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Guest Message by DevFuse

chuck-a-luck

11 replies to this topic

#1 BMAD

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Posted 03 July 2013 - 03:36 AM

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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#2 Barcallica

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Posted 06 July 2013 - 09:54 AM

Spoiler for

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#3 BMAD

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Posted 06 July 2013 - 09:02 PM

Spoiler for

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

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#4 Barcallica

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Posted 08 July 2013 - 02:50 AM

Spoiler for

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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#5 BMAD

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Posted 08 July 2013 - 03:32 AM

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

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#6 k-man

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Posted 08 July 2013 - 03:53 PM

Spoiler for my answer

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#7 plasmid

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Posted 09 July 2013 - 03:17 PM

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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#8 k-man

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Posted 09 July 2013 - 03:45 PM

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.

Spoiler for

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#9 plasmid

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Posted 10 July 2013 - 07:25 AM

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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#10 bonanova

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Posted 10 July 2013 - 07:27 AM

I agree with plasmid (and Barcallica):

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

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