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# chuck-a-luck

## Question

The dice game chuck-a-luck is played in some casinos. The rules are extremely simple. You bet \$1 on a number from 1 to 6. Three dice are rolled. If your number does not appear on any dice, you lose your bet. If the number appears once, you recover your bet. If it appears twice or three times, you get \$2 or \$3, respectively (these amounts include your original \$1 bet). What are your odds?

For bonus points, assume there are five additional friends and each person, including yourself from the original problem (so six people total), bets on a different number. Now what are the odds of winning against the house (for each better separately)?

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For every round you have chance to get average 1/2\$. More detailed, you have chance to lose 5/8, chance to recover 15/56, chance to win 1\$ 5/56, to win 2\$ is 1/56

Bonus: You guys will collectively lose 3\$ every round to the house

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For every round you have chance to get average 1/2\$. More detailed, you have chance to lose 5/8, chance to recover 15/56, chance to win 1\$ 5/56, to win 2\$ is 1/56

Bonus: You guys will collectively lose 3\$ every round to the house

How much does a player stand to lose per round they play?

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For every round you have chance to get average 1/2\$. More detailed, you have chance to lose 5/8, chance to recover 15/56, chance to win 1\$ 5/56, to win 2\$ is 1/56

Bonus: You guys will collectively lose 3\$ every round to the house

How much does a player stand to lose per round they play?

I didnt quite understand the question. But Curtis Jackson??

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I am unfamiliar with Curtis Jackson, but thanks to Google, I get your reference No. Not 50 cents.

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If you're playing alone against the house then on average the house wins approximately 58 cents per game. Your odds of losing this game are 125/216, the chance of breaking even is 25/72 and the chances of winning are 5/72 for \$1 and 1/216 for \$2.

When playing with 5 friends betting on every number then the house collects \$6 every turn and pays \$3 every turn, so the expected loss is 50 cents per game for every player.

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How did you get 58 cents from that? Using the same numbers, I came up with 50 cents also.

Just to make sure, since I'm getting the same answer, is the net outcome for the player:

Lose \$1 if no dice are your number

Break even if one die is your number

Gain \$1 if two dice are your number

Gain \$2 if three dice are your number?

It could still be in the house's favor if all of the non-losing options net the player \$1 more; that is, they get their ante back and the number of dice they match is the number of dollars of profit.

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How did you get 58 cents from that? Using the same numbers, I came up with 50 cents also.

Just to make sure, since I'm getting the same answer, is the net outcome for the player:

Lose \$1 if no dice are your number

Break even if one die is your number

Gain \$1 if two dice are your number

Gain \$2 if three dice are your number?

It could still be in the house's favor if all of the non-losing options net the player \$1 more; that is, they get their ante back and the number of dice they match is the number of dollars of profit.

There is a difference in whether you're playing alone against the house or 6 people are playing. When 6 people are playing and every number has a bet on it, then the house pays \$3 every turn, since every die has a winner. When you're playing alone, some rolls will not be paid out at all. House odds are 125/216 in this case, which is equivalent to 0.578703703703...

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When playing alone, the house's odds of winning \$1 might be 125/216, but they also have non-zero odds of losing money to the player (and more than \$1). So expected winnings by the house would be lower than 125/216.

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I agree with plasmid (and Barcallica):

(25/72)+(10/72)+3/16 = (15/56)+(10/56)+3/56 = 0.5

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I see my mistake now and agree too. The expected winnings should be 50 cents in both cases.

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216 possible outcomes (6x6x6)

Let's assume you betted on 3.

1 outcome of all 3's (earn \$3)

15 outcomes with two 3's (earn \$2)

75 outcomes with one 3 (earn \$1)

125 outcomes without any 3's (lose \$1)

then we have:

(1 x \$3 + 15 x \$2 + 75 x \$1 -125 x \$1) /126 = -17/126 ~ -\$.08 per play

Now what about the bonus question?

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