### Brain Teasers

# Rolling the Dice

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

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?

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?

### Hint

What will happen if there are 6 gamblers, each of whom bet on a different number?### 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

good one to warm up my brain this morning. Makes sense!

Very good, the hint helped. Keep them rollin'.

Very good, the hint helped. Keep them rollin'.

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.

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.

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.

AH HA! just kidding it threw me off too. But keep them coming

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.

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

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!

but the intention was good

do the math and have fun!

What did you mean by "the amount you bet"?

That was good

It was nice.

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...

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...

OOOPS .... sorry..... didn't read the next post.... SORRY... SORRYYY

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.

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

kinda hard... not really good at these... idk! good anyway!!

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.

Great puzzle. I like the way you explained it. Can I use it?

Good stuff

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

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

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

PS> I could be wrong so please point out the flaw in my logic if you see it.

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.

foraneagle2: I didn't check your logic, but I think its ok. Excel probably rounded. 1.87e-16 is really tiny.

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.

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.

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.

@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

@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

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.

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.

[-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

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

Jul 22, 2012

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

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

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

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

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

The current official solution is a little bit inconvincing because it only evaluates one possibility - that all 6 gamblers are betting different numbers. What you have to evaluate is also the remaining 6^6 - 1 possibilities to see if they each also produce the same (fair) outcome, it isn't that intuitive to see that these will each or in totality result in net fair outcome as the math is somewhat changing in each case. Instead, it is easier to start with just one gambler because you have to evaluate only 6 possibilities and it is easy to see that all 6 will have identical results because you are only picking one digit, so you can simply evaluate one possibility of this, so lets say a gambler comes and bets on 1. We can ignore all the possibilities of rolling 3 dice and evaluate probabilities because we are only interested in whether fair or not so we don't need these unnecessary divisions by 6^3 as it is easy to see this does not change the outcome of whether the game is fair or not. Number of possibilities below:

None (no die shows "1") = 5 x 5 x 5

One (any one die shows "1") = 3C1 x 5 x 5

two (any two dies show "1") = 3C2 x 5

three (all three show "1") = 1

So - $1 * 5^3 + $1 * 3*5*5 + $3 * 3*5 + $5 * 1 = 0, hence fair. Some people in the comments did it this way also but I thought I will write this more clearly and also why this way of doing the calculation is more intuitive (less prone to doubt unlike the official answer).

None (no die shows "1") = 5 x 5 x 5

One (any one die shows "1") = 3C1 x 5 x 5

two (any two dies show "1") = 3C2 x 5

three (all three show "1") = 1

So - $1 * 5^3 + $1 * 3*5*5 + $3 * 3*5 + $5 * 1 = 0, hence fair. Some people in the comments did it this way also but I thought I will write this more clearly and also why this way of doing the calculation is more intuitive (less prone to doubt unlike the official answer).

So so teaser and an old one at that. A fifty-fifty probability quiz where you have to choose yes or no. I got it right but do not understand it.

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