Balls in a Jar
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.
You're walking down a street, when you see some people gambling on the roadside.
The dealer says:
"There are 10 black balls and 10 white balls in this jar. You are blindfolded and you randomly pick 10 balls. If 5 of the balls you picked are black and the other 5 are white, you lose $2. Otherwise you win $1."
Is the gamble fair?
HintYou know that, in most gambles, the dealer wins more than he loses. If not, why does he hold the gamble?
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Answer
The number of combinations to take 10 balls out of 20 is 20!/(10!)^2=184756.
The number of combinations to take 5 balls of each color is (10!/(5!)^2)^2=63504.
This means in 184756 games you lose 63504 games and win 121252 games in average, totaling a loss of $5756. In other words, you lose about $0.03116 per game in average.
Therefore, the gamble is not fair.
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Comments
redraptor50
Nov 11, 2005
 Very well done, I was going in a different direction in a sense, liked it a lot and keep these coming 
bookworm91
Nov 11, 2005
 good one. i didnt get it (i just guessed, hey 50/50 chance rite?) but had fun doin it. 
pistons_32_rip
Nov 11, 2005
 i thought u had 50% of a chance oh well im dumb and i got 2 get used 2 it 
rkaaland
Nov 11, 2005
 Guess I was just too tired to do the math on this one, although I know it wouldn't make sense for a gamble to set himself up for a loss, so I guessed!!! great one though may have to try it on someone myself although with my luck I would probably lose the first 10 
Princeford93
Nov 12, 2005
 That is a very suggestive title 
zonarita
Nov 12, 2005
 Had the answer. I was glad you told me how it worked mathematically. 
tjc_13
Nov 13, 2005
 i feel dumb i thought well i dont know what i thought 
jls900
Nov 14, 2005
 ahhh you got me 
SPUTNIK2
Jan 20, 2006
 very witty and a clever teaser thanks 
Cariapat
Feb 02, 2006
 I loved this so much going to have nightmares about my probability class taken from the infamous Robert Waller!
Keep em coming.
cariapat 
horses_rock101
Feb 17, 2006
 that was so easy!!!! 
paul726
Feb 26, 2006
 I don't think most gambling casinos pay out much more than 97 cents on the dollar, and perhaps much less. I'd say the gambler was fair, as he ought "earn" something for his work. Anyway, good calculations! 
YVAU
Mar 02, 2006
 The math is good but you don't even have to worry about the formula. You are going to pick 10 balls out of the 20. You have an equal chance that you come back with (0 black, 10 white) or (1 black, 9 white) or (2 black, 8 white), etc. Out of these 20 possibilities, 2 are (5b,5w) or (5w,5b). So 1 in 10 is 5b,5w while 9 in 10 is something else. So you lose $2 10% of the time and make $1 90% of the time. Take the bet, if you play 100 times you'll make $70. 
YVAU
Mar 02, 2006
 I screwed up in the previous post: use what I said to understand the mechnism of unfairness but my math is incorrect (oiverly simplistic, sorry). The math in the original answer is correct. 
mrfatbat
Mar 26, 2006
 Looks like I have to back to my textbooks and learn all about permutations and combinations before I do any more of these probability problems 
Cridol
May 02, 2006
 that was easy and fun.i got it.it's not fair.am i right?lol of course! 
Psychic_Master
May 11, 2006
 Cool one! 
opqpop
Sep 28, 2010
 I don't like problems where I have to use a calculator...
Nevertheless, the approach to this problem is very basic if you're comfortable with counting. 
silverhand
Jul 21, 2011
 A bit confused again... why are we squaring the combination for 5 balls of one color out of 10. If 10!/(5!)^2 is the number of combinations of picking 5 black balls... by default isn't this the same set of combinations for 5 white since you are either picking a white or a black? I manage to get most of this questions with a bit of thought but this one seems to elude me. Thanks 
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