Maths Question

[NealRaymond2]

[quote]NealRaymond2 wrote:

[quote]NealRaymond2 wrote:

[quote]Spock81 wrote:
Is there some kind of math equation someone can make up for me so I can calculate the probability of someone having a certain letter in a facebook scrabble game? For instance, if I have an S and want to set myself up for a TW score for my NEXT turn, how can I figure out the likelihood my opponent will have an S as well, or maybe a blank???[/quote]

The following assumes you are playing a two-way game of scrabble, with the standard English-language distribution of letters and blanks.

Standard English-language scrabble has 100 tiles total, of which 2 are blanks and 4 are ‘S’.

Let P = number of tiles already played (on the board).
Let R = number of tiles in your rack.
Let Q = number of tiles in your opponent’s rack.
Let B = total number of blanks you can see, including the ones on the board and the ones in your rack.
Let S = total number of ‘S’ you can see, including the ones on the board and the ones in your rack.
Let us use * to represent multiplication, e.g. 10*3 = 30
Let us use ** to represent exponent, e.g. 10**3 = 1,000

(100-P-R) is the number of tiles that are either on your opponent’s rack or still in the bag.

100-4-2 = 94 is the total number of tiles in the game that are not ‘S’ and not blank.

(94-P-R) would be the total number of tiles that are either on your opponent’s rack or in the bag, that are not ‘S’ and not blank, if it were the case that no blanks and no ‘S’ were on the board or in your rack.

(94-P-R+S+B) is the total number of tiles that are either on your opponent’s rack or in the bag, that are not ‘S’ and not blank, taking into account that S ‘S’ and B blanks are either in your rack or on the board.

(94-P-R+S+B)/(100-P-R) is the probability that any particular one tile in your opponent’s rack is not ‘S’ and not blank.

((94-P-R+S+B)/(100-P-R))**Q = ((94-P-R+S+B)**Q)/((100-P-R)**Q) is the probability that every single one of the (Q) tiles in your opponent’s rack is not ‘S’ and not blank.

1-(((94-P-R+S+B)**Q)/((100-P-R)**Q))

= (((100-P-R)**Q)-((94-P-R+S+B)**Q))/((100-P-R)**Q)

is the probability that at least one of the tiles in your opponent’s rack is an ‘S’ or a blank.

There is a possibility that I might have added something where I should have subtracted, or made some other slip-up. But if you follow the above reasoning process through and make corrections where needed, the end result should be the correct formula for determining the probability that at least one of your opponent’s tiles is a blank or an ‘S’.[/quote]

If you need a rough formula that is quicker and easier to use, I think you can derive that formula as follows:

The formula above will always come out to be less than Q*(6-S-B)/(100-P-R) . How much less will depend on the size of P (the number of letters played so far). You might consider doing four actual calculations, using

(((100-P-R)**Q)-((94-P-R+S+B)**Q))/((100-P-R)**Q)

and

Q*(6-S-B)/(100-P-R)

dividing the former result by the latter result for each of the following cases:

1. Q=R=7 ; P=9.5 ; S=2 ; B=1
2. Q=R=7 ; P=29.5 ; S=2 ; B=1
3. Q=R=7 ; P=49.5 ; S=2 ; B=1
4. Q=R=7 ; P=69.5 ; S=2 ; B=1

Then use the following formula as a “rough approximation” of the probability that your opponent has at least one ‘S’ or blank:

If the number of tiles on the board is 0 through 19 then let X = the result found from #1 above.
If the number of tiles on the board is 20 through 39 then let X = the result found from #2 above.
If the number of tiles on the board is 40 through 59 then let X = the result found from #3 above.
If the number of tiles on the board is 60 through 79 then let X = the result found from #4 above.

(Find the 4 versions of ‘X’ using the instructions above before starting a game where you want to use the following approximation.)

