Rolling the Dice
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.
A gambler goes to bet. The dealer has 3 dice, which are fair, meaning that the chance that each face shows up is exactly 1/6.
The dealer says: "You can choose your bet on a number, any number from 1 to 6. Then I'll roll the 3 dice. If none show the number you bet, you'll lose $1. If one shows the number you bet, you'll win $1. If two or three dice show the number you bet, you'll win $3 or $5, respectively."
Is it a fair game?
HintWhat will happen if there are 6 gamblers, each of whom bet on a different number?
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Answer
It's a fair game. If there are 6 gamblers, each of whom bet on a different number, the dealer will neither win nor lose on each deal.
If he rolls 3 different numbers, e.g. 1, 2, 3, the three gamblers who bet 1, 2, 3 each wins $1 while the three gamblers who bet 4, 5, 6 each loses $1.
If two of the dice he rolls show the same number, e.g. 1, 1, 2, the gambler who bet 1 wins $3, the gambler who bet 2 wins $1, and the other 4 gamblers each loses $1.
If all 3 dice show the same number, e.g. 1, 1, 1, the gambler who bet 1 wins $5, and the other 5 gamblers each loses $1.
In each case, the dealer neither wins nor loses. Hence it's a fair game.
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Comments
AZTTT
Oct 13, 2005
 good one to warm up my brain this morning. Makes sense! 
smarty_blondy
Oct 13, 2005
 Very good, the hint helped. Keep them rollin'. 
smarty_blondy
Oct 13, 2005
 Very good, the hint helped. Keep them rollin'. 
(user deleted)
Oct 13, 2005
 I disargee as to being a fiar game, If only one person plays, which is the feel of the problem to start with. There are 216 possible out comes from throughing 3 dies. 1/216 of winning $5, 5/216 of winning $3, and 25/216 of winning $1, but 185/216 of losing $1. Which means on average you lose $0.648 a game, you are better off playing cratchers. 
(user deleted)
Oct 13, 2005
 Opps, I forgot something, I'm wrong, it's actually perfectly fair, better them scratchers.
1/216 to win $5, 15/216 to win $3, 75/216 to win $1, and 125/216 to lose. Which averages to $0 per game. 
redraptor50
Oct 13, 2005
 AH HA! just kidding it threw me off too. But keep them coming 
NTaliesen
Oct 13, 2005
 No not fair. Every good game favours the house. If it doesn't no matter how fun the game is no one gets to play it long. Every game has upkeep costs. With no profitability the gambling will stop when the roller sstarves to death. 
trumpetsgirl
Oct 14, 2005
 Also it depends on how much a bet is if it is 3 dollors or something like that the outcome goes stright to the house 
schatzy228
Oct 18, 2005
 yes, the amount you bet was missing, if $1 bet, it is a fair game
but the intention was good
do the math and have fun!

shenqiang
Oct 21, 2005
 What did you mean by "the amount you bet"? 
Spark
Oct 22, 2005
 That was good 
chidam11
Oct 25, 2005
 It was nice. 
nuccha
Oct 27, 2005
 smurfdew... i disagree. kiss is actually right. great job kiss, I used another way to solve and was amazed that the expected value of a single player is actually zero (meaning, the game is fair).
To smurfdew...
There is 1 (1x1x1x(3C0)) way to win $5, 15 (1x1x5x(3C1)) ways to win $3 and 75 (1x5x5x(3C2)) ways to win $1. There only 125 (5x5x5x(3C3))ways to lose $1.
You forgot to factorin combination... 
nuccha
Oct 27, 2005
 OOOPS .... sorry..... didn't read the next post.... SORRY... SORRYYY 
lareed
Oct 27, 2005
 i liked it makes sense. but some of you are right. in gambling the house always wins. one way or another. fuuunn! keep them coming. 
paintballa944
Nov 06, 2005
 I liked it a lot and that is that. (I'm not getting into all that mathematics stuff.) 
scotchtape10
Nov 24, 2005
 kinda hard... not really good at these... idk! good anyway!! 
mr_brainiac
Jan 02, 2006
 Worrying about the house's profitabilty is irrelevant, no one said this was an actual game in a casino, only a hypothetical question about the odds of this particular scenario. 
