Game Thread: The Life and Times of Atto Ngoogol

1) Cards have no memory, so the time the player got an eight is immaterial to the odds of getting sixes the other two times. Each of the other two times it was 50% to get a six, so cumulatively the odds of getting a six twice is 25%.

2) This is solved with a series of simple facts.
P (cards that score points) = 150/3 = 50 [one third score points]
BP (blue cards that score points) = 50/2 = 25 [half of scoring are blue]
Z (cards with no points) = 150-P = 100 [remaining cards]
BZ (blue cards no points) = z*0.3 = 100*0.3 = 30 [30% of non-scoring are blue]
B (blue cards) = BP + BZ = 50+30 = 80

BS (blue card scores) = BP/B = 50/80 = 62.5%

3) It is 16 times more likely to have won on Tuesday than another day.
T = 16d / 16d+d
T = 16/17 = 94% (rounded to a %)

Note: on question 3 there is a pitfall. If you consider the odds over a week, playing one game per day, it is a 22/1024 chance of getting a royal fizbon and 16/22 chance of that being on a Tuesday. But the question asks about the odds in 1 game which is the answer given above.

1) Cards have no memory, so the time the player got an eight is immaterial to the odds of getting sixes the other two times. Each of the other two times it was 50% to get a six, so cumulatively the odds of getting a six twice is 25%.
I think the 8 matters, because it specifies that he receives and 8 one game, then a 6, then another six. Yes it's 25% for any event of 6 coming up twice in a row but that also includes instances of a 666, 766, and 866. Each are independent games or events so it is 1/4*1/2*1/2= 1/16 or 6.25%

2) This is solved with a series of simple facts.
P (cards that score points) = 150/3 = 50 [one third score points]
BP (blue cards that score points) = 50/2 = 25 [half of scoring are blue]
Z (cards with no points) = 150-P = 100 [remaining cards]
BZ (blue)
BS (blue card scores) = BP/B = 50/80 = 62.5%
The question is "what is the probability that a player who has a blue card could score a point with it?" So we need to calculated the percentage of point card blues out of total blues. 30% of Z or 30 are also blue. Total blues are 25 point plue + 30 non-point blue = 55 ... 25/55= 45.5%

3) It is 16 times more likely to have won on Tuesday than another day.
T = 16d / 16d+d
T = 16/17 = 94% (rounded to a %)

Note: on question 3 there is a pitfall. If you consider the odds over a week, playing one game per day, it is a 22/1024 chance of getting a royal fizbon and 16/22 chance of that being on a Tuesday. But the question asks about the odds in 1 game which is the answer given above.
With this one it's best to assume that there are 64 hands dealt per game which happen each day. So there would be one royal fizbon on every tuesday and one every 16 days wed-mon. With the six non tuesdays in a week there is a 6/16 chance of having a royal fizbon occur in a one week period or a 37.5% chance occurring it's inverse being 2 2/3 times more likely it occurred on a tuesdsay than any other day.

1) Cards have no memory, so the time the player got an eight is immaterial to the odds of getting sixes the other two times. Each of the other two times it was 50% to get a six, so cumulatively the odds of getting a six twice is 25%.

2) This is solved with a series of simple facts.
P (cards that score points) = 150/3 = 50 [one third score points]
BP (blue cards that score points) = 50/2 = 25 [half of scoring are blue]
Z (cards with no points) = 150-P = 100 [remaining cards]
BZ (blue) (blue cards no points) = z*0.3 = 100*0.3 = 30 [30% of non-scoring are blue]
B (total blue) = 25+30 = 55
BS (blue card scores) = BP/B = 25/55 = 45.45% (approximately)
credit to rhawn: The question is "what is the probability that a player who has a blue card could score a point with it?" So we need to calculated the percentage of point card blues out of total blues. 30% of Z or 30 are also blue. Total blues are 25 point plue + 30 non-point blue = 55 ... 25/55= 45.5%

3) It is 16 times more likely to have won on Tuesday than another day.
T = 16d / 16d+d
T = 16/17 = 94% (rounded to a %)

Note: on question 3 there is a pitfall. If you consider the odds over a week, playing one game per day, it is a 22/1024 chance of getting a royal fizbon and 16/22 chance of that being on a Tuesday. But the question asks about the odds in 1 game which is the answer given above.

puzzle said:

If he got an eight in one of the games, what is the chance he got sixes the other two times?