XQ(6-S-B)/(100-P-R)

– should then (I think) be a very rough approximation of the probability of your opponent having at least one ‘S’ or one blank: less accurate but perhaps more practical to use on-the-fly than the more accurate –

(((100-P-R)**Q)-((94-P-R+S+B)**Q))/((100-P-R)**Q)[/quote]

In order to modify all of the above to suit a letter other than ‘S’:

• Substitute the quantity of that letter plus another 2 for the blanks in place of “6” in all of the formulae above.
• Subtract the quantity of that particular letter from 100; subtract another 2 to account for the blanks; substitute that result in place of “94” in all of the formulae above.

In order to modify all of the above to suit any letter:

• Add this variable definition: “Let Y be the quantity of this letter found in the standard English-language version of Scrabble.”
• Substitute Y in place of 4 in all of the above.
• Substitute (Y+2) in place of 6 in all of the above.
• Substitute (100-(Y+2)) or (100-Y-2) in place of 94 in all of the above.
• Substitute “this letter” for all references to ‘S’ in all of the above.
• Make all other substitutions as are logically necessary, if I missed any.

Enjoy!

[bignate]

so…when the fuck did scrabble get so confusing?

[pgtips]

[quote]NealRaymond2 wrote:

[quote]NealRaymond2 wrote:

[quote]NealRaymond2 wrote:

[quote]Spock81 wrote:
Is there some kind of math equation someone can make up for me so I can calculate the probability of someone having a certain letter in a facebook scrabble game? For instance, if I have an S and want to set myself up for a TW score for my NEXT turn, how can I figure out the likelihood my opponent will have an S as well, or maybe a blank???[/quote]

The following assumes you are playing a two-way game of scrabble, with the standard English-language distribution of letters and blanks.

Standard English-language scrabble has 100 tiles total, of which 2 are blanks and 4 are ‘S’.

Let P = number of tiles already played (on the board).
Let R = number of tiles in your rack.
Let Q = number of tiles in your opponent’s rack.
Let B = total number of blanks you can see, including the ones on the board and the ones in your rack.
Let S = total number of ‘S’ you can see, including the ones on the board and the ones in your rack.
Let us use * to represent multiplication, e.g. 10*3 = 30
Let us use ** to represent exponent, e.g. 10**3 = 1,000

(100-P-R) is the number of tiles that are either on your opponent’s rack or still in the bag.

100-4-2 = 94 is the total number of tiles in the game that are not ‘S’ and not blank.

(94-P-R) would be the total number of tiles that are either on your opponent’s rack or in the bag, that are not ‘S’ and not blank, if it were the case that no blanks and no ‘S’ were on the board or in your rack.

(94-P-R+S+B) is the total number of tiles that are either on your opponent’s rack or in the bag, that are not ‘S’ and not blank, taking into account that S ‘S’ and B blanks are either in your rack or on the board.

(94-P-R+S+B)/(100-P-R) is the probability that any particular one tile in your opponent’s rack is not ‘S’ and not blank.

((94-P-R+S+B)/(100-P-R))**Q = ((94-P-R+S+B)**Q)/((100-P-R)**Q) is the probability that every single one of the (Q) tiles in your opponent’s rack is not ‘S’ and not blank.

1-(((94-P-R+S+B)**Q)/((100-P-R)**Q))

= (((100-P-R)**Q)-((94-P-R+S+B)**Q))/((100-P-R)**Q)

is the probability that at least one of the tiles in your opponent’s rack is an ‘S’ or a blank.

