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An approximation will do nicely if an accurate answer to this question isn't possible. If I use the 98 letter tiles from the English-language version of a Scrabble game (omitting the 2 blank tiles), what is the probability of selecting 7 random letters that make a 7-letter word found in an English language dictionary?
For this exercise, it can include proper names, but not words with hyphens, apostrophes etc.
The letter distribution is as follows:
JackofOz 03:26, 28 March 2007 (UTC)
Which English language dictionary? 202.168.50.40 04:08, 28 March 2007 (UTC)
Lets start with the brute force method, you will want a very fast computer amd alot of time for this:
First, Get a list of the 7 letter words that are valid. (TWL06 in my case) Sort the letters in each word- ie courage would be acegoru. Lets call that an "anagram". Also count the number of each letter in the word. If the word is impossible "POPPERS" (too many Ps), toss the word out.
Second, Generate each of all the possible racks that can occur. There should be 100*99*98*97*96*95*94 or 80,678,106,432,000 possible racks. Think of it as nearly 100 to the 7th. As each rack is generated, get the anagram of the rack and discard it if none of the word anagrams matches it. If a word anagram matches it, add one to your counter of "bingo" racks.
At the end, you will have a count of the number of bingo racks that you can divide by the number of possible racks. That is the # of successes and the number of trials. Divide Success by trials for probability.
If you are not into waiting around that long, you could apply some optimization and only generate unique racks plus a number of how many ways that rack can be generated - ie if the rack is ADLSJKQ there would be 9 ways to choose an A, 4 ways to choose a D, 4 ways to choose an L... or 9*4*4*4*1*1*1 or 576 ways this same rack could occur. This will certainly speed up the process because you are comparing your words to a smaller set of racks.
More complicated than the above example is racks with repeated letters, like 3 Es for instance. Then you have to put in the number of ways you can choose 3 of the 12 Es, That would be 12*11*10 or 1320. So, if the rack was ADEEELS, it would be 9*4*(12*11*10)*4*4 or 760320 ways.
So now you get to the point where you say, Eureeka!, I can see that all I really have to do is run each valid word through and see how many racks can be generated that make that word! Way better, because then I dont have to mess around with generating all those racks. So, what you do is make an anagram of each possible word and then remove the duplicate anagrams. For example "RESHOOT", "SHOOTER", and "HOOTERS" all have the same anagam, "EHOORST". Only need one of those. Then calculate the number of possible ways to rack EHOORST - 12*2*(8*7)*6*4*6 = 193536. (Probability of getting a bingo so you can play HOOTERS is 193536/80,678,106,432,000 or 0.000000002398866). Anyway, sum the number of racks for each unique letter sorted word and you have the number of successes. The number of trials is 80,678,106,432,000. Successes over trials is the probability of a bingo.
MagRobot
I am stuck with the following problem involving polynomials. (this is not a HW question but a teaser by my sister's math teacher that's been bugging me for a while: How can someone proceed with such a problem.
The reaminder of p(x) when divided by x+1 is 2. The reaminder of P(X) when divided by x-2 is 4. What is the reaminder p(x) by (x+1).(x-2) (the product)?
When i first read the problem i directly thought of modular arithematics,but i don't think the solution involves modular arithematics because my sister is still in high school and she is not familiar with it.
Hisham1987 15:38, 28 March 2007 (UTC)
Here's a hint... denote your polynomial as and see what happens. Cheers -- Hirak 99 21:40, 28 March 2007 (UTC)
I am not familiar with all the proper wordings and notation, so hopefully the subject isn't too confusing.
What I would like to learn is of a method to calculate the following kind of problem:
I have a series of 1,000 "events" For each such event, the outcome will be HEAD with 24.5% certainty, and TAIL with 75.5% certainty. The outcome for each event is independent of any "history".
1. What is the likelihood that
2) What is the most likely longest chain of HEADs?
3) Given the above certainities, how many events have to occur for the probability of having a chain of HEADs 7 or longer being .75 or greater?
NB: I do not wish for the answers to these particular questions, as they are just examples I just made up, but rather learn of the reasoning behind solving such problems. If the methods used are fairly simple, such as a few formulas where I'd put in the different values, I would still greatly appreciate any links to articles on the underlying math. 81.20.147.194 17:37, 28 March 2007 (UTC)
5+6+5+4+7+9+54+3+54+56+67+76+56+56+67+76+67+67+76+67+54+54+54+54+54+45*3 —Preceding unsigned comment added by 86.129.225.138 ( talk • contribs)
What is the ith root of i? —Preceding unsigned comment added by 202.168.50.40 ( talk • contribs)
Mathematics desk | ||
---|---|---|
< March 27 | << Feb | March | Apr >> | March 29 > |
Welcome to the Wikipedia Mathematics Reference Desk Archives |
---|
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages. |
An approximation will do nicely if an accurate answer to this question isn't possible. If I use the 98 letter tiles from the English-language version of a Scrabble game (omitting the 2 blank tiles), what is the probability of selecting 7 random letters that make a 7-letter word found in an English language dictionary?
