Maths Question

Dr_Matt581 · June 22, 2012, 2:33pm

[quote]yolo84 wrote:

[quote]Dr.Matt581 wrote:
[/quote]

Out of interest, what are you academic credentials? Apologies if you have mentioned them elsewhere.[/quote]

I have mentioned them in another thread but it was a giant thread about an old forest so it was easy to miss that post. I did my undergrad at what is now known as the University of Moscow where I majored in Mathematics and Physics, then I moved to the US not long after the fall of the Soviet Union and went to Carnegie Mellon University where I got my PhD in nuclear physics, which these days is more commonly known as particle physics.

I also got an Msc in mathematics. I had to fulfill most of the degree requirements while getting my PhD so I just went ahead and did it. My early research interests were about the decay mechanisms of long-lived radionuclides, but these days I am doing more theoretical work involving quark-gluon plasmas.

IamMarqaos · June 22, 2012, 2:36pm

[quote]Iron Dwarf wrote:

[quote]Professor X wrote:

The PX is silent.
[/quote]

Hell, THAT’S rare around here.

;)[/quote]

Only after the Apocalypse!

IamMarqaos · June 22, 2012, 2:38pm

[quote]Dr.Matt581 wrote:

[quote]lnname wrote:
for everyone that wanted a maths question, here is one.

given
y=f(g(x))/h(g(x))
where f and h are linear functions of g(x)

what is
d^[1] y/ dx^[2]

?
[/quote]

What class are you taking that you would be asked to do this? This is a very difficult problem to solve since it involves multiple compositions, but it is doable with a lot of time and effort. I am assuming that you have been given this problem for a class so I will not do a full proof for you, but I can give you a general idea on what you need to do. Plus, I do not want to work out a full proof. I did plenty of that in college. First, you must decide on how to begin. This starts out as a product rule problem,you should start by using the General Leibniz Rule, which is:

d^[3] y/ dx^[4] = [sigma k=0 to n] (binomial coefficient)(f^k)((h^-1)^n-k)

but you are going to have to take into account that both f and h are both compositions of the function g(x), which means that you are going to need to expand both f^k and (h^-1)^n-k using Faa di Bruno’s Formula. It is too tedious to type out Faa di Bruno’s Formula so you will have to look it up. It should be in your textbook or notes somewhere. Or just google it, it is a very commonly used formula so you will not have any trouble finding it. I hope this helps.

[/quote]

Not afraid to admit I developed a bit of a man crush. Just a bit.

n ↩︎
n ↩︎
n ↩︎
n ↩︎

yolo84 · June 22, 2012, 2:38pm

[quote]Dr.Matt581 wrote:
I have mentioned them in another thread but it was a giant thread about an old forest so it was easy to miss that post. I did my undergrad at what is now known as the University of Moscow where I majored in Mathematics and Physics, then I moved to the US not long after the fall of the Soviet Union and went to Carnegie Mellon University where I got my PhD in nuclear physics, which these days is more commonly known as particle physics.

I also got an Msc in mathematics. I had to fulfill most of the degree requirements while getting my PhD so I just went ahead and did it. My early research interests were about the decay mechanisms of long-lived radionuclides, but these days I am doing more theoretical work involving quark-gluon plasmas.[/quote]

Interesting stuff, thanks.

Ulty · June 22, 2012, 2:59pm

Dr_Matt581 · June 22, 2012, 2:42pm

[quote]yolo84 wrote:

[quote]Derek542 wrote:

[quote]yolo84 wrote:

[quote]Dr.Matt581 wrote:
[/quote]

Out of interest, what are you academic credentials? Apologies if you have mentioned them elsewhere.[/quote]

Well looking at his name I would go with PHD. [/quote]

Thanks for your help!

I meant more along the lines of particular areas, what is he currently teaching etc.[/quote]

As for what I am teaching, at the undergrad level I usually teach one of the introductory physics physics for scientists and engineers, intermediate and advanced E&M, and intro to quantum mechanics. The graduate courses that I teach are usually various field theory and particle physics courses depending on the semester and how many students are interested in taking a particular class that I usually teach. Right now I am also completely revising the curriculum for our mathematical and computational methods in physics classes and when I am finished with that, I will take over teaching those classes and stop teaching E&M.

