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Rolling the Dice

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


Puzzle ID:#26344
Fun:*** (2.84)
Difficulty:** (2.05)
Submitted By:kissA****
Corrected By:Shadows




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?

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Oct 13, 2005

good one to warm up my brain this morning. Makes sense!
Oct 13, 2005

Very good, the hint helped. Keep them rollin'.
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.
Oct 13, 2005

AH HA! just kidding it threw me off too. But keep them coming
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.
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
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!
Oct 21, 2005

What did you mean by "the amount you bet"?
Oct 22, 2005

That was good
Oct 25, 2005

It was nice.
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 factor-in combination...
Oct 27, 2005

OOOPS .... sorry..... didn't read the next post.... SORRY... SORRYYY
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.
Nov 06, 2005

I liked it a lot and that is that. (I'm not getting into all that mathematics stuff.)
Nov 24, 2005

kinda hard... not really good at these... idk! good anyway!!
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.
Feb 25, 2006

Great puzzle. I like the way you explained it. Can I use it?
Mar 26, 2006

Good stuff
Aug 09, 2006

Always a nice change of pace to see one that isn't the most obvious anwser
Oct 19, 2006

Excel tells me there is an expected payout of $-1.8735E-16.

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+1Y-X
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.
Oct 05, 2007

foraneagle2: I didn't check your logic, but I think its ok. Excel probably rounded. 1.87e-16 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 (216-123=93) $93. Is this a fair game for the player I do not agree. You win 1 -$5 14 - $3 76 - $1.
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.
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 (216-123=93) $93. Is this a fair game for the player I do not agree. You win 1 -$5 14 - $3 76 - $1.


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
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.
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.


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


(-125 + 25 + 15 + 5)/216

= -80/216= -0.37 or -37 cents.

The game isn't fair
Jul 22, 2012

@Fishbulb if only 1 shows up you win $1 instead of $3. Otherwise you're correct. Regardless, the game is unfair.
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
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
Jul 23, 2013

Yup, sorry.... I wrote it too quickly... it's indeed a fair game with 0 expected win.

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