For this exercise, it can include proper names, but not words with hyphens, apostrophes etc.
The letter distribution is as follows:
JackofOz 03:26, 28 March 2007 (UTC)
Which English language dictionary? 202.168.50.40 04:08, 28 March 2007 (UTC)
Lets start with the brute force method, you will want a very fast computer amd alot of time for this:
First, Get a list of the 7 letter words that are valid. (TWL06 in my case) Sort the letters in each word- ie courage would be acegoru. Lets call that an "anagram". Also count the number of each letter in the word. If the word is impossible "POPPERS" (too many Ps), toss the word out.
Second, Generate each of all the possible racks that can occur. There should be 100*99*98*97*96*95*94 or 80,678,106,432,000 possible racks. Think of it as nearly 100 to the 7th. As each rack is generated, get the anagram of the rack and discard it if none of the word anagrams matches it. If a word anagram matches it, add one to your counter of "bingo" racks.
At the end, you will have a count of the number of bingo racks that you can divide by the number of possible racks. That is the # of successes and the number of trials. Divide Success by trials for probability.
If you are not into waiting around that long, you could apply some optimization and only generate unique racks plus a number of how many ways that rack can be generated - ie if the rack is ADLSJKQ there would be 9 ways to choose an A, 4 ways to choose a D, 4 ways to choose an L... or 9*4*4*4*1*1*1 or 576 ways this same rack could occur. This will certainly speed up the process because you are comparing your words to a smaller set of racks.
More complicated than the above example is racks with repeated letters, like 3 Es for instance. Then you have to put in the number of ways you can choose 3 of the 12 Es, That would be 12*11*10 or 1320. So, if the rack was ADEEELS, it would be 9*4*(12*11*10)*4*4 or 760320 ways.
So now you get to the point where you say, Eureeka!, I can see that all I really have to do is run each valid word through and see how many racks can be generated that make that word! Way better, because then I dont have to mess around with generating all those racks. So, what you do is make an anagram of each possible word and then remove the duplicate anagrams. For example "RESHOOT", "SHOOTER", and "HOOTERS" all have the same anagam, "EHOORST". Only need one of those. Then calculate the number of possible ways to rack EHOORST - 12*2*(8*7)*6*4*6 = 193536. (Probability of getting a bingo so you can play HOOTERS is 193536/80,678,106,432,000 or 0.000000002398866). Anyway, sum the number of racks for each unique letter sorted word and you have the number of successes. The number of trials is 80,678,106,432,000. Successes over trials is the probability of a bingo.
MagRobot
I am stuck with the following problem involving polynomials. (this is not a HW question but a teaser by my sister's math teacher that's been bugging me for a while: How can someone proceed with such a problem.
The reaminder of p(x) when divided by x+1 is 2. The reaminder of P(X) when divided by x-2 is 4. What is the reaminder p(x) by (x+1).(x-2) (the product)?
When i first read the problem i directly thought of modular arithematics,but i don't think the solution involves modular arithematics because my sister is still in high school and she is not familiar with it.
Hisham1987 15:38, 28 March 2007 (UTC)
Here's a hint... denote your polynomial as and see what happens. Cheers -- Hirak 99 21:40, 28 March 2007 (UTC)
I am not familiar with all the proper wordings and notation, so hopefully the subject isn't too confusing.
What I would like to learn is of a method to calculate the following kind of problem:
I have a series of 1,000 "events" For each such event, the outcome will be HEAD with 24.5% certainty, and TAIL with 75.5% certainty. The outcome for each event is independent of any "history".
1. What is the likelihood that
2) What is the most likely longest chain of HEADs?
3) Given the above certainities, how many events have to occur for the probability of having a chain of HEADs 7 or longer being .75 or greater?
NB: I do not wish for the answers to these particular questions, as they are just examples I just made up, but rather learn of the reasoning behind solving such problems. If the methods used are fairly simple, such as a few formulas where I'd put in the different values, I would still greatly appreciate any links to articles on the underlying math. 81.20.147.194 17:37, 28 March 2007 (UTC)
5+6+5+4+7+9+54+3+54+56+67+76+56+56+67+76+67+67+76+67+54+54+54+54+54+45*3 —Preceding unsigned comment added by 86.129.225.138 ( talk • contribs)
What is the ith root of i? —Preceding unsigned comment added by 202.168.50.40 ( talk • contribs)