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General discussionGOG Mafia Sign-Up: Game #62 (Mafia/Werewolf - an online social deduction game, new players welcome!)(159 posts)(159 posts)
(159 posts)
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This is my favourite topic
Posted September 26, 2019
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RedFireGaming: Joe said we could boot this game to 15 players if there's enough interest, and I remember GameRager saying he could join if the signup phase lasted long enough.
I'd like to see myself more in the lines of a filler of Joe's [game's] empty slot. My participation alone doesn't justify the need to find more people to fill in the additional slots it brings with it.
Posted September 26, 2019
I still say we draft Bookwyrm.
Posted September 26, 2019
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JoeSapphire: In this scenario I don't have a twelve sided dice handy, so I'm using two dice to randomly determine a seat.
In this scenario I would take offence if somebody thought I would confuse a random selection method from 12 places with a skewed selection method from 11 places, so please factor that into your calculations.
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ZFR: Assuming that regardless of the number of dice you use, you get a number in range [1,12] with a uniform probability, then:

In essence, you need to roll until you hit an empty seat (the fact that the person replaces someone in a seat that's already taken doesn't matter).
The average (mean, not median) number of times, it takes to roll an empty seat is the inverse of probabililty (proof below for those interested). So when there are k empty seats, it takes an average of 12/k times to roll for one.

Therefore the answer is:
Sum(n: 12 to 1)(12/n) = 12/12 + 12/11 + 12/10 + ... + 12/2 + 12/1 ≈ 35.74
Neat! I got it done in 24 rolls (meaning 48 dice were rolled but that's a technicality)
Looks like I beat maths.


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JoeSapphire: While we wait to find out who the last sign-up is I have a problebility for you:

If I have 12 seats that need to be filled, and every time someone rocks up to sit in a seat I roll two dice to randomly determine a seat for that person. If someone is sat in a seat that is occupied, the occupator is displaced, and I roll again to determine them a new seat. What is the average number of times that I will roll a dice to seat twelve people?
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Lifthrasil: Two. After rolling twice, you'll be fed up and just assign seats arbitrarily.
Incorrect


It looks like we have enough players. I haven't heard from MindCollapse but I'll try and put something together so that GameRager and dedoporno can both get involved and hopefully begin the game over the weekend.


Can I get a "Woohoo"?
Posted September 26, 2019
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JoeSapphire: Can I get a "Woohoo"?
WOOHOO!
Posted September 26, 2019
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JoeSapphire: Can I get a "Woohoo"?
Yoohoo?
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JoeSapphire: I haven't heard from MindCollapse
We haven't heard anything from the other spambot either... :(
Post edited September 26, 2019 by SirPrimalform
Posted September 26, 2019
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JoeSapphire: Can I get a "Woohoo"?
Can you?
Posted September 26, 2019
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ZFR: Assuming that regardless of the number of dice you use, you get a number in range [1,12] with a uniform probability, then:

In essence, you need to roll until you hit an empty seat (the fact that the person replaces someone in a seat that's already taken doesn't matter).
The average (mean, not median) number of times, it takes to roll an empty seat is the inverse of probabililty (proof below for those interested). So when there are k empty seats, it takes an average of 12/k times to roll for one.

Therefore the answer is:
Sum(n: 12 to 1)(12/n) = 12/12 + 12/11 + 12/10 + ... + 12/2 + 12/1 ≈ 35.74
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JoeSapphire: Neat! I got it done in 24 rolls (meaning 48 dice were rolled but that's a technicality)
Looks like I beat maths.
35.74 is average (mean), not median. The median is much lower than that. Meaning you're much more often going to get a value less than 35.74 than a value greater than. But, since there is no uppper limit, you can have rare large values that bring the mean up.

For example, for a single dice, it takes an average of 6 rolls to roll a 6. But to get the median x:
(5/6)^x = 1/2

x = -1 / log2(5/6) ≈ 3.8

Which means that on average 50% of the time you need 3.8 rolls or less and 50% of the time you get more than that.

Or, you can look at it this way: even though the average number of times it takes to roll 6 is 6, but about 66.5% (roughly two thirds) of the people who try it are going to "beat maths" and get it in a smaller number of rolls.
Posted September 26, 2019
ADD: Of course since the distribution of rolls is defined on integers (you can't roll 3.8 times), the median is going to be 4, and not 3.8.
Posted September 27, 2019
I cannot play until after the weekend at the earliest, so I say we push for 15!

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ZFR: You don't actually have to do that. You can easily simulate a uniform distribution (1-12) with two normal d6 dice.

e.g. multiply the result of first dice by 2. Then use the second dice to either subtract 1 from the result or leave it as it is (if the second dice rolls 1-3 you subtract 1 from the first dice's result multiplied by 2). So:

First dice determines one of those groups:

(1,2)
(3,4)
(5,6)
(7,8)
(9,10)
(11,12)

Second dice determines if it's the first or second number from each group that's chosen.

With two d6 dice rolls, you can simulate a random uniform distribution in any range up to 36.
Thanks! I hadn't considered that possibility.
Posted September 27, 2019
While we wait:

The unbeatable duo want to determine whom to send to do the NK. Problem is, *someone* doesn't want to use random.org; he loves rolling dice instead. He says that if he rolls an even number then the prettier half of the duo does the NK, and if he rolls an odd number then the smarter half will do it. Problem is, the only dice he has is clearly unbalanced: some numbers are more likely to come than others. Fortunately, the smarter half of the unbeatable duo devices a clever method of using the unbalanced dice to randomly pick one of them with a 50% probability. How?

(This actually has a proper mathematical solution).
Posted September 27, 2019
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Microfish_1: I cannot play until after the weekend at the earliest, so I say we push for 15!
I still say we draft Bookwyrm.
Posted September 27, 2019
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dedoporno: I see a new (to me at least) player has expressed an interest in what seems to be the last free spot but in case they decide otherwise or you guys end up needing one last push do let me know.
If you don't want to play and the new guys get spots already i'll take it gladly.

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RedFireGaming: Joe said we could boot this game to 15 players if there's enough interest, and I remember GameRager saying he could join if the signup phase lasted long enough.
If Joe adds more spots or a spot opens up I will join. I do little between posting to the few active threads and hosting the rare contest here anyways.

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JoeSapphire: It looks like we have enough players. I haven't heard from MindCollapse but I'll try and put something together so that GameRager and dedoporno can both get involved and hopefully begin the game over the weekend.

Can I get a "Woohoo"?
What about a Yahooo-ooo-ooo?
Post edited September 27, 2019 by GameRager
Posted September 27, 2019
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JoeSapphire: It looks like we have enough players. I haven't heard from MindCollapse but I'll try and put something together so that GameRager and dedoporno can both get involved and hopefully begin the game over the weekend.

Can I get a "Woohoo"?
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GameRager: What about a Yahooo-ooo-ooo?
What about a Woo?
Posted September 27, 2019
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GameRager: What about a Yahooo-ooo-ooo?
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Lifthrasil: What about a Woo?
That works too.

Also an aside: I love that game series....nice pic....gonna have to swipe a copy.
Posted September 27, 2019
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Lifthrasil: What about a Woo?
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GameRager: That works too.

Also an aside: I love that game series....nice pic....gonna have to swipe a copy.
PS2 or PS3 version? I love it too.
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This is my favourite topic
Community
General discussionGOG Mafia Sign-Up: Game #62 (Mafia/Werewolf - an online social deduction game, new players welcome!)(159 posts)(159 posts)
(159 posts)