Probability of multiple ranged values within multiple events
January 25th, 2012 at 8:07:58 PM
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Ran across your website after spending the past couple of hours reaching out for help... thought it showcased your math genius very well! (obviously, thats why Im here)
So onwards...To make things simple, imagine you have 5 100 sided dice and you roll them all at once.
Im trying to solve (and understand the math behind!) the probability of having both a value between 75 and 100 and also a value between 86 and 100 showing up within those 5 rolls...
I think this answer lies somewhere between using the inclusion-exclusion principle and combination probability... but im completely stumped how to go about it! Hoping you can help-- this is a bit complex.
Thanks so much for your time either way!
So onwards...To make things simple, imagine you have 5 100 sided dice and you roll them all at once.
Im trying to solve (and understand the math behind!) the probability of having both a value between 75 and 100 and also a value between 86 and 100 showing up within those 5 rolls...
I think this answer lies somewhere between using the inclusion-exclusion principle and combination probability... but im completely stumped how to go about it! Hoping you can help-- this is a bit complex.
Thanks so much for your time either way!
January 25th, 2012 at 8:12:24 PM
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Do the other three dice have to be less than 75?
Someday, joor goin' to see the name of Googie Gomez in lights and joor goin' to say to joorself, "Was that her?" and then joor goin' to answer to joorself, "That was her!" But you know somethin' mister? I was always her yuss nobody knows it! - Googie Gomez
January 25th, 2012 at 8:15:50 PM
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Hey thanks for the prompt response s2dbaker!
No, the other values make no difference. Im merely trying to find the probablity that within the 5 rolled dice, there will be at least one die roll between the values of 75 and 100 and 86 and 100. The only stipulation is that both of these ranges be realized within the 5 rolls.
No, the other values make no difference. Im merely trying to find the probablity that within the 5 rolled dice, there will be at least one die roll between the values of 75 and 100 and 86 and 100. The only stipulation is that both of these ranges be realized within the 5 rolls.
January 26th, 2012 at 6:43:59 AM
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Assuming that 'Between' is inclusive of 75, 86 and 100, you're talking about a 26 in 100 chance and a 15 in 100 chance. The other three chances are 100 out of 100. With five dice there are 10 combinations of two in five. To skip to the meat of it, your chances are. 26 times .15 = .039 times the ten possible combinations = 39 chances in 100.
There are some bonus bets that pay less than that :)
Edit: on second thought it's probably better to calculate the chances of it 'not' happening. In that case, what's left is a 32.82% chance which is quite different from 39.
There are some bonus bets that pay less than that :)
Edit: on second thought it's probably better to calculate the chances of it 'not' happening. In that case, what's left is a 32.82% chance which is quite different from 39.
Someday, joor goin' to see the name of Googie Gomez in lights and joor goin' to say to joorself, "Was that her?" and then joor goin' to answer to joorself, "That was her!" But you know somethin' mister? I was always her yuss nobody knows it! - Googie Gomez
January 30th, 2012 at 7:08:58 AM
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You are right that sometimes it's easier to work out the chances of something not happening (e.g. throwing a six somewhere in n dice being 1 - 5/6^n) but in this case I think it's easier to think of 5 events each being a success or failure and the chances of getting at least N successes. In this case T=5 (five rolls) N=3 (three successes) P = 15% or 26%.
(a) WWWWW : Pr(5 wins) = P^5
(b) WWWWL : Pr(4 wins)*Pr(1 loss)*5 = P^4*(1-P)^1*5 (Note the multiply by 5 as any of the dice could be the failure)
(c) WWWLL : Pr(3 wins)*(Pr(2 losses)*[5*4/2] = P^3 * (1-P)^2 * 10 (Similarly there are 10 combos of the two dice that fail).
etc
Then double check your match that total Pr(5)..Pr(0) is 1.
So chances of at least 3 successes if (a)+(b)+(c).
(a) WWWWW : Pr(5 wins) = P^5
(b) WWWWL : Pr(4 wins)*Pr(1 loss)*5 = P^4*(1-P)^1*5 (Note the multiply by 5 as any of the dice could be the failure)
(c) WWWLL : Pr(3 wins)*(Pr(2 losses)*[5*4/2] = P^3 * (1-P)^2 * 10 (Similarly there are 10 combos of the two dice that fail).
etc
Then double check your match that total Pr(5)..Pr(0) is 1.
So chances of at least 3 successes if (a)+(b)+(c).
