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Subject: Need help with math... rss

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Jack Darwid
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We are doing final playtests for Adventure of D now and need help...

The Power Deck has 4 types of cards: STR, AGL, INT and NoType. Each type has Power of 1,2,3,4 and 5. There are 3 identical sets of these cards (for example: there are three [INT 4] cards). So there are 4x5x3=60 cards in the Power Deck.

Every Turn I draw 8 cards from the Power Deck. So when I draw these 8 cards:

1. What is the probability that one of my card is AGL 5?
(my answer is: (3/60) x 8 = 24/60 = 40%, am I right ?)

2. What is the probability that at least two of my cards are INT?

Thanks a lot!

JackD
 
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Josh P.
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Dr. Pete wrote:
As a general rule, probabilities don't simply add together. You're assuming that drawing two cards makes the probability twice as likely as drawing one card, but this is not true because you're always drawing two distinct cards, and so these events are not independent.


http://mathforum.org/library/drmath/view/56532.html
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J. Atkinson
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It's been a while since I messed around with probability and game theory, so here goes..

1.
First card, 3/60, 2nd 2/59, and 3rd, 1/58 since you only have 3 of AGL 5. Each card you draw diminishes the pool of available cards, so you multiply them together. They're dependent on each other.

3/60 x 2/59 x 1/58 = 6/205320 = 0.0029%


2.
15/60 x 14/59 = 210 / 3540 = 5.9 %

Correct me if I'm wrong. I had to shake the cobwebs off.


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David Molnar
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JackDarwid wrote:
We are doing final playtests for Adventure of D now and need help...

The Power Deck has 4 types of cards: STR, AGL, INT and NoType. Each type has Power of 1,2,3,4 and 5. There are 3 identical sets of these cards (for example: there are three [INT 4] cards). So there are 4x5x3=60 cards in the Power Deck.

Every Turn I draw 8 cards from the Power Deck. So when I draw these 8 cards:

1. What is the probability that one of my card is AGL 5?
(my answer is: (3/60) x 8 = 24/60 = 40%, am I right ?)

2. What is the probability that at least two of my cards are INT?

Thanks a lot!

JackD


First, ignore math advice from websites that confuse independence with disjointness.

The way to solve most problems like this [ie, having the phrase "at least" in them] is to first find the probability of the event not happening.

Assuming we are drawing the first 8 cards from the deck...

a) The probability of not getting an AGL5 is the number of combinations of 8 cards drawn from 57, divided by the number of combinations of 8 cards drawn from 60. In Excel, this'd be COMBIN(57,8)/COMBIN(60,8). From my TI-89, this is 1105/1711. So the probability of getting at least one AGL5 is 1 minus this, or 606/1711, which is about 35%.

b) For "at least two", I'm going to count the number of ways to get no INTs, and the number of ways to get exactly one INT. Since there are 45 non-INT cards, there are COMBIN(45,8) ways to get no INTs. To get exactly one INT, there are 15 possibilities for that one card, and then COMBIN(45,7) different combinations for the other seven, non-INT cards.
So all told, COMBIN(45,8)+15*COMBIN(45,7) ways to not get at least two INTs. [COMBIN(45,8)+15*COMBIN(45,7)]/COMBIN(60,8)= 1810601/5168931. The probability of getting at least two INTs is then 3358330/5168931, or about 65%. (The answers to your two questions adding up to 100% is a complete coincidence and only due to rounding.)
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'Bernard Wingrave'
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JackDarwid wrote:
1. What is the probability that one of my card is AGL 5?

Assuming you're interested in the probability of a hand that has at least one AGL5, I think it works like this:

P(hand with at least one AGL5)

= 1 - P(hand with no AGL5s).

=1-[(57/60)(56/59)...(50/53)]

which comes out to about 35%
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David Molnar
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bwingrave wrote:
JackDarwid wrote:
1. What is the probability that one of my card is AGL 5?

Assuming you're interested in the probability of a hand that has at least one AGL5, I think it works like this:

P(hand with at least one AGL5)

= 1 - P(hand with no AGL5s).