Derek542 · June 22, 2012, 2:42pm

[quote]yolo84 wrote:

[quote]Derek542 wrote:

[quote]yolo84 wrote:

[quote]Dr.Matt581 wrote:
[/quote]

Out of interest, what are you academic credentials? Apologies if you have mentioned them elsewhere.[/quote]

Well looking at his name I would go with PHD. [/quote]

Thanks for your help!

I meant more along the lines of particular areas, what is he currently teaching etc.[/quote]

Always here to help.

DBCooper · June 22, 2012, 5:17pm

I have a math question. It’s more of a probability question than whatever gibberish the OP posted.

I’m trying to come up with a formula for winning at acey-deucey. The game is simple. Two cards are dealt face up and a player has to bet whether the third card will have a numerical value between the first two. The first ace dealt face up is chosen as either high or low and the second one and any ace falling in the middle is a high ace. Players can bet anything up to the size of the pot and can pass on any possibility they want, like when a 9 and a jack are dealt out. The deck is used until no cards remain, reshuffled and the game continues, sort of like blackjack.

What I want to know is if it is possible to use a formula to determine my odds at any given point in the game of the next card being between the first and second card. Obviously, simply memorizing every card that comes out is the best way, but that isn’t feasible with a fast dealer. And everyone I gamble with knows I’m an expert card counter when it comes to blackjack so they try to deal so fast that it’s hard for me to remember each card this way.

Now, given that when counting cards 2 thru 6 equals a positive value of one, 7-9 equals zero and 10-A equals a negative one value. When the count in the deck is relatively high (depending on how many decks are used) the player’s odds go up. WIth acey-deucey the odds are against me when I bet with anything less than a differential of 7.

If the assumption is made that I wouldn’t bet with a differential of less than 7 if I knew nothing about any of the cards that have already come out, is there a way to use the blackjack card-counting system to win at acey-deucey? Or is the only way to gain an edge to simply memorize the cards as they come? I understand that if there IS a formula for deciphering these odds independent of memorizing cards that it would not be feasible to calculate these odds while I’m playing. But I’m just curious. I admit, I’m no mathematician and my interest in statistics and probability has more to do with sabermetrics than anything else.

Derek542 · June 22, 2012, 5:23pm

[quote]DBCooper wrote:
I have a math question. It’s more of a probability question than whatever gibberish the OP posted.

I’m trying to come up with a formula for winning at acey-deucey. The game is simple. Two cards are dealt face up and a player has to bet whether the third card will have a numerical value between the first two. The first ace dealt face up is chosen as either high or low and the second one and any ace falling in the middle is a high ace. Players can bet anything up to the size of the pot and can pass on any possibility they want, like when a 9 and a jack are dealt out. The deck is used until no cards remain, reshuffled and the game continues, sort of like blackjack.

What I want to know is if it is possible to use a formula to determine my odds at any given point in the game of the next card being between the first and second card. Obviously, simply memorizing every card that comes out is the best way, but that isn’t feasible with a fast dealer. And everyone I gamble with knows I’m an expert card counter when it comes to blackjack so they try to deal so fast that it’s hard for me to remember each card this way.

Now, given that when counting cards 2 thru 6 equals a positive value of one, 7-9 equals zero and 10-A equals a negative one value. When the count in the deck is relatively high (depending on how many decks are used) the player’s odds go up. WIth acey-deucey the odds are against me when I bet with anything less than a differential of 7.

If the assumption is made that I wouldn’t bet with a differential of less than 7 if I knew nothing about any of the cards that have already come out, is there a way to use the blackjack card-counting system to win at acey-deucey? Or is the only way to gain an edge to simply memorize the cards as they come? I understand that if there IS a formula for deciphering these odds independent of memorizing cards that it would not be feasible to calculate these odds while I’m playing. But I’m just curious. I admit, I’m no mathematician and my interest in statistics and probability has more to do with sabermetrics than anything else.[/quote]

No

bignate · June 22, 2012, 5:42pm

[quote]DBCooper wrote:
I have a math question. It’s more of a probability question than whatever gibberish the OP posted.

I’m trying to come up with a formula for winning at acey-deucey. The game is simple. Two cards are dealt face up and a player has to bet whether the third card will have a numerical value between the first two. The first ace dealt face up is chosen as either high or low and the second one and any ace falling in the middle is a high ace. Players can bet anything up to the size of the pot and can pass on any possibility they want, like when a 9 and a jack are dealt out. The deck is used until no cards remain, reshuffled and the game continues, sort of like blackjack.

