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?
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[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.