Purple Lottery 1

[jerryloo71]

In too!

[pcad]

Wednesday May 22, 2013
Lotto numbers for the
28th round of the Purple lottery

1 - 10 - 23 - 35 - 39 - 45 - 46

Still out.

Only two common numbers today. The rest are in the lower fifty percent by frequency seen in this lottery.

Eight numbers are needed to survive this round. Of the four combinations that worked for round 27, only one combo works for round 28.

The chances for surviving round 29 is now 1 in 7. Good luck on Saturday.

[maztec]

Wednesday May 22, 2013
Lotto numbers for the
28th round of the Purple lottery

1 - 10 - 23 - 35 - 39 - 45 - 46

Still out.

I am waiting for numbers to be drawn that make you suddenly back in.

[pcad]

True, it sounds a bit weird, but I'll be back in when the numbers that finish this first lottery come up.

Maybe I'm misinterpreting quantum cows statistics, but with 1 in 7 odds I'd think we'd have finished this lottery already.

[maztec]

True, it sounds a bit weird, but I'll be back in when the numbers that finish this first lottery come up.

Maybe I'm misinterpreting quantum cows statistics, but with 1 in 7 odds I'd think we'd have finished this lottery already.

Yes, you are. Plus those stats were based on a presumption of how many people entered the lottery and with how many numbers each. May be much larger than those assumptions, which changes the odds.

[pcad]

That may be. My understanding of quantumcow's post on 9/25/12 refers to the probability that you'll "win" or match at least one number on any given round.

I took the result of (34*33*32*31*30*29*28) / (49*48*47*46*45*44*43) and subtracted it from 1.

This is 0.9373739 or a 94% chance of matching at least one number if you have selected 15 numbers.

For each additional round you just keep multiplying by the probability of matching a single round. In other words the probability for surviving a round is p^n where p = 0.937379 and n is the number of rounds.

For round 28, the probability of matching a number in all 28 rounds is 0.937379^28 = 0.1635. Which is roughly 1/6. Similar to rolling a particular number with a die.

For a lottery to end there has to be fewer than 4 people playing. If the only way to win is to roll a specific number with a die it shouldn't take long to whittle down the players to less than 4.

At 29 rounds the odds are 1/7. Even worse, and the odds continue to shrink to 1/8 for round 31 and 1/9 in round 33.

Where am I incorrect?

[stet]

Results of the past don't influence the present or the future. The only result of the past that matters is if you survived or not. For everyone still in the lottery they have a 93.73739% of surviving to through the next round. And that applies to every round.

[pcad]

True. But we're looking at the odds of an event with an increasing number of trials. A coin toss is simple. The probability of a particular result are always 0.50. What are the odds of getting heads twice in a row, 0.5 x 0.5 = 0.25. The same goes for getting heads 28 times in a row, 0.50^28 = 0.000000003725 . In the case of the lottery the odds for any one round is 0.94 (if you have 15 numbers). Just substitute 0.94 for the 0.50 . The odds for any one round remains the same, 94%.

[jerryloo71]

In....

[maztec]

Ah, there is where you are wrong.

Every week every person remaining, with the right number of numbers, has a 93% chance of making it to the next week. That means for 7 out of every 100 draws you might be booted.

Now, here is the thing, if you want to say what percentage chance of making it X more weeks you have, you do %^w (w=weeks) and you know your odds of going that far. However, each time you reach the next week, if you want to calculate the odds of winning that week on its own, you're still at 93%.

And that 7/100 odd of losing stays at 7/100 per week. Yo do not cumulate, so it does not become 14/100 or even 10/100 for the next week, it stays at 7/100 for each week chance of losing.