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Owlcatisback [3147591] Level 15
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- Does CE affect general success rate in crime 2.0
- Questions about the expected value of lotteries
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Next source word: [background color=var(--bbc-body-bg-color)]Flamingo[/background]
Looking for a 10* pub/ amusement park
Long term if possible
Can join any time
10/10 EE
10/10 Addiction Mitigation
Rehab regularly
Active daily
Manual labor 6,798
Intelligence 35,825
Endurance 52,156
Total 94,779
Thx!
Source Word: CHOREOGRAPHY
Recent archived posts
Hello everyone,
I searched Q&A and found a similar question back on 17/03/24. There is only one answer, which suggests a mechanic similar to crime 1.0.
It's mentioned in the wiki page of crime 1.0. Yet in crime 2.0 only OC is explicitly included.
I was just wondering if it's a de facto norm or something.
Thx!
Looking for a 10* pub/ amusement park
Long term if possible
Can join any time
10/10 EE
10/10 Addiction Mitigation
Rehab regularly
Active daily
Manual labor 6,798
Intelligence 35,825
Endurance 52,156
Total 94,779
Thx!
Looking for 7* + music store
- Manual labor 809
- Intelligence 16,336
- Endurance 37,513
- Total 54,658
10/10 EE
Active daily
Thanks :)
Hi there
Looking for a 7*+ candle shop
Active daily and rehab frequently
I can join as early as this Saturday
Current working stats:
Manual labor 559Intelligence 12,688Endurance 30,086Total 43,333
Thanks :)
b2: Number of tickets from lucky vouchers
s: Total number of tickets sold overall
p: Price of a single ticket
d:discounted rate
EV: (b1+b2) / s * p( s -(b1+b2*d)) -p(b1+b2*d) (1 - (b1+b2) / s)
= (b1+b2)*p - (b1+b2)*p*(b1+b2*d)/s -p(b1+b2*d) + p(b1+b2*d)(b1+b2)/s
=(b1+b2)*p -p(b1+b2*d)
=p(b1+b2-b1-b2*d)
=p*(b2-b2*d)
=b2*p*(1-d)
I double checked this using python and got the same result.
[image: i.postimg.cc]
So back to your equation. I think it's correct but only when all of the tickets are from the lucky vouchers. If using both tokens and vouchers, then "amount bought" mentioned above is incorrect, as that would means number of tickets you have purchased with tokens and number of tickets from the lucky vouchers.
Using both tokens and vouchers or either one, the simplified EV is b2 * p * (1-d) ,aka
Number of tickets from the lucky vouchers * Price of a single ticket * the discounted value.
It's funny that it's has nothing to do with the tokens though. For examples, the EV of Lucky Shot Lotto would be the same between using 1000 tokens + 2 vouchers and 10 tokens + 2 vouchers. If you get 15% off from both vouchers compared to the standard 1 million(100 tickets that cost 10k each), then no matter how many tokens you use for tickets, the EV is still 15%*1millon*2, which is 300k.
If you get them from different prices, then it looks like this:
[image: i.postimg.cc]
If the assumptions and calculation above are correct though. Anyway, thanks for bringing it up. And I think It could be even more interesting once you know there has an positive EV and you can apply the Kelly Criterion. But that's off-topic for now.
Once again, I couldn't find this specific topic while searching, so please bear with me if you've come across these questions repeatedly. Thank you in advance for any thoughts on this topic.
My question is this: Does the lottery have an expected value of zero?
If so, then buying lottery tickets is the same as playing RR, but with much higher volatility, and daily limit on tokens for sure. That being said, getting in a cruise line company for extra casino tokens seems good in this way, as in a long run, you don't lose anything regarding the 0 EV but have a higher chance of winning than others.
For the calculation part, the result seems intuitive, but just in case someone needs it or if I made any mistakes or redundancies...
b :Number of tickets you bought
s :Total number of tickets sold overall
p:Price of a ticket
chance of winning = b / s
chance of losing = 1 -b / s
value of winning = p * s -b * p
value of losing = -b * p
EV: b / s * (p * s - b * p) -b * p * (1-b / s )
= b * p - b * p + b ^ 2 * p / s - b ^ 2 * p / s
= 0
Source Word: FETCH
Speaking of speed broadly, it actually reminds me of some memes . I wonder how many things have been become billions of times faster than they used to be xD.
[image: www.physicsforums.com]
[image: www.physicsforums.com]
Source Word: CHOREOGRAPHY
Disclaimer:
Background music, sound effects, and stock footage used in this video were obtained from royalty-free sources. A complete list of the links is available upon OP's request.
Source Word: CONSPICUOUS
Source word: SERENDIPITY
Source Word: CANDLESTICK
I think I also got a royal flush when I was still in the newbie room :D
- Best hand:10♥ J♥ Q♥ K♥ A♥