What I want to know is if it is possible to use a formula to determine my odds at any given point in the game of the next card being between the first and second card. Obviously, simply memorizing every card that comes out is the best way, but that isn’t feasible with a fast dealer. And everyone I gamble with knows I’m an expert card counter when it comes to blackjack so they try to deal so fast that it’s hard for me to remember each card this way.

Now, given that when counting cards 2 thru 6 equals a positive value of one, 7-9 equals zero and 10-A equals a negative one value. When the count in the deck is relatively high (depending on how many decks are used) the player’s odds go up. WIth acey-deucey the odds are against me when I bet with anything less than a differential of 7.

If the assumption is made that I wouldn’t bet with a differential of less than 7 if I knew nothing about any of the cards that have already come out, is there a way to use the blackjack card-counting system to win at acey-deucey? Or is the only way to gain an edge to simply memorize the cards as they come? I understand that if there IS a formula for deciphering these odds independent of memorizing cards that it would not be feasible to calculate these odds while I’m playing. But I’m just curious. I admit, I’m no mathematician and my interest in statistics and probability has more to do with sabermetrics than anything else.[/quote]

i believe this would require many different chance nodes and a shit load of different branches to model the start of every possible game, followed by the probabilities of each chance for each situation following that. It can be done, but it sounds worse than mapping out chess which is essentially doable in the same fashion

Dr_Matt581 · June 22, 2012, 6:52pm

[quote]DBCooper wrote:
I have a math question. It’s more of a probability question than whatever gibberish the OP posted.

I’m trying to come up with a formula for winning at acey-deucey. The game is simple. Two cards are dealt face up and a player has to bet whether the third card will have a numerical value between the first two. The first ace dealt face up is chosen as either high or low and the second one and any ace falling in the middle is a high ace. Players can bet anything up to the size of the pot and can pass on any possibility they want, like when a 9 and a jack are dealt out. The deck is used until no cards remain, reshuffled and the game continues, sort of like blackjack.

What I want to know is if it is possible to use a formula to determine my odds at any given point in the game of the next card being between the first and second card. Obviously, simply memorizing every card that comes out is the best way, but that isn’t feasible with a fast dealer. And everyone I gamble with knows I’m an expert card counter when it comes to blackjack so they try to deal so fast that it’s hard for me to remember each card this way.

Now, given that when counting cards 2 thru 6 equals a positive value of one, 7-9 equals zero and 10-A equals a negative one value. When the count in the deck is relatively high (depending on how many decks are used) the player’s odds go up. WIth acey-deucey the odds are against me when I bet with anything less than a differential of 7.

If the assumption is made that I wouldn’t bet with a differential of less than 7 if I knew nothing about any of the cards that have already come out, is there a way to use the blackjack card-counting system to win at acey-deucey? Or is the only way to gain an edge to simply memorize the cards as they come? I understand that if there IS a formula for deciphering these odds independent of memorizing cards that it would not be feasible to calculate these odds while I’m playing. But I’m just curious. I admit, I’m no mathematician and my interest in statistics and probability has more to do with sabermetrics than anything else.[/quote]

I thought acey-deucey was a form of backgammon, not a card game. Anyway, to answer your question yes, it is possible but the formula would be very complex and completely useless in an actual game environment. There is an entire branch of applied mathematics that deals with gambling, but nothing really useful has come out of it in years.

A good strategy is to just think the basic probability through. If the difference between the value of the two cards is large, the probability of the third card is falling between them is high and it is a safe bet, if it is not then it is not a safe bet. If you have to actually guess the value of the card, then it is going to start at between a 4 and 8 % chance of guessing right (for a single deck), but that will change as more cards are dealt and discarded. If you can do a halfway decent job of keeping track of the cards that have been dealt then you may be able to improve your odds a bit.

NealRaymond2 · June 23, 2012, 3:25am

[quote]Dr.Matt581 wrote:

[quote]DBCooper wrote:
I have a math question. It’s more of a probability question than whatever gibberish the OP posted.

I’m trying to come up with a formula for winning at acey-deucey. The game is simple. Two cards are dealt face up and a player has to bet whether the third card will have a numerical value between the first two. The first ace dealt face up is chosen as either high or low and the second one and any ace falling in the middle is a high ace. Players can bet anything up to the size of the pot and can pass on any possibility they want, like when a 9 and a jack are dealt out. The deck is used until no cards remain, reshuffled and the game continues, sort of like blackjack.

