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## Balls in a Jar

Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.

 Puzzle ID: #26847 Fun: (2.46) Difficulty: (2.17) Category: Probability Submitted By: shenqiang Corrected By: Dave

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?

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 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