Click to expand...

The part in red is a red herring, often asked in statistics questions to mislead the solver into calculating odds on permutations that include that game. Which is why I said "cards have no memory" -- the probability in the other two hands is not affected by the one where an eight came up.

OK maybe the trick in Q1 made me think Q3 was also a trick. I could go either way. Maybe robbiecon knows, having just had a course on it? It's been over 15 years for me...

1. Yes, the 3 cards have no correlation with each other. Nor does it matter what else has been drawn, given that we just have a probability for each card draw, and not how many cards there are.
So 1/2 *1/2 = 1/4 = 25%
We could have been told he had gotten a 6, 7 or 8 in one of the games, it wouldn't have mattered. There is a type of probability, given A, find B, but A and B here have no effect on each other.

2.
Number of cards: 150
Number of card that score points: 50
Number of blue cards that score points : 25
Number of blue cards that don't score points: 30

Total blue cards: 55

Pr(Blue card scores points) = 25/55 = 45.5% roughly.

Yep, this is just a basic read the question, and get the facts. Nothing too tricky.

3.
Tuesday = 16x
other 6 days = 1x each. (x = Pr(getting a royal fizbon)

Total possibilities for the week, 22, given one hand.

Therefore Pr(Prince beaten by a royal fizbon on Tuesday) = 16/22 = 72.72%

Yep, DaveShack I do believe you are right with your add on. We are asked what is the probability of it happening on Tuesday. Not necessarily what is the probability as opposed to another individual day. So yes, my interpretation is what's the odds of this event happening on a Tuesday.

1. Yes, the 3 cards have no correlation with each other. Nor does it matter what else has been drawn, given that we just have a probability for each card draw, and not how many cards there are.
So 1/2 *1/2 = 1/4 = 25%
We could have been told he had gotten a 6, 7 or 8 in one of the games, it wouldn't have mattered. There is a type of probability, given A, find B, but A and B here have no effect on each other.

2.
Number of cards: 150
Number of card that score points: 50
Number of blue cards that score points : 25
Number of blue cards that don't score points: 30

Total blue cards: 55

Pr(Blue card scores points) = 25/55 = 45.5% roughly.

Yep, this is just a basic read the question, and get the facts. Nothing too tricky.

3.
Tuesday = 16x
other 6 days = 1x each. (x = Pr(getting a royal fizbon)

Total possibilities for the week, 22, given one hand.

Therefore Pr(Prince beaten by a royal fizbon on Tuesday) = 16/22 = 72.72%

1. 1/2 *1/2 = 1/4 = 25%

2.
Pr(Blue card scores points) = 25/55 = 45.5% roughly.

3.
Tuesday = 16x
other 6 days = 1x each. (x = Pr(getting a royal fizbon)

Total possibilities for the week, 22, given one hand.

Therefore Pr(Prince beaten by a royal fizbon on Tuesday) = 16/22 = 72.72%

1. 1/2 *1/2 = 1/4 = 25%

2.
Pr(Blue card scores points) = 25/55 = 45.5% roughly.

3.
Tuesday = 16x
other 6 days = 1x each. (x = Pr(getting a royal fizbon)

Total possibilities for the week, 22, given one hand.

Therefore Pr(Prince beaten by a royal fizbon on Tuesday) = 16/22 = 72.72%

1. 1/2 *1/2 = 1/4 = 25%

2.
Pr(Blue card scores points) = 25/55 = 45.5% roughly.

3.
Tuesday = 16x
other 6 days = 1x each. (x = Pr(getting a royal fizbon)

Total possibilities for the week, 22, given one hand.

