3 ms·
Here is a chart for probabilities for starting value: https://docs.google.com/spreadsheets/d/e/2PACX-1vThljkK2nUILt_emMyPpFbdwvgVYpeaGsgPIswKopeFcQsofetYjKqojF
by Lockal 2y ago
Here is a chart for probabilities for starting value:
https://docs.google.com/spreadsheets/d/e/2PACX-1vThljkK2nUILt_emMyPpFbdwvgVYpeaGsgPIswKopeFcQsofetYjKqojFCLn-w_kM4PCZczKhIcsQ7i/pubchart?oid=412606124&format=interactive https://docs.google.com/spreadsheets/d/e/2PACX-1vThljkK2nUIL...
I find it interesting: it is definitely symmetrical, but I did not expect that in the final result 1/98 could be more important as a starting value, while 2-17/82-97 are not used at all.
- gukoff 2y agoThis really depends on the pure strategies that you choose. The initial set of strategies wasn't very diverse and compensated for the binary search "weaknesses" on the ends of the spectrum by sometimes guessing 1 and 98. But after adding some more pure strategies to the set, we've got a far better mixed strategy that prefers the numbers between 28-70 as the first pick: https://github.com/gukoff/ballmer_puzzle#winning-strategy https://github.com/gukoff/ballmer_puzzle#winning-strategy
- Lockal 2y agoO, wow, post got update! > Avg win if Ballmer chooses randomly: $0.16247848000093376 > Win if Ballmer chooses adversarially: $0.14657033010415976 So the goal is to find a set of strategies where adversarial avg win == random avg win? Or these numbers will never be equal?