Talk:Ice - Revision history

Tomsod: /* Melting ice */ more perma-ice odds

2020-01-08T19:58:10Z

‎Melting ice: more perma-ice odds

Tomsod: i did the math

2018-07-11T14:52:44Z

i did the math

Qazmlpok: Created page with "==Melting ice== The way the timeout is determined is kinda odd, so I don't know what the actual probabilities are. If I'm not misreading the source: *Timer is initialized to 5..."

2016-01-03T19:07:37Z

Created page with "==Melting ice== The way the timeout is determined is kinda odd, so I don't know what the actual probabilities are. If I'm not misreading the source: *Timer is initialized to 5..."

New page

==Melting ice==
The way the timeout is determined is kinda odd, so I don't know what the actual probabilities are. If I'm not misreading the source:
*Timer is initialized to 50 (MIN_ICE_TIME) if not already set. It's already set if this is an ice bolt travelling over existing (thawing) ice rather than plain water.
*There is a 1 in 2000 (MAX_ICE_TIME) chance that the current duration of 50 will be used as the duration.
*If this is not the case, increment the duration. There is a 1 in 1999 chance that the current duration (51) will be used.
*And so on; up to 1950 trials of "1 in (2000 - duration + 50)", with the final one being "1 in 50"
*If none of these trials pass, the ice is made permanent.
-- [[User:Qazmlpok|Qazmlpok]] ([[User talk:Qazmlpok|talk]]) 19:07, 3 January 2016 (UTC)

← Older revision		Revision as of 19:58, 8 January 2020
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	-- [[User:Qazmlpok\|Qazmlpok]] ([[User talk:Qazmlpok\|talk]]) 19:07, 3 January 2016 (UTC)		-- [[User:Qazmlpok\|Qazmlpok]] ([[User talk:Qazmlpok\|talk]]) 19:07, 3 January 2016 (UTC)
	: If that is correct, the chance of permanent ice forming after one zap is 1999/2000 * 1998/1999 * 1997/1998 * ... * 50/51 * 49/50 = 49/2000 (about 2.5% chance), and the chance of duration being N is 1999/2000 * 1998/1999 * ... * (2050-N)/(2051-N) * 1/(2050-N) = 1/2000, exactly the same for each N. However, each subsequent zap will start with higher duration, so the chances of permanent ice should improve quickly with several zaps. After two zaps, the chance is already 49/2000 * (H(2000) - H(49) + 1) [H(n) being harmonic numbers], which is about 11.5%. For three zaps the calculation becomes slightly complicated, but Wolfram\|Alpha gives about 28.3% chance for permanent ice. I'm too lazy to calculate further, but you get the idea. [[User:Tomsod\|Tomsod]] ([[User talk:Tomsod\|talk]]) 14:50, 11 July 2018 (UTC)		: If that is correct, the chance of permanent ice forming after one zap is 1999/2000 * 1998/1999 * 1997/1998 * ... * 50/51 * 49/50 = 49/2000 (about 2.5% chance), and the chance of duration being N is 1999/2000 * 1998/1999 * ... * (2050-N)/(2051-N) * 1/(2050-N) = 1/2000, exactly the same for each N. However, each subsequent zap will start with higher duration, so the chances of permanent ice should improve quickly with several zaps. After two zaps, the chance is already 49/2000 * (H(2000) - H(49) + 1) [H(n) being harmonic numbers], which is about 11.5%. For three zaps the calculation becomes slightly complicated, but Wolfram\|Alpha gives about 28.3% chance for permanent ice. I'm too lazy to calculate further, but you get the idea. [[User:Tomsod\|Tomsod]] ([[User talk:Tomsod\|talk]]) 14:50, 11 July 2018 (UTC)
		+	:: Returning to this, I just recalculated the chances of permanent ice (I gave slightly wrong figures before). For one zap it's 2.45%, for two, 11.56%, for three, 28.48%, and for four I just calculated a 49.61% chance, although it's approximate and I'm not sure all four digits are significant. Perhaps I should add it to the article. [[User:Tomsod\|Tomsod]] ([[User talk:Tomsod\|talk]]) 19:58, 8 January 2020 (UTC)

← Older revision		Revision as of 14:52, 11 July 2018
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	*If none of these trials pass, the ice is made permanent.		*If none of these trials pass, the ice is made permanent.
	-- [[User:Qazmlpok\|Qazmlpok]] ([[User talk:Qazmlpok\|talk]]) 19:07, 3 January 2016 (UTC)		-- [[User:Qazmlpok\|Qazmlpok]] ([[User talk:Qazmlpok\|talk]]) 19:07, 3 January 2016 (UTC)
		+	: If that is correct, the chance of permanent ice forming after one zap is 1999/2000 * 1998/1999 * 1997/1998 * ... * 50/51 * 49/50 = 49/2000 (about 2.5% chance), and the chance of duration being N is 1999/2000 * 1998/1999 * ... * (2050-N)/(2051-N) * 1/(2050-N) = 1/2000, exactly the same for each N. However, each subsequent zap will start with higher duration, so the chances of permanent ice should improve quickly with several zaps. After two zaps, the chance is already 49/2000 * (H(2000) - H(49) + 1) [H(n) being harmonic numbers], which is about 11.5%. For three zaps the calculation becomes slightly complicated, but Wolfram\|Alpha gives about 28.3% chance for permanent ice. I'm too lazy to calculate further, but you get the idea. [[User:Tomsod\|Tomsod]] ([[User talk:Tomsod\|talk]]) 14:50, 11 July 2018 (UTC)