simple question

[mostsmooth]

the chance of rolling a total of 6 using two dice is 5/36,right? how do you figure the chance of rolling a 6 after x rolls?

[mostsmooth]

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the chance of rolling a total of 6 using two dice is 5/36,right? how do you figure the chance of rolling a 6 after x rolls?


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maybe i need to clarify better, what are the chances youd roll at least one 6 after 5 rolls?

[Copernicus]

1-(1-31/36)^5

That is 1 minus the probability that you rolled no sixes in 5 rolls.

[BruceZ]

what are the chances youd roll at least one 6 after 5 rolls?

1 - (31/36)^5 = 52.7%

That's 1 minus the probability of not getting a six 5 times in a row (C had 1 too many 1 minuses there [img]/images/graemlins/blush.gif[/img]).

Now, if you roll until you get a 6, on which roll are you most likely to get it?

[pudley4]

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Now, if you roll until you get a 6, on which roll are you most likely to get it?

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The first [img]/images/graemlins/smile.gif[/img]

[Wake up CALL]

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Now, if you roll until you get a 6, on which roll are you most likely to get it?

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The last roll of course.

[Copernicus]

Or one too few "1-"s [img]/images/graemlins/smile.gif[/img] Thanks...not sleeping enough lately trying to work and play tourneys at night!

[Bozeman]

1st?

[mostsmooth]

so the formula requires use the chances of a 6 not coming up ( 31/36) when calculating? theres no formula that uses the chance that it will come up? (5/36)?

[Bozeman]

True. You can approximate the answer as n*p (n=#of tries, p= prob. on one try) if n and p are small. Basically, this multiply counts all the times where it happens more than once.

A conceptually different way to arrive at the accurate answer is:
P= p |chance of getting it on the first try
+(1-p)*p |chance of getting it on the second try and not getting it on the first
+(1-p)^2*p |chance of getting it on the third try and not 1st or 2nd
+(1-p)^3*p |etc.
+(1-p)^4*p
=1-(1-p)^5 |as given above with a little algebra
=52.6% for p=5/36