There is a possibility that I might have added something where I should have subtracted, or made some other slip-up. But if you follow the above reasoning process through and make corrections where needed, the end result should be the correct formula for determining the probability that at least one of your opponent’s tiles is a blank or an ‘S’.[/quote]

If you need a rough formula that is quicker and easier to use, I think you can derive that formula as follows:

The formula above will always come out to be less than Q*(6-S-B)/(100-P-R) . How much less will depend on the size of P (the number of letters played so far). You might consider doing four actual calculations, using

(((100-P-R)**Q)-((94-P-R+S+B)**Q))/((100-P-R)**Q)

and

Q*(6-S-B)/(100-P-R)

dividing the former result by the latter result for each of the following cases:

1. Q=R=7 ; P=9.5 ; S=2 ; B=1
2. Q=R=7 ; P=29.5 ; S=2 ; B=1
3. Q=R=7 ; P=49.5 ; S=2 ; B=1
4. Q=R=7 ; P=69.5 ; S=2 ; B=1

Then use the following formula as a “rough approximation” of the probability that your opponent has at least one ‘S’ or blank:

If the number of tiles on the board is 0 through 19 then let X = the result found from #1 above.
If the number of tiles on the board is 20 through 39 then let X = the result found from #2 above.
If the number of tiles on the board is 40 through 59 then let X = the result found from #3 above.
If the number of tiles on the board is 60 through 79 then let X = the result found from #4 above.

(Find the 4 versions of ‘X’ using the instructions above before starting a game where you want to use the following approximation.)

XQ(6-S-B)/(100-P-R)

– should then (I think) be a very rough approximation of the probability of your opponent having at least one ‘S’ or one blank: less accurate but perhaps more practical to use on-the-fly than the more accurate –

(((100-P-R)**Q)-((94-P-R+S+B)**Q))/((100-P-R)**Q)[/quote]

In order to modify all of the above to suit a letter other than ‘S’:

• Substitute the quantity of that letter plus another 2 for the blanks in place of “6” in all of the formulae above.
• Subtract the quantity of that particular letter from 100; subtract another 2 to account for the blanks; substitute that result in place of “94” in all of the formulae above.

In order to modify all of the above to suit any letter:

• Add this variable definition: “Let Y be the quantity of this letter found in the standard English-language version of Scrabble.”
• Substitute Y in place of 4 in all of the above.
• Substitute (Y+2) in place of 6 in all of the above.
• Substitute (100-(Y+2)) or (100-Y-2) in place of 94 in all of the above.
• Substitute “this letter” for all references to ‘S’ in all of the above.
• Make all other substitutions as are logically necessary, if I missed any.

Enjoy!
[/quote]

= 11?

[NealRaymond2]

[quote]pgtips wrote:

[quote]NealRaymond2 wrote:

[quote]NealRaymond2 wrote:

[quote]NealRaymond2 wrote:

[quote]Spock81 wrote:
Is there some kind of math equation someone can make up for me so I can calculate the probability of someone having a certain letter in a facebook scrabble game? For instance, if I have an S and want to set myself up for a TW score for my NEXT turn, how can I figure out the likelihood my opponent will have an S as well, or maybe a blank???[/quote]

The following assumes you are playing a two-way game of scrabble, with the standard English-language distribution of letters and blanks.

Standard English-language scrabble has 100 tiles total, of which 2 are blanks and 4 are ‘S’.

Let P = number of tiles already played (on the board).
Let R = number of tiles in your rack.
Let Q = number of tiles in your opponent’s rack.
Let B = total number of blanks you can see, including the ones on the board and the ones in your rack.
Let S = total number of ‘S’ you can see, including the ones on the board and the ones in your rack.
Let us use * to represent multiplication, e.g. 10*3 = 30
Let us use ** to represent exponent, e.g. 10**3 = 1,000

(100-P-R) is the number of tiles that are either on your opponent’s rack or still in the bag.

100-4-2 = 94 is the total number of tiles in the game that are not ‘S’ and not blank.

(94-P-R) would be the total number of tiles that are either on your opponent’s rack or in the bag, that are not ‘S’ and not blank, if it were the case that no blanks and no ‘S’ were on the board or in your rack.

(94-P-R+S+B) is the total number of tiles that are either on your opponent’s rack or in the bag, that are not ‘S’ and not blank, taking into account that S ‘S’ and B blanks are either in your rack or on the board.