What I want to know is if it is possible to use a formula to determine my odds at any given point in the game of the next card being between the first and second card. Obviously, simply memorizing every card that comes out is the best way, but that isn’t feasible with a fast dealer. And everyone I gamble with knows I’m an expert card counter when it comes to blackjack so they try to deal so fast that it’s hard for me to remember each card this way.

Now, given that when counting cards 2 thru 6 equals a positive value of one, 7-9 equals zero and 10-A equals a negative one value. When the count in the deck is relatively high (depending on how many decks are used) the player’s odds go up. WIth acey-deucey the odds are against me when I bet with anything less than a differential of 7.

If the assumption is made that I wouldn’t bet with a differential of less than 7 if I knew nothing about any of the cards that have already come out, is there a way to use the blackjack card-counting system to win at acey-deucey? Or is the only way to gain an edge to simply memorize the cards as they come? I understand that if there IS a formula for deciphering these odds independent of memorizing cards that it would not be feasible to calculate these odds while I’m playing. But I’m just curious. I admit, I’m no mathematician and my interest in statistics and probability has more to do with sabermetrics than anything else.[/quote]

I thought acey-deucey was a form of backgammon, not a card game. Anyway, to answer your question yes, it is possible but the formula would be very complex and completely useless in an actual game environment. There is an entire branch of applied mathematics that deals with gambling, but nothing really useful has come out of it in years.

A good strategy is to just think the basic probability through. If the difference between the value of the two cards is large, the probability of the third card is falling between them is high and it is a safe bet, if it is not then it is not a safe bet. If you have to actually guess the value of the card, then it is going to start at between a 4 and 8 % chance of guessing right (for a single deck), but that will change as more cards are dealt and discarded. If you can do a halfway decent job of keeping track of the cards that have been dealt then you may be able to improve your odds a bit.
[/quote]

I think the odds shift mainly based on how many “middle” cards have already been dealt out vs. how many “extreme” cards have been dealt out. 2-3-4-Q-K-A could be considered “extreme”, while 5-6-7-8-9-10 could be considered “middle”. (Consider the Jack neutral, just to keep the number of “middle” cards equal to the number of “extreme” cards.) At the beginning of each new deck, mentally reset the count to zero. Every “extreme” card already dealt out is plus-one; every “middle” card already dealt out is minus-one. Bet more heavily when the count is positive (more “middle” cards left in the deck); bet more lightly when the count is negative (more “extreme” cards left in the deck). Or bet with a tighter spread between the current low card and the current high card when the count is positive and the current low and high cards straddle mostly “middle” cards; reserve betting for a wider spread when the count is negative.

Somebody with more knowledge of the intricacies of probability and/or knowledge of how to run computer simulations could probably work out what spread should be the betting cutoff for any given “count”, or maybe work out a chart of low-high combinations to bet on or not bet on depending on the “count”.

Again: this is analogous to the high-low blackjack card-counting system, but middle-extreme because that is what makes the difference in the odds for acey-deucey.

DBCooper · June 23, 2012, 3:25am

[quote]NealRaymond2 wrote:

[quote]Dr.Matt581 wrote:

[quote]DBCooper wrote:
I have a math question. It’s more of a probability question than whatever gibberish the OP posted.

I’m trying to come up with a formula for winning at acey-deucey. The game is simple. Two cards are dealt face up and a player has to bet whether the third card will have a numerical value between the first two. The first ace dealt face up is chosen as either high or low and the second one and any ace falling in the middle is a high ace. Players can bet anything up to the size of the pot and can pass on any possibility they want, like when a 9 and a jack are dealt out. The deck is used until no cards remain, reshuffled and the game continues, sort of like blackjack.

What I want to know is if it is possible to use a formula to determine my odds at any given point in the game of the next card being between the first and second card. Obviously, simply memorizing every card that comes out is the best way, but that isn’t feasible with a fast dealer. And everyone I gamble with knows I’m an expert card counter when it comes to blackjack so they try to deal so fast that it’s hard for me to remember each card this way.

Now, given that when counting cards 2 thru 6 equals a positive value of one, 7-9 equals zero and 10-A equals a negative one value. When the count in the deck is relatively high (depending on how many decks are used) the player’s odds go up. WIth acey-deucey the odds are against me when I bet with anything less than a differential of 7.