Jimbo
Feb 25, 2006
 Great puzzle. I like the way you explained it. Can I use it? 
mrfatbat
Mar 26, 2006
 Good stuff 
Dedrik
Aug 09, 2006
 Always a nice change of pace to see one that isn't the most obvious anwser 
foraneagle2
Oct 19, 2006
 Excel tells me there is an expected payout of $1.8735E16.
Chance of none being the number: X =(5/6)*(5/6)*(5/6)
Chance of one: Y =(1/6)*(5/6)*(5/6)*3
Chance of two = Z =(1/6)*(1/6)*(5/6)*3
Chance of three = T =(1/6)*(1/6)*(1/6)
Expected Value: 5T+3Z+1YX 
foraneagle2
Oct 20, 2006
 PS> I could be wrong so please point out the flaw in my logic if you see it. 
(user deleted)
Mar 19, 2007
 I am not sure wether it is fair or not . The puzzle it self never mentions that there are 6 gamblers and the answer is based on that fact. 
pzakielarz
Oct 05, 2007
 foraneagle2: I didn't check your logic, but I think its ok. Excel probably rounded. 1.87e16 is really tiny. 
(user deleted)
Oct 15, 2009
 There are 216 possible rolls (6^3). Now if you would bet on 1 for all of these rolls you would only win $123. So you would lose (216123=93) $93. Is this a fair game for the player I do not agree. You win 1 $5 14  $3 76  $1. 
opqpop
Jan 26, 2010
 Great brainteaser. I first did it without the hint via probability, but when I looked at the hint, boy did I feel defeated. I really hope someday I can be someone who can come up with that beautiful solution. 
opqpop
Sep 11, 2010
 There are 216 possible rolls (6^3). Now if you would bet on 1 for all of these rolls you would only win $123. So you would lose (216123=93) $93. Is this a fair game for the player I do not agree. You win 1 $5 14  $3 76  $1.
@mjoshua72
Your calculations are wrong. You win 1 $5, 15 $3, and 75 $1 which totals to $125 won. The amount you lose is the $1 * (number of rolls where you didn't win) = $1 (216  1  15  75) = $125 
fishbulb
May 22, 2011
 P(none of the 3 show your number) = (5/6)^3
P(1 die has your number) = (1/6)(5/6)(5/6)(3)
P(2 dice have your number) = (1/6)(1/6)(5/6)(3)
P(3 dice have your number) = (1/6)(1/6)(1/6)
EV = 125/216 + 75/216 + 45/216 + 5/216 = 0
Fair game. 
83457mode
Jul 22, 2012
 It's not a fair game. You have to multiple the probability by the respective payout to determine the expected value.
Probability that you don't get any correct.
[1](5/6)^3
Probability one of the dice matches
+ [1](1/6)(5/6)^2
Probability two dice match
+ 3(5/6)(1/6)^2
Probability three dice match
[5](1/6)^3
(125 + 25 + 15 + 5)/216
= 80/216= 0.37 or 37 cents.
The game isn't fair 
83457mode
Jul 22, 2012
 @Fishbulb if only 1 shows up you win $1 instead of $3. Otherwise you're correct. Regardless, the game is unfair. 
lbertolotti
Jul 22, 2013
 I Agree it's unfair... the hint about dealer with 6 players is misleading; there is only a correct way to solve the puzzle, and is to compute the EXPECTED WIN of the player : if game is fair, it must be 0: a very simple computation returns (as already written by others) 37 cents as expected win : to say, if I AND THE DEALER will continue to play, in the long run I will lose my money by sure 
spikethru4
Jul 22, 2013
 The calculations above forget that there are three combinations each of a singleton or a pair.
p(0) = (5/6)^3 = 125/216
p(1) = 3*(1/6)*(5/6)^2 = 75/216
p(2) = 3*(5/6)*(1/6)^2 = 15/216
p(3) = (1/6)^3 = 1/216
The expected payout is, therefore:
e = 5*p(3) + 3*p(2) + 1*p(1)  1*p(0)
....= 5 + 45 + 75  125
....= 0 
lbertolotti
Jul 23, 2013
 Yup, sorry.... I wrote it too quickly... it's indeed a fair game with 0 expected win. 
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