(94-P-R+S+B)/(100-P-R) is the probability that any particular one tile in your opponent’s rack is not ‘S’ and not blank.

((94-P-R+S+B)/(100-P-R))**Q = ((94-P-R+S+B)**Q)/((100-P-R)**Q) is the probability that every single one of the (Q) tiles in your opponent’s rack is not ‘S’ and not blank.

1-(((94-P-R+S+B)**Q)/((100-P-R)**Q))

= (((100-P-R)**Q)-((94-P-R+S+B)**Q))/((100-P-R)**Q)

is the probability that at least one of the tiles in your opponent’s rack is an ‘S’ or a blank.

There is a possibility that I might have added something where I should have subtracted, or made some other slip-up. But if you follow the above reasoning process through and make corrections where needed, the end result should be the correct formula for determining the probability that at least one of your opponent’s tiles is a blank or an ‘S’.[/quote]

If you need a rough formula that is quicker and easier to use, I think you can derive that formula as follows:

The formula above will always come out to be less than Q*(6-S-B)/(100-P-R) . How much less will depend on the size of P (the number of letters played so far). You might consider doing four actual calculations, using

(((100-P-R)**Q)-((94-P-R+S+B)**Q))/((100-P-R)**Q)

and

Q*(6-S-B)/(100-P-R)

dividing the former result by the latter result for each of the following cases:

1. Q=R=7 ; P=9.5 ; S=2 ; B=1
2. Q=R=7 ; P=29.5 ; S=2 ; B=1
3. Q=R=7 ; P=49.5 ; S=2 ; B=1
4. Q=R=7 ; P=69.5 ; S=2 ; B=1

Then use the following formula as a “rough approximation” of the probability that your opponent has at least one ‘S’ or blank:

If the number of tiles on the board is 0 through 19 then let X = the result found from #1 above.
If the number of tiles on the board is 20 through 39 then let X = the result found from #2 above.
If the number of tiles on the board is 40 through 59 then let X = the result found from #3 above.
If the number of tiles on the board is 60 through 79 then let X = the result found from #4 above.

(Find the 4 versions of ‘X’ using the instructions above before starting a game where you want to use the following approximation.)

XQ(6-S-B)/(100-P-R)

– should then (I think) be a very rough approximation of the probability of your opponent having at least one ‘S’ or one blank: less accurate but perhaps more practical to use on-the-fly than the more accurate –

(((100-P-R)**Q)-((94-P-R+S+B)**Q))/((100-P-R)**Q)[/quote]

In order to modify all of the above to suit a letter other than ‘S’:

• Substitute the quantity of that letter plus another 2 for the blanks in place of “6” in all of the formulae above.
• Subtract the quantity of that particular letter from 100; subtract another 2 to account for the blanks; substitute that result in place of “94” in all of the formulae above.

In order to modify all of the above to suit any letter:

• Add this variable definition: “Let Y be the quantity of this letter found in the standard English-language version of Scrabble.”
• Substitute Y in place of 4 in all of the above.
• Substitute (Y+2) in place of 6 in all of the above.
• Substitute (100-(Y+2)) or (100-Y-2) in place of 94 in all of the above.
• Substitute “this letter” for all references to ‘S’ in all of the above.
• Make all other substitutions as are logically necessary, if I missed any.

Enjoy!
[/quote]

= 11?[/quote]

No. The smallest number would be 0; the largest number would be 1. It’s probability.

[NealRaymond2]

[quote]NealRaymond2 wrote:

[quote]NealRaymond2 wrote:

[quote]NealRaymond2 wrote:

[quote]Spock81 wrote:
Is there some kind of math equation someone can make up for me so I can calculate the probability of someone having a certain letter in a facebook scrabble game? For instance, if I have an S and want to set myself up for a TW score for my NEXT turn, how can I figure out the likelihood my opponent will have an S as well, or maybe a blank???[/quote]

The following assumes you are playing a two-way game of scrabble, with the standard English-language distribution of letters and blanks.