If the assumption is made that I wouldn’t bet with a differential of less than 7 if I knew nothing about any of the cards that have already come out, is there a way to use the blackjack card-counting system to win at acey-deucey? Or is the only way to gain an edge to simply memorize the cards as they come? I understand that if there IS a formula for deciphering these odds independent of memorizing cards that it would not be feasible to calculate these odds while I’m playing. But I’m just curious. I admit, I’m no mathematician and my interest in statistics and probability has more to do with sabermetrics than anything else.[/quote]

I thought acey-deucey was a form of backgammon, not a card game. Anyway, to answer your question yes, it is possible but the formula would be very complex and completely useless in an actual game environment. There is an entire branch of applied mathematics that deals with gambling, but nothing really useful has come out of it in years.

A good strategy is to just think the basic probability through. If the difference between the value of the two cards is large, the probability of the third card is falling between them is high and it is a safe bet, if it is not then it is not a safe bet. If you have to actually guess the value of the card, then it is going to start at between a 4 and 8 % chance of guessing right (for a single deck), but that will change as more cards are dealt and discarded. If you can do a halfway decent job of keeping track of the cards that have been dealt then you may be able to improve your odds a bit.
[/quote]

I think the odds shift mainly based on how many “middle” cards have already been dealt out vs. how many “extreme” cards have been dealt out. 2-3-4-Q-K-A could be considered “extreme”, while 5-6-7-8-9-10-J could be considered “middle”. At the beginning of each new deck, mentally reset the count to zero. (Or maybe to some number other than zero to adjust for the 6:7 ratio of “middle” cards to “extreme” cards, but I am not sure what the exact starting number should be. But in any case, starting at zero and counting from there should be good enough to improve your odds over betting blind.) Every “extreme” card already dealt out is plus-one; every “middle” card already dealt out is minus-one. Bet a little more heavily when the count is high (more “middle” cards left in the deck); bet a little more lightly when the count is low (more “extreme” cards left in the deck). Or bet with a slightly tighter spread between the current low card and the current high card when the count is high and the current low and high cards straddle mostly “middle” cards; reserve betting for a somewhat wider spread when the count is low.

Somebody with more knowledge of the intricacies of probability and/or knowledge of how to run computer simulations could probably work out what the starting count should actually be to take into account the 6:7 ratio, and what spread should be the cutoff for any given count.

Again: this is analogous to the high-low blackjack card-counting system, but middle-extreme because that is what makes the difference in the odds.[/quote]

Yes, I think you may be on to something with the whole extreme vs middle card thing. It would be pretty hard to count cards even during acey-deucey, but it would be even harder to memorize each individual card as they come. So I think you are on the right track by assigning at least a value to the “extreme” end of the possibilities and the “middle”, or non-extreme, values. At least this way one could have a working knowledge of their odds even if they weren’t sure of every single card that had come out.

I suppose the best I’m looking for here is a formula that approaches some sort of approximation, sort of like how in statistics there are formulas that can ascertain whether or not a certain hypothesis is within, say, 95% certainty as it pertains to the population based on a sample survey or whatever. I don’t think there is actually a formula that can hold true from one random game to the next that will give me completely accurate odds from card to card.

But as far as this whole middle vs. extreme value thing goes, I think this may be the way to “count” the cards, rather than the traditional blackjack style where there is a positive, negative and zero value for the cards. At least with a middle/extreme value set I would have a reasonable working knowledge of the odds when the differential of plus-seven or more falls in about the middle of the overall range of values available. Like, if it were a plus-seven differential and the higher card out there was a high ace and the low one was a seven instead of a ten and a three.

I don’t know. I have a feeling that even a formula that derives an approximation of one’s odds isn’t possible without the ability to remember each individual card that comes out, even if one doesn’t have to remember the suits, which is immaterial in acey-deucey anyways.

And to whoever thought acey-deucey was a backgammon game: it’s actually a variation of both backgammon AND poker. I don’t know anything about the backgammon variation, but I just recently took up backgammon and I’m already hooked so I’d love to know a thing or two about how to play acey-deucey backgammon.

In fact, I propose that this entire thread be devoted from here on out to games of chance and the mathematical odds behind each game. That is, if it’s alright with everyone else here.

DBCooper · June 23, 2012, 3:40am

TO NealRaymond:

I think that with the extreme vs middle value thing we are looking at a mean and standard deviation situation in which we assign x to each possible value and use: the sum of {x times P(x)}= the mean and the square root of (the sum of {(x-m)squared multiplied by P(x)}=the standard deviation. If we assign values greater than the third Quartile and less then the third Quartile the “middle” or neutral value and the values lying outside of the first and third Quartiles the extreme values, we will be closer to the final formula sought.