Therefore Pr(Prince beaten by a royal fizbon on Tuesday) = 16/22 = 72.72%

Not meant to be overly long or difficult, but didn't quite work out I guess. As I once again remarked to the dead I should have made this one into a possible "trick question" short of puzzle, by having a bit of writeup in the original backstory something about Tuesdays being a day of religious observance, gambling and games of chances outlawed, and so on... but I digress.

One of the most useful things to learn in basic statistics, or at least I think it's cool enough, is Bayes' Theorem. Now, not every problem here really required it or was the most conducive to having to use it but interesting nonetheless, if you want to learn something easy to learn but useful to know in the long run. (part two was the most obvious/easy to work though, especially if I hadn't given the total number of cards I guess but still not so bad)

What you'll be asking now is what part everyone got wrong then...so, you all mostly had the answers to 2 and 3 correct. It was meant to be interpreted that there's a 1/7 chance of any game having been on a Tuesday as opposed to another day of the week but that seemed to work itself out in discussion. 2 you all got right again.

So problem one has a rather different answer. I could see one way there might have been an unclear alternative interpretation perhaps, a bit on that in a moment, but 25% is definitely not correct. The answer I had was 12/37.

To start here's an example problem, or roughly a more famous problem you may have heard of that illustrates the point.

You know a family with three children each a couple of years apart in school. If two of the children are boys, what is the probability the third child is a girl?

Note this is quite different from: The eldest and youngest child are boys, what is the probability the middle child is a girl?

Same thing basically applied here, just different probabilities and outcomes all around. You could write out all the cases, and find that 12/37 times when one of the games drew an 8 you'd draw 2 other sixes. I suppose I didn't make it quite perfectly clear in that cases of two 8's or three 8's would also count, if you eliminated those and assumed "exactly one game got an 8" you might get a slightly different answer, but still not 1/4.

By Bayes' Theorem with this you could work out a bunch of different cases for comparison if you got curious though again it would be simple enough to lay out all the possible permutations. Though anyway:

1/2 of the time you'd get two 6's drawn for any three such games as in the problem.
(3/4)^3 or 27/64 is how often you won't get a single 8, so getting at least one 8 is 37/64.

and then it's a 3/8 chance that if you got 6's in two games, you got an 8 in the other. So then you can determine the inverse, (3/8)*(1/2) / (37/64) is 12/37.

Or case-by-case basis



So yeah, that first one probably was "trickier" but then again some of you were right on in that your thinking for, say, #3 should also apply. More of a minor puzzle this time anyway though as you know from the Strength bonus/penalties anywahy, and again we're not going to be on math all the time

Grof Mouthbreather 7 - Seon, TheLastOne36, Jarrema, Zack, robbiecon, taillesskangaru, TheLastOne36
Otto von Ventiari 7 3 - dcmort93, Autolycus, Seon, TheLastOne36, Jarrema, Love, Duke Blackstone dcmort93, Autolycus, Love
Yarai 4- taillesskangaru, DaveShack, Duke Blackstone, robbiecon
abstaining - Darth Caesar

Attacks occurred after 4 more hours today

First Nominee: Otto von Ventiari
Second Nominee: Grof Mouthbreather

Keen Sword
Otto von Ventiari 6 - Autolycus, robbiecon, DaveShack, Jarrema, taillesskangaru, rhawn

Shovel
Ceroz Molten - Autolycus
Grof Mouthbreather - Jarrema
Honorius Cummerbund 3 - robbiecon, taillesskangaru, rhawn

Fire Elemental Core
Irobal Sebermerse 3 - robbiecon, taillesskangaru, rhawn
Yarai 2- Autolycus, Jarrema

Potion Kit
Anna 3 - Autolycus, DaveShack, Jarrema
Lo Hwang Halfblood 3 - robbiecon, taillesskangaru, rhawn
TIED VOTE

Smooth Stone
Anna 2 - DaveShack, rhawn
Elvsh Darkfoot 2- Autolycus, Jarrema
Honorius Cummerbund 2- robbiecon, taillesskangaru
TIED VOTE

Autolycus was Grof Mouthbreather, an Evil-aligned orc and Ally of Atto Ngoogol!