Welcome Guest

NetConnect

Internet Domains, simple and cheap

Find a domain name:

Home

The time is now 1:59 pm
You last visited December 6, 2013, 1:58 pm
All times shown are
Eastern Time (GMT-5:00)

Odds for Calif scratcher Power 77 (#732)?

Topic closed. 2 replies. Last post 1 year ago by mathhead.

 Page 1 of 1

United States
Member #130800
July 25, 2012
67 Posts
Offline
 Posted: July 25, 2012, 3:47 am - IP Logged

I believe I know how to calculate the odds for the Calif scratcher Power 77 (#732).  But I am surprised by my conclusions.  I would appreciate it another person skilled in "lottery math" would double-check my work.

I would like to include a pointer to the Calif Lottery webpage for that scratcher, but I am not sure if this forum's rules permit it.  For now, I will summarize the relevant information.

This is a "match key number" game with 5 chances to match a winning number.  Prizes include a ticket or any of several cash amounts.  According to the webpage, the "overall odds" are 1 in 4.83, in contrast to the "cash odds" of 1 in 9.34.  The webpage has a break-down of each prize, the number of "wins" (outcomes that win that prize), and the individual odds (rounded). [1]

Based on that information, I have calculated the "overall odds" (probability) to be exactly 6,359,552 / 30,720,000, which is about 20.7017%.  I interpret that to be the probability of matching the winning number for each of the 5 chances.

Based on that interpretation, I believe that the probability of matching the winning number at least once on the ticket is 1 - (1-20.7017%)^5, which is about 68.64%.

And if we purchase 7 tickets (\$7), the probability of winning something is about 99.97%:  1 - (1-20.7017%)^35.

I am not saying that "winning something" would make the investment profitable.  That "something" might be only \$2 or a ticket (\$1 value).  But I am still struck by the near-certainty of at least one win with 7 tickets.

Am I right?

-----

[1] I should note that these numbers are based on the number of wins and total outcomes at the beginning of the game.  I don't believe the Calif Lottery provides sufficient information for us to calculate the current odds as the game has progressed.  They do provide information about the number of prizes claimed to-date.  Based on that, we know that about 64.66% of the prizes have been claimed, and the numerator (available prizes) is now 2,247,280.  But that does not tell us how many total outcomes are represented by that number.  So we cannot determine the denominator.  Of course, we might guess that it is also reduced by 64.66%.  But in that case, the "overall odds" would be unchanged.

Jacksonville Florida
United States
Member #23018
October 6, 2005
493 Posts
Offline
 Posted: July 25, 2012, 9:46 am - IP Logged

The odds are a simple 1 in 4.83 ... statistically, on average and rounding up, if someone buys 5 tickets, 4 will be losers and 1 will win at least a free ticket. Sometimes you win more and sometimes you win less than the average.

The number of "chances" to match a winner number on the ticket is irrelevant to the odds of the game. When games are created, the game rules dictate how many winners and what amounts will be created. For example, if a game will have 1000 winners of \$100, we can make the \$100 winning tickets a combination of one number match for \$100, five numbers match for \$20 each, a 2x doubler symbol for \$50, and so on.

Tickets with phrases like "Win up to 10 times" or "Tripler" or other multiplers are simply marketing tools to enhance the perception of the game and do not change the odds of the game.

The amounts to be won are already pre-determined and these multi-win phrases are simply alternate ways to create a set winning amount.

United States
Member #130800
July 25, 2012
67 Posts
Offline
 Posted: July 25, 2012, 1:33 pm - IP Logged

duckman wrote:

``The number of "chances" to match a winner number on the ticket is irrelevant to the odds of the game.  When games are created, the game rules dictate how many winners and what amounts will be created.  For example, if a game will have 1000 winners of \$100, we can make the \$100 winning tickets a combination of one number match for \$100, five numbers match for \$20 each, a 2x doubler symbol for \$50, and so on.``

Aha!  Thanks so much for that insight.  I can see now that you are right.

For this game (#732), the CA lottery publishes odds for only the following cash prizes:  \$77, 20, 10, 7, 3 and 2, as well as free ticket.

I believe you are saying that those are predetermined sums of the 5 possible matches (or exactly one "ticket" match in this game), one for each ticket.  Thus, if a ticket has a match paying \$20, it will not have any other matches.  And if a ticket has matches for \$10, \$7 and \$3, it will not have any other matches; and it will be counted as a \$20 prize for the purpose of computing odds.

I have a ticket [1] that has two matches for \$1 each.  But \$1 is not listed as a prize in that game's table of odds.  \$2 is listed as the minimum cash prize, which can be the sum of 2 \$1 matches.  QED.

That makes much more sense mathematically now.  Thanks.

-----

[1] FYI, the ticket that I have is for different game that has ended.  So its table of odds does not appear on the CA lottery website -- at least, I have not found a way to get to it.  But your comment motivated me to try a Google search.  And that did find the CA lottery webpage with that game's table of odds.  So now I can see the error in my way of thinking about how the outcomes are determined.

 Page 1 of 1