# Lottery probability ! emergency !

• Dec 3rd 2008, 08:55 AM
xmenymenzmen
Lottery probability ! emergency !
We are given a question that states as follows for mega millions lottery

REPEAT HITS LOTTERY TIPS:
In the 5/56 (chose 5 out of 1 to 56 without repititions) MEGA Millions, on average, one lottery number will be a repeat hit from the
last drawing every other drawing or 42 percent of the time.

thats the information given.

Now the question is .. how did they calculate 42 % ? Also how true this claim is ?

I figured it has something to do with conditional probability but m not able to figure out how ? Any help will be appreciated ! project due today ! (Headbang)
• Dec 3rd 2008, 09:33 AM
CaptainBlack
Quote:

Originally Posted by xmenymenzmen
We are given a question that states as follows for mega millions lottery

REPEAT HITS LOTTERY TIPS:
In the 5/56 (chose 5 out of 1 to 56 without repititions) MEGA Millions, on average, one lottery number will be a repeat hit from the
last drawing every other drawing or 42 percent of the time.

thats the information given.

Now the question is .. how did they calculate 42 % ? Also how true this claim is ?

I figured it has something to do with conditional probability but m not able to figure out how ? Any help will be appreciated ! project due today ! (Headbang)

Hint: what is the probability than none of the last draw's numbers are drawn this time.

CB
• Dec 3rd 2008, 09:38 AM
xmenymenzmen
thanks for the hint
correct,

I used than use (1-(Whatever that probability is))
But when i do that .. it comes out to 44.5% probability and not 42% ..
Thats the reason i was stuck.
• Dec 3rd 2008, 09:50 AM
Soroban
Hello, xmenymenzmen!

If I read the problem correctly, I don't agree with their answer.

Quote:

In the 5/56 (chose 5 out of 1 to 56 without repititions) MEGA Millions, on average,
one lottery number will be a repeat hit from a previous drawing 42% of the time.

There are: . ${56\choose5} \:=\:3,829,816$ possible outcomes.

The opposite of "at least one repeat" is "no repeats:.

To get no repeats, the five numbers are chosen from the other 51 numbers.
. . There are: . ${51\choose5} \:=\:2,349,060$ ways.
Hence: . $P(\text{no re{p}eats}) \;=\;\frac{2,349,060}{3,819,816} \;=\;0.614966794 \;\approx\;61.5\%$

Therefore: . $P(\text{some re{p}eats}) \;=\;100\% - 62,5\% \;=\;\boxed{38.5\%}$

If they meant exactly one repeat, the probability is even smaller.

• Dec 3rd 2008, 10:25 AM
xmenymenzmen
correct
You are right...

But it says "on average," one lottery number will be a repeat hit from the
last drawing every other drawing or 42 percent of the time. and thats what confuses me more ..

if it goes "on average" of the drawings ... than it becomes i guess a normal distribution ? and from that i guess they are getting 42% probability

But yes..thats exactly the way i thought .. but not right i suppose ..

MEGA Millions

she is the one predicting 42% and most of the other websites.
• Dec 3rd 2008, 10:33 AM
CaptainBlack
Quote:

Originally Posted by xmenymenzmen
You are right...

But it says "on average," one lottery number will be a repeat hit from the
last drawing every other drawing or 42 percent of the time. and thats what confuses me more ..

if it goes "on average" of the drawings ... than it becomes i guess a normal distribution ? and from that i guess they are getting 42% probability

But yes..thats exactly the way i thought .. but not right i suppose ..

MEGA Millions

she is the one predicting 42% and most of the other websites.

But almost all I see there is crap.

If you are betting just one set of 5 from 56 every choice has an equal probability of winning. In which can you are better off choosing a set that others will avoid (since then you are less likely to share the prize in the event that you win).

CB
• Dec 3rd 2008, 11:27 AM
xmenymenzmen
thanks !
you know what .. that actually helps..

Our project is to either support or stand against the claim ..

I think i will take your advice .. it is kind of crap.

Thanks again.