lottery probability
What is my chance of winning in the round two lottery? What was my chance of winning in the first round? I thought it would be interesting to know, so I calculated some probabilities.
The probability of winning in this lottery changes with each draw, since the winner and all of their tickets is removed. Plus, you get 90 chances to win (87 with the shipping disaster buffer). So I'm most interested in the probability of winning by round 87. That can be calculated using a set of nested equations:
probability at the start =
(number of tickets held) / (total number of tickets)
probability in the next round =
(probability of winning in the first round) + (1 - probability of winning in the first round)*((number of tickets held) / (total number of tickets remaining))
... and so on until... probability in round 87 =
(probability of winning by round 86) + (1 - probability of winning by round 86)*((number of tickets held) / (total number of tickets remaining))
Hence, the number of tickets held and the number of tickets remaining is needed. I can't tell you how many tickets you have and I don't know the number of tickets removed in each round, but I can estimate based on the bins Jason provided in post 560 and plot the probability of winning for a range of tickets held.
Now, the fine print. I had to make some assumptions. First, I only know a rough distribution of the number of tickets held by each participant and I don't know how many tickets are removed by each draw. But I do know the total number of tickets at the start and end of round 1 and the number of participants. For the actual round 1 lottery, I can use the mean number of tickets removed per round (5965 tickets removed / 87 rounds) to calculate the actual probability that you would have won by round 87 in the first lottery (round1.actual in the plot). But I'm interested in the probability of winning in round two, where I don't know the number of tickets removed yet....
So I simulated the draws for round one (round1.est on the plot) and round two (round2.est) by creating 10,000 fake entry pools that meet the bins provided by Jason and drew 87 winners from each. Given the bins, the distribution of tickets held must have been skewed toward the low end to give the total number of tickets, so I had to weight the draws toward smaller numbers of tickets to avoid needing a cluster server. This means my simulations may not sample the true distribution, but should get close. Then I used the mean number of tickets removed and standard deviation to plot the estimated probabilies on the plot. Indeed, my simulation for round one slightly underestimates the real data.
However, the estimate for round one closely matches the actual data, suggesting 1) my approximation is OK and 2) that there is little reason to suspect shenanigans.
The estimate for round two reflects the obvious conclusion: your chances of winning in round two are better. For those like me in the 10-99 ticket bin, they look pretty good! Anyway, estimate your number of tickets, find your dot on the plot, and have fun!
What is my chance of winning in the round two lottery? What was my chance of winning in the first round? I thought it would be interesting to know, so I calculated some probabilities.
The probability of winning in this lottery changes with each draw, since the winner and all of their tickets is removed. Plus, you get 90 chances to win (87 with the shipping disaster buffer). So I'm most interested in the probability of winning by round 87. That can be calculated using a set of nested equations:
probability at the start =
(number of tickets held) / (total number of tickets)
probability in the next round =
(probability of winning in the first round) + (1 - probability of winning in the first round)*((number of tickets held) / (total number of tickets remaining))
... and so on until... probability in round 87 =
(probability of winning by round 86) + (1 - probability of winning by round 86)*((number of tickets held) / (total number of tickets remaining))
Hence, the number of tickets held and the number of tickets remaining is needed. I can't tell you how many tickets you have and I don't know the number of tickets removed in each round, but I can estimate based on the bins Jason provided in post 560 and plot the probability of winning for a range of tickets held.
Now, the fine print. I had to make some assumptions. First, I only know a rough distribution of the number of tickets held by each participant and I don't know how many tickets are removed by each draw. But I do know the total number of tickets at the start and end of round 1 and the number of participants. For the actual round 1 lottery, I can use the mean number of tickets removed per round (5965 tickets removed / 87 rounds) to calculate the actual probability that you would have won by round 87 in the first lottery (round1.actual in the plot). But I'm interested in the probability of winning in round two, where I don't know the number of tickets removed yet....
So I simulated the draws for round one (round1.est on the plot) and round two (round2.est) by creating 10,000 fake entry pools that meet the bins provided by Jason and drew 87 winners from each. Given the bins, the distribution of tickets held must have been skewed toward the low end to give the total number of tickets, so I had to weight the draws toward smaller numbers of tickets to avoid needing a cluster server. This means my simulations may not sample the true distribution, but should get close. Then I used the mean number of tickets removed and standard deviation to plot the estimated probabilies on the plot. Indeed, my simulation for round one slightly underestimates the real data.
However, the estimate for round one closely matches the actual data, suggesting 1) my approximation is OK and 2) that there is little reason to suspect shenanigans.
The estimate for round two reflects the obvious conclusion: your chances of winning in round two are better. For those like me in the 10-99 ticket bin, they look pretty good! Anyway, estimate your number of tickets, find your dot on the plot, and have fun!
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I didn't follow most of this, its been a long time since I studied probability, but thanks for the analysis! I love quantitative analysis.
Can additional people enter round 2 that didn't enter in round 1?
Can you confirm that (1 - probability of winning in round n) is the (prob of losing in round n)
Yes.
See...
https://www.diyaudio.com/forums/pas...y-vfet-lottery-discussion-22.html#post6602945
"...there will be another window opened for additional entries"
Well that changes things. Probability for around two no longer accurate.
Just means that if you didn't win in round 1, you probably still won't win in round 2
We held all the orders to perform verification. It's my understanding most orders have now shipped.
As someone who loves graphs, I am in awe of your post and charts. Bravo!
You have been appointed the position of official lottery statistician. I'll be in touch
I had been saying there will be a window for stragglers to enter round 2, but considering the situation... Hmm... My primary goal to have all people from round 1 enter round 2 was to reduce everyone's stress (it wasn't in order to not hold a second entry window). If you entered round 1, you don't need to do anything else, just relax.
To be fair, as far as I am concerned, whatever I have said must be followed through on. If don't, people will get upset. I am happy to state right now though that I will not make any fanfare and only super keen members who are following the pages and info will become aware of the second window.
No. You're good
Your order has been taken, and showing at our 3PL as "In Process", along withall the others the remaining 50% or so that haven't shipped yet.
Our 3PL has been struggling a bit lately, and doing stuff like asking them to hold and release orders so we can check them throws a spanner in their works. In the future we're moving to a different kind of arrangement where orders are staged before being submitted to the 3PL giving a lot more flexibility for order verification of high value or important orders like these. Right now it's kind of like asking a container ship to stop and start, with a lot of lag.
EDIT: About half the orders shipped yesterday, half still to go, so I would assume tomorrow.
Your order has been taken, and showing at our 3PL as "In Process", along with
Our 3PL has been struggling a bit lately, and doing stuff like asking them to hold and release orders so we can check them throws a spanner in their works. In the future we're moving to a different kind of arrangement where orders are staged before being submitted to the 3PL giving a lot more flexibility for order verification of high value or important orders like these. Right now it's kind of like asking a container ship to stop and start, with a lot of lag.
EDIT: About half the orders shipped yesterday, half still to go, so I would assume tomorrow.
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