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  #1  
Old 02-06-2005, 05:20 PM
XChamp XChamp is offline
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Default Probability of Rolling a 5 before a 7

You have 2 fair dice. What is the probability that you roll a 5 before you roll a 7?

I answered it this way, although I am not confident with this answer.

On average you will roll a 5 once every ln(2)/ln(9/8)=5.88 tries. I found this by solving (8/9)^x = 1/2 since probability of rolling a 5 is 1/9.

I then found the odds of rolling a 7 in approx 5.88 tries which is (5/6)^5.88= 0.342

so the probability of rolling a 5 before a 7 is 34.2%


See anything wrong with this?
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  #2  
Old 02-06-2005, 05:59 PM
tylerdurden tylerdurden is offline
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Default Re: Probability of Rolling a 5 before a 7

2 to 3. There are six ways to roll a seven, and four ways to roll a 5.
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  #3  
Old 02-06-2005, 06:16 PM
XChamp XChamp is offline
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Default Re: Probability of Rolling a 5 before a 7

I don't understand. Can you elaborate?

I just realized that in my first post i meant to say "probability of rolling", not "odds of rolling".
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  #4  
Old 02-06-2005, 06:48 PM
777 777 is offline
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Default Re: Probability of Rolling a 5 before a 7

buyy scarnes compleat guide to gambling the probability bible he will tell you verything you need to know
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  #5  
Old 02-06-2005, 07:39 PM
TomCollins TomCollins is offline
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Default Re: Probability of Rolling a 5 before a 7

There are 4 ways to roll a 5, and 6 ways to roll a 7.

So you will roll a 5 first 40% of the time.
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  #6  
Old 02-06-2005, 08:44 PM
RocketManJames RocketManJames is offline
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Default Re: Probability of Rolling a 5 before a 7

You have a bag of 36 marbles...

6 are labeled "7" and 4 are labeled "5" and the rest are labeled "I don't care."

What is the probability of choosing a "5" before choosing a "7"? You can simply remove the 26 marbles labeled "I don't care," because they only help to slow down this game of choosing a "5" or "7" marble.

So, now you have 10 marbles in there. What's the chance that you choose a "5" before a "7" in this simplified situation?

-RMJ
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  #7  
Old 02-06-2005, 10:32 PM
eobmtns eobmtns is offline
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Default Re: Probability of Rolling a 5 before a 7

There are 36 possible results from a roll of two dice, 1-1, 2-1, 1-2 .... 5-6, 6-5, 6-6. The possible ways to make a 7 are 1-6, 6-1, 2-5, 5-2, 3-4, 4-3 (6 ways). The possible ways to make a 5 are 1-4, 4-1, 2-3, 3-2 (4 ways). So, of these 10 possible results, the probability of rolling a 5 before a 7 is 4 in 10 or 0.4.
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  #8  
Old 02-07-2005, 12:01 PM
alThor alThor is offline
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Default Re: Probability of Rolling a 5 before a 7

[ QUOTE ]
What is the probability that you roll a 5 before you roll a 7?

[/ QUOTE ]

On the last roll of this "game" you will have rolled either a 5 or a 7, so you can rephrase the question as follows: Given I rolled a 5 or rolled a 7, what is the probability I rolled a 5? Now use the definition of conditional probability to find the answer. (This is an alternate explanation of the marbles analogy someone else posted.)

[ QUOTE ]
On average you will roll a 5 once every ln(2)/ln(9/8)=5.88 tries. I found this by solving (8/9)^x = 1/2 since probability of rolling a 5 is 1/9.

[/ QUOTE ]

You are confusing "average" with median. The probability of rolling 5 is 5/36, so on average, you will roll 5 five times every 36 rolls, or one time every 36/5=7.2 rolls. Instead, you found that P(roll 5 in no more than 5.88 tries) = 50%, which is a median statistic. (And 1/9 is the wrong probability anyway.)

alThor
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  #9  
Old 02-07-2005, 12:17 PM
XChamp XChamp is offline
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Default Re: Probability of Rolling a 5 before a 7

Ok this makes more sense now. thanks

I still get 4 ways to roll a 5, however. [1,4][4,1][2,3][3,2]
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