Of course, this is all based on a normal distribution and I think the distribution could be become distorted as the actual game itself processes. Then I think something along the lines of a binomial distribution formula is in order, like radical n/(n-x)radical multiplied by radicals (P to the Xth power multiplied by Q to the n-xth power).

NealRaymond2 · June 23, 2012, 5:11am

[quote]DBCooper wrote:
TO NealRaymond:

I think that with the extreme vs middle value thing we are looking at a mean and standard deviation situation in which we assign x to each possible value and use: the sum of {x times P(x)}= the mean and the square root of (the sum of {(x-m)squared multiplied by P(x)}=the standard deviation. If we assign values greater than the third Quartile and less then the third Quartile the “middle” or neutral value and the values lying outside of the first and third Quartiles the extreme values, we will be closer to the final formula sought.

Of course, this is all based on a normal distribution and I think the distribution could be become distorted as the actual game itself processes. Then I think something along the lines of a binomial distribution formula is in order, like radical n/(n-x)radical multiplied by radicals (P to the Xth power multiplied by Q to the n-xth power).[/quote]

That I don’t know one way or the other. Except that in terms of middle vs. extreme: I think second and third quartiles would be “middle” while first and fourth quartiles would be “extreme”.

I think people who could learn to track precise odds for a fast-moving game of blackjack or acey-deucey in their heads would be extremely rare. I think people who could learn to keep a count of extreme vs. middle cards dealt and apply the count to adjust the cutoff spread for betting, assuming a lot of practice, would not be all that rare. I also suspect that memorizing a chart of high-low pairs with differences of 5, 6, 7, 8, 9 and what the minimum extreme-vs.-middle count should be to bet on each would also be something that a lot of people could do, assuming a substantial investment of time and effort, similar to how people memorize the strategy chart for when to hit, stand, split, etc. in blackjack and what the adjustments are for some hands based on hi-lo counts.

I understand the extreme-middle system would not provide accurate odds for each individual hand in acey-deucey, but I don’t think any realistically usuable counting system provides accurate odds for each individual hand in blackjack, either. The best that a realistically usable counting system can do is make the reckoning of the odds for each individual hand substantially better on average than it would be if the cards already dealt were completely unaccounted for. And I believe the “on average” is important: I think any counting system that is realistically usuable will make the approximation of the odds worse instead of better for certain individual hands after certain sets of cards have already been dealt: because the particular cards that were actually dealt will not have the same effect on the odds for certain hands as the average of all cards that might have been dealt to produce the same count. That will be true for both blackjack and for acey-deucey, because any counting system that is realistically usuable equates different cards that do not actually interact with all other combinations of cards in the same manner. So again, the best we can hope for is to make the reckoning of the odds substantially better on average than it would be if we were completely blind to the history of the previously-dealt cards.

(I like my edited version above where I suggested making the Jack neutral better than my original version, so that the two types of cards have a 1:1 ratio. Another way to make the two types of cards have a 1:1 ratio might be to make the 5 of hearts and Jack of hearts “extreme” while the other 5’s and Jacks are “middle”. Bear in mind that the “average effect” of middle vs. extreme cards is what we are concerned with, so arbitrary assignment of borderline cards to make the two types balance should probably be ok as long as the computer simulations to determine the strategy charts for the pairs with differences of 5, 6, 7, 8, 9 use the same definition of “extreme” vs. “middle” that will be used when playing.)

Spock81 · June 24, 2012, 5:14pm

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???

DBCooper · June 24, 2012, 5:29pm

[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]

How many S’s are there in the game? How many total letter pieces are there? How many pieces does each player get?

NealRaymond2 · June 24, 2012, 7:49pm

[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’.

NealRaymond2 · June 24, 2012, 7:53pm

[quote]DBCooper 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]

How many S’s are there in the game? How many total letter pieces are there? How many pieces does each player get?[/quote]

4 ‘S’; 2 blank; 100 total tiles. 7 tiles on each player’s rack for most of the game. But at the end of the game there can be fewer than 7 tiles on each player’s rack, which is why I did my preceding derivation and formula using Q and R instead of just 7 for the number of letters on each player’s rack.

NealRaymond2 · June 24, 2012, 8:32pm