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Old 05-27-2020, 06:51 PM   #1
ScotchOnDaRocks
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Probability Help

Hello, was hoping someone could help me with the following problems and show the work so I can learn. (This has OFC implications but I'm making it more generic so anyone can possibly help).

You have an urn with 27 objects and you randomly choose three items from the following:

2 Gold Bars
9 silver bars
3 copper coins
13 pieces of coal

1) Prob of getting at least one gold bar and at least one silver bar
2) Prob of getting at least one gold bar and at least one copper coin (but no silver bar)
3) Prob of getting no gold bars but at least one silver bar
4) Prob of getting no gold bars at least one copper coin (but no silver bar)
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Old 05-27-2020, 08:52 PM   #2
heehaww
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Re: Probability Help

1. You can get exactly 1 of each, or 2 and 1.
[16*2*9 + 2*C(9,2) + 9] / C(27,3) = 41/325

2. [13*2*3 + 2*3 + 3] / C(27,3)

3. 100% - P(no silver) - 41/325

4. P(1 copper, 2 coal) + P(2 copper, 1 coal) + P(3 copper)

Last edited by heehaww; 05-27-2020 at 09:00 PM. Reason: Dohhh
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Old 05-27-2020, 11:32 PM   #3
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Re: Probability Help

Quote:
Originally Posted by heehaww View Post
1. You can get exactly 1 of each, or 2 and 1.
[16*2*9 + 2*C(9,2) + 9] / C(27,3) = 41/325

2. [13*2*3 + 2*3 + 3] / C(27,3)

3. 100% - P(no silver) - 41/325

4. P(1 copper, 2 coal) + P(2 copper, 1 coal) + P(3 copper)
Thanks hehaw, there are different items or maybe I'm doing something wrong below?

1) Prob of getting at least one gold bar and at least one silver bar

1 gold, 1 silver, 1 copper =2*9*3 =54
2 gold, 1 silver =2*1*9 =18
1 gold, 2 silver =2*9*8 = 144
1 gold, 1 silver, 1 coal =2*9*13 =234

Total from above 450

C(27,3) 2925

15.4%
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Old 05-28-2020, 02:23 AM   #4
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Re: Probability Help

Perhaps explicitly showing the cases and combinatorics may help.

Q1. There are three ways to get at least one gold (G) and at least one silver (S); copper or coal will be denoted by X.

GSX: C(2,1)*C(9,1)*C(16,1) = 2*9*16 = 288
GGS: C(2,2)*C(9,1) = 1*9 = 9
GSS: C(2,1)*C(9,2) = 2*36 = 72

Total = 288+9+72 = 369

ALL POSSIBLE DRAWS = C(27,3) = 2925

Pct = 369/2925 = 41/325 = 12.615%

-----

Q2. This is straightforward as heehaww shows.

-----

Q3. heehaww uses "logic" and a preceding result here. A "direct" approach is also possible.

Let copper & coal = X. Then we have:

SXX: C(9,1)*C(16,2) = 9*120 = 1080
SSX: C(9,2)*C(16,1) = 36*16 = 576
SSS: C(9,3) = 84

Total = 1080+576+84 = 1740

Denom = C(27,3) = 2925

-----

Q4. This is straightforward as heehaww shows.

Last edited by whosnext; 05-28-2020 at 02:47 AM.
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Old 05-28-2020, 05:18 AM   #5
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Re: Probability Help

Quote:
Originally Posted by ScotchOnDaRocks View Post
2 gold, 1 silver =2*1*9 =18
1 gold, 2 silver =2*9*8 = 144
Your mistake was using permutations (bolded, 2*1=2P2 and 9*8=9P2). Your denominator is combinations. Either use combos/combos or perms/perms. (If the latter, you'd also have to multiply those tallies by 3 to account for all 3! arrangements.)
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Old 05-28-2020, 09:46 AM   #6
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Re: Probability Help

Quote:
Originally Posted by heehaww View Post
Your mistake was using permutations (bolded, 2*1=2P2 and 9*8=9P2). Your denominator is combinations. Either use combos/combos or perms/perms. (If the latter, you'd also have to multiply those tallies by 3 to account for all 3! arrangements.)
Thanks, if the numerator is correct, I can see it more easily using permutations and would like to go that route going forward

But I'm sorry, how do I adjust? Can you show the full work for Number 1 doing permutations and I'll see if I can match the others.
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Old 05-28-2020, 10:19 AM   #7
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Re: Probability Help

Quote:
Originally Posted by whosnext View Post
Perhaps explicitly showing the cases and combinatorics may help.

