Calculating odds

Reply #12

Tucker Black — Sun, 08 Mar 2020 10:34:09 GMT

If the terminal picks 5 numbers 7 times independently, then the probability of winning the jackpot is very slightly less, because it's possible to get the jackpot numbers on more than one line. You share the prize with yourself, not collect twice. The difference is about 10 millionths of a percent. If the terminal checks the previous lines (if all 5 numbers are a match, toss the numbers and run the RNG again) then it makes no difference.

For self-picks, it makes no difference whether or not y... [ More ]

Reply #11

rcbbuckeye — Fri, 28 Feb 2020 17:25:11 GMT

There actually have been 2 step jackpot winners that pick their own numbers. Most winners are qp's, but not all.

Reply #10

Todd — Thu, 20 Feb 2020 00:16:57 GMT

In order to know if it was better to buy quick picks or your own numbers, you would first have to know the percentage of tickets for the game that were sold as quick picks.

If 90% of the tickets sold over the course of the game were quick picks, then that does not mean it is better to buy quick picks. It just means that since so few people chose their own numbers that it is only a slim chance that one such person would win.

However, if only 50% of the tickets sold were quick picks, then y... [ More ]

Reply #9

Big Joey — Thu, 20 Feb 2020 00:07:12 GMT

With 5/35 matrix ball games, it's best to get store terminal purchased quick picks rather than selecting your own numbers. The odds are better with store purchased quick picks it seems. For example, the Texas Lottery has Two Step which is a double 5/35 matrix, and no one has ever won the jackpot by selecting their own numbers, all jackpot winners have been won with store purchased quick picks.

Reply #8

Todd — Wed, 19 Feb 2020 21:40:41 GMT

That is highly misleading. The odds expressed in one mathematical expression (which is what he wanted) are 7 in 324,000. Which is the same as (drumroll please) 1 in 46,285.7.

Put another way:

Please recommend how to express the odds of winning if I purchase 5 tickets for a game with odds of 1 in 10. Would you express it as:

Ticket #1: 1 in 10

Ticket #2: 1 in 9

Ticket #3: 1 in 8

Ticket #4: 1 in 7

Ticket #5: 1 in 6

OR

1 in 2 (reduced from 5 in 10)

Mathematics works th... [ More ]

Reply #7

Stack47 — Wed, 19 Feb 2020 20:51:17 GMT

you can also look at the odds at different times, and that's what raven did

I was looking at specific seven lines; in other words each of the 35 numbers are used and thought the question is do those seven lines have better odds than seven QPs. The simplest approach is doing the same math as you and others did, but that doesn't show an odds bias either way. Because each number only appears on 1 line, the odds should be the total number of combinations each number appears in and each number ap... [ More ]

Reply #6

KY Floyd — Wed, 19 Feb 2020 17:37:24 GMT

, but it depends on how odd are expressed.

You can express the odds in different ways, but you can also look at the odds at different times, and that's what raven did.

Before the drawing there's a chance you'll win and a chance you'll lose, and the chances are based on the odds of the game and how many tickets you have. For a 5/35 game there are 324,632 possible combinations, so every ticket has the same 1 in 324,632 chance. Ticket # 1 has a 1 in 324,632 chance, ticket # 2 has a 1 in 324,... [ More ]

Reply #5

Stack47 — Tue, 18 Feb 2020 18:57:52 GMT

, but it depends on how odd are expressed. The most common way is to divide the number of chances (324,630) by the number of your chances (7). Some will say the odds are 46,376 to 1 and others will say there is 1 chance for every 46,376 chances.

If the odds are expressed as chances to win compared to chances to lose than the odds for all seven tickets are expressed as 1 in 324,626 because only one of those seven chances can match all 5 numbers.

IMO, if you're going to play seven lines, us... [ More ]

Reply #4

Raven62 — Tue, 18 Feb 2020 00:43:51 GMT

Ticket #1: 1 in 324,632

Ticket #2: 1 in 324,631

Ticket #3: 1 in 324,630

Ticket #4: 1 in 324,629

Ticket #5: 1 in 324,628

Ticket #6: 1 in 324,627

Ticket #7: 1 in 324,626

Reply #3

AlanHogan — Tue, 18 Feb 2020 00:36:04 GMT

That's what I thought was just second guessing myself. Thank you

Reply #2

Todd — Mon, 17 Feb 2020 23:25:26 GMT

It's easy. 7 tickets is 7 chances out of 324,000. It's a simple fraction to reduce. 324,000 / 7 = 1 in 46,285.7.

Reply #1

Stack47 — Mon, 17 Feb 2020 22:09:43 GMT

Regardless of how the odds or chances are expressed, by using all 35 numbers, you have slightly reduced the odds. Obviously you need to match 3 numbers to win a prize but by using each of the 35 number, the odds of matching three numbers is slightly better than 1 in 75.

Another way to look at using all 35 numbers is the odds against correctly placing the 35 numbers in order into your 1 if 5 wheel.

Calculating odds

AlanHogan — Mon, 17 Feb 2020 20:20:05 GMT

I play the MassCash 5/35 and I usually play 7 combos which cover all the numbers once. How do I find out what the odds are reduced to by playing 7 combos? The game odds are 324,000/1 to hit a grand prize for one ticket but how do I figure out what I have reduced my odds to by playing 7,14,35,etc combos? Thanks players