I cant manage to solve this

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Helldrake
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I cant manage to solve this

Postby Helldrake » Fri Mar 26, 2010 4:24 am UTC

I know this is a really really lame problem, but somehow the solution is escaping me.

here it is:

By throwing 100 times a 1000-sided (from 000 to 999) dice... what is the probability of getting at least one triple number (000,111,222,333,444,etc).

(first post btw...yay!)

afk2011
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Re: I cant manage to solve this

Postby afk2011 » Fri Mar 26, 2010 4:49 am UTC

The probability of getting at least one triple number in 1000 throws is
1 - Q

Where Q is the probability of NOT getting a triple number at all in any throws. This should be enough to help you, since the probability of getting or not getting a triple number on a single throw is pretty trivial.

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joshz
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Re: I cant manage to solve this

Postby joshz » Sat Mar 27, 2010 12:11 am UTC

How many different ways can you get a triple number?
What's the probability of that happening one one specific throw?
How do you combine probabilities of independent events?
You, sir, name? wrote:If you have over 26 levels of nesting, you've got bigger problems ... than variable naming.
suffer-cait wrote:it might also be interesting to note here that i don't like 5 fingers. they feel too bulky.

Helldrake
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Re: I cant manage to solve this

Postby Helldrake » Sat Mar 27, 2010 12:35 am UTC

is it 1-((1/100)^100)?

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Kurushimi
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Re: I cant manage to solve this

Postby Kurushimi » Sat Mar 27, 2010 12:53 am UTC

No. That would be the probability of not getting a triple 100 times in a row.

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joshz
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Re: I cant manage to solve this

Postby joshz » Sat Mar 27, 2010 12:55 am UTC

That's not what I get.
Instead of using the 1- (it's generally useful, but unnecessary in this case), look at what I posted above.
You, sir, name? wrote:If you have over 26 levels of nesting, you've got bigger problems ... than variable naming.
suffer-cait wrote:it might also be interesting to note here that i don't like 5 fingers. they feel too bulky.

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Qaanol
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Re: I cant manage to solve this

Postby Qaanol » Sat Mar 27, 2010 2:04 am UTC

If there’s more than one way to do a job, then any particular one of them is not “necessary”, but it is quite possible that one could be simpler than the rest. In this case, I believe it is easiest to find the probability of the describe outcome not occurring, and subtract that from 1, as afk2011 describes.
wee free kings

Helldrake
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Re: I cant manage to solve this

Postby Helldrake » Sat Mar 27, 2010 3:42 am UTC

i never studied probability (i'll start this year). I dont know how to combine probabilities of independant events.

what do you get joshz?

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PM 2Ring
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Re: I cant manage to solve this

Postby PM 2Ring » Sat Mar 27, 2010 4:22 am UTC

Helldrake wrote:i never studied probability (i'll start this year). I dont know how to combine probabilities of independant events.

what do you get joshz?

Ok. So I guess this isn't homework. Hence the usual rule about not doing people's homework for them doesn't apply. But we'd still prefer to give you hints so you can discover the answer for yourself.

What answers do you have for the other two questions joshz asked?

As for combining the probabilities of independant events, think about a regular 6 sided dice & a coin. If you toss the dice & the coin (once), what's the probability of getting 1 on the dice and heads on the coin?

Helldrake
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Re: I cant manage to solve this

Postby Helldrake » Sat Mar 27, 2010 4:43 am UTC

How many different ways can you get a triple number? 10

What's the probability of that happening one one specific throw? 1/100

PM 2Ring wrote:
Helldrake wrote:i never studied probability (i'll start this year). I dont know how to combine probabilities of independant events.

what do you get joshz?

Ok. So I guess this isn't homework. Hence the usual rule about not doing people's homework for them doesn't apply. But we'd still prefer to give you hints so you can discover the answer for yourself.

What answers do you have for the other two questions joshz asked?

As for combining the probabilities of independant events, think about a regular 6 sided dice & a coin. If you toss the dice & the coin (once), what's the probability of getting 1 on the dice and heads on the coin?


How many different ways can you get a triple number? 10

What's the probability of that happening one one specific throw? 1/100

And the coin and the dice is 1/12 so i guess the formula is (1/6 x 1/2) but this leads me to the same solution... ((1/100)^100) unless there's something im missing

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PM 2Ring
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Re: I cant manage to solve this

Postby PM 2Ring » Sat Mar 27, 2010 4:54 am UTC

Helldrake wrote:
PM 2Ring wrote:What answers do you have for the other two questions joshz asked?

As for combining the probabilities of independant events, think about a regular 6 sided dice & a coin. If you toss the dice & the coin (once), what's the probability of getting 1 on the dice and heads on the coin?


How many different ways can you get a triple number? 10

Correct.

What's the probability of that happening one one specific throw? 1/100

Correct.

And the coin and the dice is 1/12 so i guess the formula is (1/6 x 1/2)

Correct.

but this leads me to the same solution... ((1/100)^100) unless there's something im missing


No, (1/100)^100 is the probability of rolling the 1000-sided dice 100 times and getting a triple on every roll. And (1 - 1/100) is the probability of not getting a triple on a single roll.

Spoiler:
(1 - 1/100)^100 is the probability of not getting a triple in any of 100 rolls, so 1 - (1 - 1/100)^100 is the probability of getting at least one triple in 100 rolls.
1 - (1 - 1/100)^100 ~= .63396765872677

Helldrake
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Re: I cant manage to solve this

Postby Helldrake » Sat Mar 27, 2010 5:05 am UTC

I had to look at the spoiler, but i understood perfectly once i saw the answer. thanks a lot.

Btw, i wouldnt have asked if it was homework...

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joshz
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Re: I cant manage to solve this

Postby joshz » Sat Mar 27, 2010 10:09 am UTC

Or, what I did: (1/10)^100 (1/10 probability of getting a triple, then 100 rolls)

EDIT: that's wrong. I thought you meant all triples. It'd be 100(1/10) probability, I think. 1/10 prob. of getting it each roll and 100 rolls. But that's greater than 1, so it's probably wrong.

EDIT 2: oops. 1/100 probability, not 1/10. I blame the lack of sleep.
You, sir, name? wrote:If you have over 26 levels of nesting, you've got bigger problems ... than variable naming.
suffer-cait wrote:it might also be interesting to note here that i don't like 5 fingers. they feel too bulky.


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