Standard English-language scrabble has 100 tiles total, of which 2 are blanks and 4 are ‘S’.

Let P = number of tiles already played (on the board).
Let R = number of tiles in your rack.
Let Q = number of tiles in your opponent’s rack.
Let B = total number of blanks you can see, including the ones on the board and the ones in your rack.
Let S = total number of ‘S’ you can see, including the ones on the board and the ones in your rack.
Let us use * to represent multiplication, e.g. 10*3 = 30
Let us use ** to represent exponent, e.g. 10**3 = 1,000

(100-P-R) is the number of tiles that are either on your opponent’s rack or still in the bag.

100-4-2 = 94 is the total number of tiles in the game that are not ‘S’ and not blank.

(94-P-R) would be the total number of tiles that are either on your opponent’s rack or in the bag, that are not ‘S’ and not blank, if it were the case that no blanks and no ‘S’ were on the board or in your rack.

(94-P-R+S+B) is the total number of tiles that are either on your opponent’s rack or in the bag, that are not ‘S’ and not blank, taking into account that S ‘S’ and B blanks are either in your rack or on the board.

(94-P-R+S+B)/(100-P-R) is the probability that any particular one tile in your opponent’s rack is not ‘S’ and not blank.

((94-P-R+S+B)/(100-P-R))**Q = ((94-P-R+S+B)**Q)/((100-P-R)**Q) is the probability that every single one of the (Q) tiles in your opponent’s rack is not ‘S’ and not blank.

1-(((94-P-R+S+B)**Q)/((100-P-R)**Q))

= (((100-P-R)**Q)-((94-P-R+S+B)**Q))/((100-P-R)**Q)

is the probability that at least one of the tiles in your opponent’s rack is an ‘S’ or a blank.

There is a possibility that I might have added something where I should have subtracted, or made some other slip-up. But if you follow the above reasoning process through and make corrections where needed, the end result should be the correct formula for determining the probability that at least one of your opponent’s tiles is a blank or an ‘S’.[/quote]

If you need a rough formula that is quicker and easier to use, I think you can derive that formula as follows:

The formula above will always come out to be less than Q*(6-S-B)/(100-P-R) . How much less will depend on the size of P (the number of letters played so far). You might consider doing four actual calculations, using

(((100-P-R)**Q)-((94-P-R+S+B)**Q))/((100-P-R)**Q)

and

Q*(6-S-B)/(100-P-R)

dividing the former result by the latter result for each of the following cases:

1. Q=R=7 ; P=9.5 ; S=2 ; B=1
2. Q=R=7 ; P=29.5 ; S=2 ; B=1
3. Q=R=7 ; P=49.5 ; S=2 ; B=1
4. Q=R=7 ; P=69.5 ; S=2 ; B=1

Then use the following formula as a “rough approximation” of the probability that your opponent has at least one ‘S’ or blank:

If the number of tiles on the board is 0 through 19 then let X = the result found from #1 above.
If the number of tiles on the board is 20 through 39 then let X = the result found from #2 above.
If the number of tiles on the board is 40 through 59 then let X = the result found from #3 above.
If the number of tiles on the board is 60 through 79 then let X = the result found from #4 above.

(Find the 4 versions of ‘X’ using the instructions above before starting a game where you want to use the following approximation.)

XQ(6-S-B)/(100-P-R)

– should then (I think) be a very rough approximation of the probability of your opponent having at least one ‘S’ or one blank: less accurate but perhaps more practical to use on-the-fly than the more accurate –

(((100-P-R)**Q)-((94-P-R+S+B)**Q))/((100-P-R)**Q)[/quote]

In order to modify all of the above to suit a letter other than ‘S’:

• Substitute the quantity of that letter plus another 2 for the blanks in place of “6” in all of the formulae above.
• Subtract the quantity of that particular letter from 100; subtract another 2 to account for the blanks; substitute that result in place of “94” in all of the formulae above.