Q1. There are three ways to get at least one gold (G) and at least one silver (S); copper or coal will be denoted by X.

GSX: C(2,1)*C(9,1)*C(16,1) = 2*9*16 = 288
GGS: C(2,2)*C(9,1) = 1*9 = 9
GSS: C(2,1)*C(9,2) = 2*36 = 72

Total = 288+9+72 = 369

ALL POSSIBLE DRAWS = C(27,3) = 2925

Pct = 369/2925 = 41/325 = 12.615%

-----

Q2. This is straightforward as heehaww shows.

-----

Q3. heehaww uses "logic" and a preceding result here. A "direct" approach is also possible.

Let copper & coal = X. Then we have:

SXX: C(9,1)*C(16,2) = 9*120 = 1080
SSX: C(9,2)*C(16,1) = 36*16 = 576
SSS: C(9,3) = 84

Total = 1080+576+84 = 1740

Denom = C(27,3) = 2925

-----

Q4. This is straightforward as heehaww shows.
Thanks, this helps, maybe I can do this using combinations

I'm just so bad with combinatorics, it never takes
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Old 05-28-2020, 12:08 PM   #8
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Re: Probability Help

Thanks to both of you, I understand completely and my numbers reconcile exactly to what I expected.

FYI explicitly showing it helped alot
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Old 05-28-2020, 12:14 PM   #9
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Re: Probability Help

The following is solely a personal observation ...

Without intending to restart any methodological wars, for poker probability questions I use Combinations C(X,Y) something like 99% of the time. In very very few situations do I find it necessary to utilize Permutations P(X,Y) or other methods.

Essentially having just one tool in the toolkit has the advantage of never being confused as to which tool to use. Of course, someone could argue that using just one tool is inefficient.

Of course, feel free to post any further questions. Even if you are just looking for confirmation of results you have already derived.
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Old 05-28-2020, 12:45 PM   #10
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Re: Probability Help

Quote:
Originally Posted by whosnext View Post
The following is solely a personal observation ...

Without intending to restart any methodological wars, for poker probability questions I use Combinations C(X,Y) something like 99% of the time. In very very few situations do I find it necessary to utilize Permutations P(X,Y) or other methods.

Essentially having just one tool in the toolkit has the advantage of never being confused as to which tool to use. Of course, someone could argue that using just one tool is inefficient.

Of course, feel free to post any further questions. Even if you are just looking for confirmation of results you have already derived.
Yes, I think I agree. And it's quite nice that I think I have this concept down solid! Thanks, I'll probably swing by sometime in the future for another application for a confirmation that once seemed difficult but is probably much easier now.
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Old 05-28-2020, 04:01 PM   #11
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Re: Probability Help

I'm the opposite: moarr tools

Sometimes I use permutations when using a calculator without an nCr function because it's fewer keystrokes when I don't have to divide by a factorial. C(n,r) = nPr / r!

Or occasionally they're more efficient regardless, for instance the chance of an unpaired flop is 48*44/51/50, which is nicer than doing C(13,3)*4 / C(52,3)

Quote:
how do I adjust?
GSX has to be multiplied by 3!
For GSS, by using perms you already counted the 2! ways to arrange the S's, so now you just need to multiply by the 3 ways to place the G among the S's. Same thing for GGS.
Then for the denominator use 27*26*25

After those adjustments, there is a factor of 3! present in both the numerator and denominator.

Having to make those multiplications is the drawback of permutations, which is why combos are usually simpler (when order doesn't matter), esp when you have the nCr function available (or the "choose" function in a google search or wolfram|alpha).

Some people like to multiply fractions together (compound probabilities) without worrying about combos or perms. Those same multiplications are necessary when doing that, because fractions are algebraically the same as dividing perms by perms. For instance:

P(GSS) = 3(2*9*8) / (27*26*25)
or P(GSS) = 3(2/27)(9/26)(8/25)
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