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Good Probably and combinations Question - Yahtzee Game

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Good Probably and combinations Question - Yahtzee Game

by faraz_jeddah » Sat Sep 07, 2013 10:18 pm
In the game of Yahtzee, a "full house" occurs when a player rolls five fair dice and the result is two dice showing one number and the other three all showing another (three of a kind). What is the probability that a player will roll a full house on one roll of each of the three remaining dice if that player has already rolled a pair on the first two?

A - 1/3
B - 5/54
C - 5/72
D - 5/216
E - 1/1296

OA to be posted later.

A "Good" roll would be
If AA (1,1 or 2,2 or 3,3..) is rolled on the first two tries then the remaining 3 rolls can be BCD

So Good combinations would be 5 x 4 x 3

Probability = Good scenario / Total scenario

I am not sure how to calculate the denominator.

Is it 6 x 6 x 6 /3! ?
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by [email protected] » Sun Sep 08, 2013 1:10 am
Hi faraz_jeddah,

This is a quirky probability question Here's one way to solve this problem:

We're told that the first 2 dice already match, so we'll call them AA. It's the other 3 dice we have to deal with. To get a "full house", we need a set of 3 and a pair (the pair we start with COULD be part of the set of 3 though). We should map out the possibilities:

AA BBB = (5/6)(1/6)(1/6) = 5/216

AA ABB = (1/6)(5/6)(1/6) = 5/216

AA BAB = (5/6)(1/6)(1/6) = 5/216

AA BBA = (5/6)(1/6)(1/6) = 5/216

Total = 20/216 = 5/54

Final Answer: B

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by faraz_jeddah » Sun Sep 08, 2013 1:19 am
Aah.. I didnt understand the question. I thought three of a kind meant 3 different numbers.
Looks like I need brush up on my poker skills.

PS: Your solution is inline with the OA.
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by vinay1983 » Sun Sep 08, 2013 1:26 am
[email protected] wrote:Hi faraz_jeddah,

This is a quirky probability question Here's one way to solve this problem:

We're told that the first 2 dice already match, so we'll call them AA. It's the other 3 dice we have to deal with. To get a "full house", we need a set of 3 and a pair (the pair we start with COULD be part of the set of 3 though). We should map out the possibilities:

AA BBB = (5/6)(1/6)(1/6) = 5/216

AA ABB = (1/6)(5/6)(1/6) = 5/216

AA BAB = (5/6)(1/6)(1/6) = 5/216

AA BBA = (5/6)(1/6)(1/6) = 5/216

Total = 20/216 = 5/54

Final Answer: B

GMAT assassins aren't born, they're made,
Rich

Dice with alphabets!Never heard of it. Sigh!maybe as faraz has opined, even I need to brush my knowledge about various games played in the US :?

The GMAT playing games with us is not enough I suppose. But Rich, I need further explanation, maybe you have to 'hand hold" me through this!
You can, for example never foretell what any one man will do, but you can say with precision what an average number will be up to!
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by faraz_jeddah » Sun Sep 08, 2013 2:16 am
[email protected] wrote:
AA BBB = (5/6)(1/6)(1/6) = 5/216
Rich one more Q - why 5/6? Isnt each roll independent of the other?

Edit: Never mind I figured it out.
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by faraz_jeddah » Sun Sep 08, 2013 2:25 am
vinay1983 wrote:
Dice with alphabets!Never heard of it. Sigh!maybe as faraz has opined, even I need to brush my knowledge about various games played in the US :?

The GMAT playing games with us is not enough I suppose. But Rich, I need further explanation, maybe you have to 'hand hold" me through this!
Hehe. There are no dice with alphabets. The A mentioned is to represent any number between 1-6.

So the rolls could look like

1,1, 1,2,2
1,1, 2,1,2
1,1, 2,2,1
1,1, 2,2,2

2,2, 1,1,1
2,2, 2,1,1
2,2, 1,2,1
2,2, 1,1,2

etc....

as long as we have 2 A's and 3 B's
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by [email protected] » Sun Sep 08, 2013 1:07 pm
Hi vinay1983,

In my explanation, the letter A would represent one of the six numbers on a die. Thus AA would be either 11, 22, 33, etc. B represents one of the OTHER possible numbers.

So, if we had 11 222 ----that would be a full house. So would 55 333 etc.

The "math" behind this question is based on whether a roll would "match" or "not match" a number we already had.

In the first part of the calculation:

AA BBB

The AA has already been established, so we need 3 numbers that are all equal to one another and NOT equal to A.

If A = 1, then B could be anything BUT 1 (2, 3, 4, 5, or 6)

On the first B, there's a 5/6 chance for that to happen.

Once we get to the second B though, we need it to match the first B. There's only a 1/6 chance for that to happen.

The same applies to the third B: only a 1/6 chance for it to match the first and second.

So, for that specific situation to occur, we have (5/6)(1/6)(1/6) = 5/216 chance

The other 3 situations require you to do the math in a slightly different order, but the math still works out the same.

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