In order to modify all of the above to suit any letter:

• Add this variable definition: “Let Y be the quantity of this letter found in the standard English-language version of Scrabble.”
• Substitute Y in place of 4 in all of the above.
• Substitute (Y+2) in place of 6 in all of the above.
• Substitute (100-(Y+2)) or (100-Y-2) in place of 94 in all of the above.
• Substitute “this letter” for all references to ‘S’ in all of the above.
• Make all other substitutions as are logically necessary, if I missed any.

Enjoy!
[/quote]

Oops: I’m pretty sure I made an incorrect assumption regarding the X multiplier for the simplified-approximation version of the formula. I assumed that the ratio of –

(((100-P-R)**Q)-((94-P-R+S+B)**Q))/((100-P-R)**Q)

– to –

Q*(6-S-B)/(100-P-R)

will only vary significantly based on P, but not vary significantly based on S and B. But thinking about it some more, I am pretty sure that is wrong. So if you want to use the approximation –

XQ(6-S-B)/(100-P-R)

– then you would actually need to calculate –

(((100-P-R)**Q)-((94-P-R+S+B)**Q))/((100-P-R)*Q)
divided by
Q(6-S-B)/(100-P-R)

– not only for different values of P, but also for different values of S and B. Something like:

1.1) Q=R=7 ; P=9.5 ; S+B=0
1.2) Q=R=7 ; P=9.5 ; S+B=1
1.3) Q=R=7 ; P=9.5 ; S+B=2
1.4) Q=R=7 ; P=9.5 ; S+B=3
1.5) Q=R=7 ; P=9.5 ; S+B=4
1.6) Q=R=7 ; P=9.5 ; S+B=5
2.1) Q=R=7 ; P=29.5 ; S+B=0
2.2) Q=R=7 ; P=29.5 ; S+B=1
2.3) Q=R=7 ; P=29.5 ; S+B=2
2.4) Q=R=7 ; P=29.5 ; S+B=3
2.5) Q=R=7 ; P=29.5 ; S+B=4
2.6) Q=R=7 ; P=29.5 ; S+B=5
3.1) Q=R=7 ; P=49.5 ; S+B=0
3.2) Q=R=7 ; P=49.5 ; S+B=1
3.3) Q=R=7 ; P=49.5 ; S+B=2
3.4) Q=R=7 ; P=49.5 ; S+B=3
3.5) Q=R=7 ; P=49.5 ; S+B=4
3.6) Q=R=7 ; P=49.5 ; S+B=5
4.1) Q=R=7 ; P=69.5 ; S+B=0
4.2) Q=R=7 ; P=69.5 ; S+B=1
4.3) Q=R=7 ; P=69.5 ; S+B=2
4.4) Q=R=7 ; P=69.5 ; S+B=3
4.5) Q=R=7 ; P=69.5 ; S+B=4
4.6) Q=R=7 ; P=69.5 ; S+B=5

If the number of tiles on the board is 0 through 19 and there are no blanks nor ‘S’ on the board nor in your rack, then let X = the result found from #1.1 above.
If the number of tiles on the board is 0 through 19 and there is only 1 blank or ‘S’ on the board or in your rack, then let X = the result found from #1.2 above.

… etc. …

XQ(6-S-B)/(100-P-R)

But there might be too many variants of X to realistically use this approximation on the fly in a game.

===============================================

If someone does not already have a good seat-of-the-pants idea of the probability that your opponent has a particular letter or blank under a particular set of circumstances, this might be the best way to improve that situation for future games:

Work through some numbers using the more accurate formula –

(((100-P-R)**Q)-((94-P-R+S+B)**Q))/((100-P-R)**Q)

(replace the “94” with the actual number of tiles in the entire game that do not match the letter you have in mind, subtracting another 2 for the blanks)

– in different sets of circumstances, and try to get an approximate “feel” for the probabilities’ overall pattern.