}}||{\vec {b}}|$. ∎ יהודה שמחה ולדמן (talk) 19:44, 14 January 2016 (UTC)