Probability question...
September 3rd, 2013 at 3:57:12 PM
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You have 150 marbles in a bag.
140 of them are green
10 of them are red
If you reach into the bag and randomly pull out 13 of them, what % of the time would all 13 of them be green?
And....go.
140 of them are green
10 of them are red
If you reach into the bag and randomly pull out 13 of them, what % of the time would all 13 of them be green?
And....go.
September 3rd, 2013 at 4:05:55 PM
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Quote: gravity89You have 150 marbles in a bag.
140 of them are green
10 of them are red
If you reach into the bag and randomly pull out 13 of them, what % of the time would all 13 of them be green?
There are COMBIN(140,13) ways to choose 13 marbles from the 140 green marbles.
There are COMBIN(150,13) ways to choose 13 marbles from the 150 green and red marbles.
The probability of pulling all green marbles is therefore COMBIN(140,13)/COMBIN(150,13) which is 0.392061792, so about 39.2%.
September 3rd, 2013 at 4:06:37 PM
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Zero.
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
September 3rd, 2013 at 4:09:22 PM
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If only 10 out of 150 are green, if you pull out 13 you will never pull out all green marbles.
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
September 3rd, 2013 at 4:09:56 PM
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Quote: DJTeddyBearIf only 10 out of 150 are green, if you pull out 13 you will never pull out all green marbles.
10 are red, 140 are green.
September 3rd, 2013 at 4:24:55 PM
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Quote: DJTeddyBearZero.
Re-read the question :)
"Then you can admire the real gambler, who has neither eaten, slept, thought nor lived, he has so smarted under the scourge of his martingale, so suffered on the rack of his desire for a coup at trente-et-quarante" - Honore de Balzac, 1829
September 3rd, 2013 at 4:26:38 PM
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Deleted
DUHHIIIIIIIII HEARD THAT!
September 3rd, 2013 at 4:43:39 PM
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Quote: gravity89You have 150 marbles in a bag.
140 of them are green
10 of them are red
If you reach into the bag and randomly pull out 13 of them, what % of the time would all 13 of them be green?
And....go.
Short answer: JB's right.
Long answer, and long way:
(140/150) * (139/149) * (138/148) * (137/147) * (136/146) * (135/145) * (134/144) * (133/143) * (132/142) * (131/141) * (130/140) * (129/139) * (128/138) = 0.3920617916597633 or 39.2%
Same thing JB did, just expressed differently. His way is better.
https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219
September 3rd, 2013 at 5:21:31 PM
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Why the question?Quote: gravity89If you reach into the bag and randomly pull out 13 of them, what % of the time would all 13 of them be green?
And....go.
The Hypergeometric Distribution
JB almost showed the complete formula for it.
(he did not really need to for your exact Q)
More Qs
How about exactly:
(same draw without replacement of 13 total)
12 greens and 1 red
or
11 greens and 2 red
or
at least 11 greens
(11,12,13)
lots of math
(do not attempt this while driving
and very large numbers doing this long hand)
the complete distribution can easily be done in Excel
or even R
here is R results
(R is FREE or online FREE)
http://www.compileonline.com/execute_r_online.php
0 to 13 greens (read across)
dhyper(0:13, 140, 10, 13)
[1] 0.000000e+00 0.000000e+00 0.000000e+00 2.445376e-13 8.375413e-11
[6] 1.025151e-08 6.150903e-07 2.060553e-05 4.110802e-04 5.024314e-03
[11] 3.761058e-02 1.666832e-01 3.981878e-01 3.920618e-01
or without e
options(scipen=999)
dhyper(0:13, 140, 10, 13)
[1] 0.000000000000000000000 0.000000000000000000000 0.000000000000000000000
[4] 0.000000000000244537599 0.000000000083754127661 0.000000010251505225729
[7] 0.000000615090313543749 0.000020605525503715599 0.000411080233799126572
[10] 0.005024313968655985153 0.037610578851081946650 0.166683247180931387588
[13] 0.398187757154447097285 0.392061791659763236861
winsome johnny (not Win some johnny)
September 3rd, 2013 at 5:46:51 PM
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Quote: 7crapsThe Hypergeometric Distribution
JB almost showed the complete formula for it.
(he did not really need to for your exact Q)
Correct, as this solution was simpler. But you could get the same result with the Excel formula =HYPGEOMDIST(13,13,140,150).
September 3rd, 2013 at 6:09:17 PM
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Yes, the beauty of functions.Quote: JBBut you could get the same result with the Excel formula =HYPGEOMDIST(13,13,140,150).
your use of COMBIN() can also be expanded so the OP can see how to answer a few of my Qs.
COMBIN(140,13) * COMBIN(10,0) / COMBIN(150,13)
answers the OP question also
(I can easily solve COMBIN(10,0) in my head without thinking,
the others require some thought)
still wondering why this question
(maybe a M&M type question)
trying to relate it to gambling
winsome johnny (not Win some johnny)
