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# In a \$10 game of chance in Las Vegas, there are two identica

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Manager
Joined: 16 Feb 2011
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In a \$10 game of chance in Las Vegas, there are two identica [#permalink]

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28 Aug 2012, 18:10
In a \$10 game of chance in Las Vegas, there are two identical bowls that contain 100 marbles. Each bowl contains black colored 95 marbles along with 5 different colored marbles: one blue, one yellow, one green, one red and one gold. You get to pick one marble randomly from each bowl. If you get a matching pair of colored marbles (blue-blue, gold-gold), you win \$10000. What is the probability that you win the prize in one play of the game?

HEre's what I tried :

Method 1 = total pairs = 100; Total combinations = 100*100 => PRob = 1/100

Method 2 (really crash landed) --

Choose 1 of the black colored balls from the first bowl and then the second one => 95C1 * 95C1
Choose 1 of blue balls => 1C1 * 1C1 ...Now this will be done for all the five colors. Hence, total = 5*1c1 * 1c1 = 5

Total = (95^2 + 5)/(100^2) = CRASHED !

Thanks

Kudos [?]: 238 [0], given: 78

Manager
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Kudos [?]: 238 [0], given: 78

Schools: ABCD
If 4 fair dice are thrown simulatenously, what is the probab [#permalink]

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28 Aug 2012, 18:19
If 4 fair dice are thrown simulatenously, what is the probability of getting at least one pair?

(I was able to compute the solution using Permutations, but I wanted to try calculating one pair and two pair probabilities)

Method1 -

One pair => there are 12/2 = 6 arrangements for a pair in XXYZ

(6*1*5*4) 6
---------------- [First two dice form a pair, the other two don't]
(6*6*6*6)

=20/36 = 10/18

Similarly, Two different pair => there are 3 arrangements for a pair in XXYY

(6*1*5*1)*3
-----------
6*6*6*6

=5/72

Two same pair (i.e. four dice with the same number) => 6/(6*6*6*6) = 1/6*6*6

Method 2 (using perm) => 1-(6*5*4*3/6*6*6*6) = 13/18

Thanks

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Kaplan GMAT Instructor
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Re: If 4 fair dice are thrown simulatenously, what is the probab [#permalink]

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28 Aug 2012, 21:56
1
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Expert's post
Hi Voodoo,

Your solution is incorrect because you forgot three-of-a-kind with a fourth different number.

For problems like this, with multiple "wins," far and away the easier solution is to find the odds of NOT getting what we want. In this case, we fail if and only if all four dice are different. So, we fail 6*5*4*3/6*6*6*6 of the time. Reducing, that gives us 5*4*3/6*6*6 --> 5*2*1/6*6 --> 5/18. And thus, we succeed 18/18 - 5/18 = 13/18 of the time. This is much easier, and much simpler, than doing all the counting of permutations!
_________________

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Re: In a \$10 game of chance in Las Vegas, there are two identica [#permalink]

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28 Aug 2012, 22:20
1
KUDOS
Expert's post
voodoochild wrote:
In a \$10 game of chance in Las Vegas, there are two identical bowls that contain 100 marbles. Each bowl contains black colored 95 marbles along with 5 different colored marbles: one blue, one yellow, one green, one red and one gold. You get to pick one marble randomly from each bowl. If you get a matching pair of colored marbles (blue-blue, gold-gold), you win \$10000. What is the probability that you win the prize in one play of the game?

HEre's what I tried :

Method 1 = total pairs = 100; Total combinations = 100*100 => PRob = 1/100

Method 2 (really crash landed) --

Choose 1 of the black colored balls from the first bowl and then the second one => 95C1 * 95C1
Choose 1 of blue balls => 1C1 * 1C1 ...Now this will be done for all the five colors. Hence, total = 5*1c1 * 1c1 = 5

Total = (95^2 + 5)/(100^2) = CRASHED !

Thanks

Note here that 'matching pair of colored marbles' implies only the 5 non-black colors. I agree that there is some ambiguity here - is black considered a colored marble or not? Well, if you are to win \$10,000 from a \$10 game, a pair of black marbles should not help you win that kind of money so I assume that we are only talking about the 5 different colored marbles.

Probability of picking a non black marble from a bowl = 5/100
Probability of picking the same color marble from the other bowl = 1/100

Probability of a matching non black pair = (5/100)*(1/100) = 5/10000 = 1/2000

On a side note, notice that the probability is against the player. If the probability of winning is 1/2000, for one's \$10, one should get \$20,000 if one wins. That is why, the house always wins and one should not gamble!
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Karishma
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Manager
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Re: If 4 fair dice are thrown simulatenously, what is the probab [#permalink]

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29 Aug 2012, 07:22
KapTeacherEli wrote:
Hi Voodoo,

Your solution is incorrect because you forgot three-of-a-kind with a fourth different number.

For problems like this, with multiple "wins," far and away the easier solution is to find the odds of NOT getting what we want. In this case, we fail if and only if all four dice are different. So, we fail 6*5*4*3/6*6*6*6 of the time. Reducing, that gives us 5*4*3/6*6*6 --> 5*2*1/6*6 --> 5/18. And thus, we succeed 18/18 - 5/18 = 13/18 of the time. This is much easier, and much simpler, than doing all the counting of permutations!

Thanks Eli! You are correct. There are four possibilities:

AABB => (6*1*5*1) * 3 = 90
AAAA => (6*1*1*1)*1 = 6
AAAB => (6*1*1*5)*4 = 120 {4 because 4!/3!}
Total = 216

Probability = 216/(6*6*6*6) = 1/6

Total probability of at least one pair = 1/6 + 10/18 (calculated above) = 13/18!!! Hurray!

My method2 is exactly same as your method! However, method1 is really fun!

Thanks again!

Kudos [?]: 238 [0], given: 78

Re: If 4 fair dice are thrown simulatenously, what is the probab   [#permalink] 29 Aug 2012, 07:22
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