=1-[(57/60)(56/59)...(50/53)]

which comes out to about 35%


Always ask a Unitarian.
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Stephen
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1. 1 - ((52x51x50)/(60x59x58)) = 35.42%

2. 1 - (8691104822400/103163592470400) - (8 x (3430699272000/103163592470400)) = 1 - 0.0842459 - 0.266040 = 64.97%

I'm not absolutely sure about these answers (which are approximate), but they look reasonable. I'd recommend drawing 20 or so hands and checking each hand for both conditions as a sanity check.
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Andrew Tullsen
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1. (3/60 + 3/59 + 3/58 + 3/57 + 3/56 + 3/55 + 3/54 + 3/53) = 42.5479387 %
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Martin Gallo
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Based on empirical evidence, it depends on who is drawing and the importance of getting one or two of those cards. For example, if I were playing and I need those cards to be effective that turn the odds of getting one are roughly 10%. Getting is about 0%. I have a friend who would probably have an 80% chance of getting two of the required cards and 100% for just one of them.

Statistics can be annoying.
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L Gravel
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I'm a little rusty, but here's my attempt.

There are "60 choose 8" or 2,558,620,845 possible hands. This is the divisor for all considerations.

The number of possible hands with one and only one of the three AGL 5 cards is:
3 x (57 choose 7) = 3 x 264,385,836 = 793,157,508.
Dividing gives us 31%.

If you meant the probability of getting at least one AGL 5 card, the easiest way is first to consider the number of hands that have no AGL 5 cards:
57 choose 8 = 1,652,411,475.
Therefore the hands that have at least one AGL 5 card are 2,558,620,845 - 1,652,411,475 = 906,209,370.
Dividing gives us 35%.

Finally re: the probability that at least two are INT,
the number of hands with no INT cards is:
45 choose 8 = 215,553,195.
The number of hands with one and only one INT card is:
15 x (45 choose 7) = 15 x 45,379,620 = 680,694,300.
Therefore the hands that have at least two INT cards are 2,558,620,845 - 215,553,195 - 680,694,300 = 1,662,373,350.
Dividing gives us 65%.

Jim
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Mike Reading
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Lol Martimer, "it depends on who is drawing and the importance of getting one or two of those cards." So true.

Thumbs up to Molnar's for pointing out =combin() in Excel! I've been using lots of =fact() until now.

Another set of thumbs to Molnar for showing a neat way of working out "at least two" (the second question). I suspect he has some actual math training! I've been running on "Probability for Dummies" and http://en.wikipedia.org/wiki/Combination...
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J. Atkinson
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bwingrave wrote:
JackDarwid wrote:
1. What is the probability that one of my card is AGL 5?

Assuming you're interested in the probability of a hand that has at least one AGL5, I think it works like this:

P(hand with at least one AGL5)

= 1 - P(hand with no AGL5s).

=1-[(57/60)(56/59)...(50/53)]

which comes out to about 35%


Yes, better. Subtract the probability that it's not there from unity! That accounts for the rest of the cards in the hand. I knew something was wrong with my answer.
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Ian Madsen
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I'm not rusty on my math, so my answers come with a money back guarantee! cool
JackDarwid wrote:
1. What is the probability that at least one of my cards is AGL 5?
(my answer is: (3/60) x 8 = 24/60 = 40%, am I right?)
(57/60) x (56/59) x (55/58) x (54/57) x (53/56) x (52/55) x (51/54) x (50/53) =~ 0.6458

1 - 0.6458 =~ 35.42%
Quote:

2. What is the probability that at least two of my cards are INT?
(45/60) x (44/59) x (43/58) x (42/57) x (41/56) x (40/55) x (39/54) x (38/53) =~ 0.0842

(45/60) x (44/59) x (43/58) x (42/57) x (41/56) x (40/55) x (39/54) x (15/53) x 8 =~ 0.2660

0.0842 + 0.2660 =~ 0.3503

1 - 0.3503 =~ 64.97%
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Jack Darwid
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Thanks for the quick answers...

So the answers are:
1. the probability to get at least one [AGL 5] is around 35%
2. the probability to get at least two INT cards is around 65%

The devil is in the details... We are playtesting AoD as many as we can and I tried to spot any game-breaker cards and try to balance everything, while preparing the final card design with T'Say.
This is a great experience (and exhausting since we do this all by ourselves and this is our first time). There is still a long road ahead... distributing+delivery(arrghh) etc etc...

That's why I want to skip the "learn again my high scholl books" about probability and want some direct anwer.

Thanks a lot! (and now back to playtesting and preparing)

JackD
 
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Robin Goodall
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In these situations, I could do the maths to work this out but I find knocking up a bit of code to do the random draw a few million times much more fun.

Hmm...makes me sound really nerdy
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Gijs van de Kuilen
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wrong answers: 2
right answers: 6
P(your statistics question will be answered correctly by a BGG-er) = .75

;)
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