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organofcorti
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 October 15, 2011, 02:46:43 PM

Just a bit more about your post, since you put a lot of thought into it:

I think you missed my point.

Firstly, I don't question your logic on the use of the Binomial distribution.

I actually didn't use the binomial distribution in the post - I only mentioned that the number of snake eyes found in N throws follows a Poisson distribution.

Quote
There are two main ways to calculate this expected value.  The first is to consider X = X_0 + ... X_9 as a random variable in it's own right.  This will be a binomial random variable (just as you describe) and you can use the known formula for the expected value of such a random variable (number of trials * probability of success; n*p).

Quote
To get the expected value of an event, we need to calculate the probability of the event and the reward for that event. In this case, if a throw is rewarded once, it will always be the same: the reward/number of throws rewarded:
Code:
B/N

To find out how probable multiple rewards are we need to determine L. Since L follows the Poisson distribution, its mean is the expected number of snake-eyes during N throws:
Code:
p*N

The expected value of the throw:
Code:
B/N*p*N = p*B

So I calculate the expected value as p*B not n*p as you have. So either we have different meaning of expected value or one of us is wrong there! I'm ok with being wrong if you can show me why.

Quote
The other way is to note that the expectation operator is linear so
E(X_0 + ... + X_9) = E(X_0) + E(X_1) + ... + E(X_9) = n*E(X_0) = n*p
Indeed, this is the main way in which the formula for the expected value of a binomial distribution is derived.

It's unnecessary to sum the series since the expected value for a throw in the game, p*B, consists of unchanging constants.

It's late for me here so I might not have followed you properly, so I'm looking forward to your reply.

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teukon
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 October 15, 2011, 04:14:32 PM

So if you threw snake eyes on the first throw of a game, you'd get \$10, and the last nine throwers of the previous game get \$10 each. Then a new game starts. In the real world, with 24 hour casinos and no end of gamblers, I think we can assume that starting seroes of games and ending a series of games would be a rare occurrence.

Ok.  Most of my original post was about what happens at the beginning and end of a "series of games".  The confusion has come from my saying "game" for your "series of games" and "round" for your "game".  Certainly there is no funny business going on shortly before or after snake-eyes.

I actually didn't use the binomial distribution in the post - I only mentioned that the number of snake eyes found in N throws follows a Poisson distribution.

Good point, I have not been reading these posts in full detail but just trying to get a feeling for the mathematical approach taken.  Your approach really is fine if you are talking about the binomial distribution but calling this a Poisson process is technically an error.  This is a discrete process, not a continuous one.  My apologies for not reading your post thoroughly.

So I calculate the expected value as p*B not n*p as you have. So either we have different meaning of expected value or one of us is wrong there! I'm ok with being wrong if you can show me why.

Ah, we're just talking about different random variables.  The expected total payout for a throw is certainly p*B.  You're right, I made an error here.
Quote
n*E(X_0) = n*p

Of course, E(X_0) is 1/n*(p*B) and hence
Code:
E(X_0 + ... + X_9) is n*1/n*(p*B) = p*B.

When focusing on how the binomial distribution is a sum of Bernoulli ones it's very easy to write "E(X_i) = p" without thinking.

Quote
It's unnecessary to sum the series since the expected value for a throw in the game, p*B, consists of unchanging constants.

It's late for me here so I might not have followed you properly, so I'm looking forward to your reply.

The sum is, in my opinion, an important part of this problem.  Without it my solution would read:

Assuming there have been at least 9 previous throws and will be at least 9 future throws, the expected value of a throw is
Code:
10 * 1/10 * (1/100 * \$100) = \$1.
Because it costs \$1 to play the expected profit from any turn is exactly \$0 and hence there is no winning strategy (possibly unless we allow ourselves throws close to the beginning or end of a series of games).

The interesting bit is in seeing why the expected value of a throw is \$1.  If someone cannot see that it really is \$1 then I doubt using words like "binomial" and "distribution" won't help.  Instead, I think the simplest reasoning is to consider how much you expect to get from each of the next 10 throws (including your own).  In each case you would be one of 10 people sharing an expected 1/100 * \$100 = \$1 reward so your expected income from the throw is \$0.10.  Summing the expected incomes of all 10 of these throws yields the expected return of \$1.
organofcorti
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 October 16, 2011, 01:08:45 AM

Ok.  Most of my original post was about what happens at the beginning and end of a "series of games".  The confusion has come from my saying "game" for your "series of games" and "round" for your "game".  Certainly there is no funny business going on shortly before or after snake-eyes.

So reading your posts in with that in mind, I found your analysis of the game start and finish boundary problems interesting. I still have more thinking to do though. It takes a while. Mills grinding slowly but exceedingly fine etc. Well, fine-ish, anyway.

Quote
Your approach really is fine if you are talking about the binomial distribution but calling this a Poisson process is technically an error.  This is a discrete process, not a continuous one.

Point taken. The 'time' between snake eyes being rolled is the number of throws, not the time taken for the throws, and the number of throws to end a game is a series of Bernoulli trials. Pooled bitcoin mining would approximate a Poisson process better. But the number of snake eyes in N throws still follows a Poisson distribution. The number of throws to result in one snake eye follows a binomial distribution. I understand you realise that but I wanted to spell it out for other readers.

Quote
The sum is, in my opinion, an important part of this problem.
Yes, from what I understand atm, you couldn't show the start/finish boundary problem without summing that way. I was assuming, as you say, as many games in the past and the future as necessary to make the start and finish of a series of games irrelevant and tbh hadn't I considered the problem at all.

Quote
The interesting bit is in seeing why the expected value of a throw is \$1.  If someone cannot see that it really is \$1 then I doubt using words like "binomial" and "distribution" won't help.
Well, no, but the conversation might interest people enough to do their own reading. I find that unless very well crafted, intuitive explanations can lead readers astray since the writer has to assume *some* level of knowledge on the part of the reader. If you assume to too little knowledge the reader gets bored and assume too much knowledge and the reader makes incorrect intuitive leaps. I can't find that happy medium so I try to make the math as simple as possible to follow.

Well, that was great! Thanks for persevering teukon. I'm sure I wasn't the only one to find that interesting.

Comp ends in only 3 days. Any other entrants?

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teukon
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 October 16, 2011, 02:15:21 AM

But the number of snake eyes in N throws still follows a Poisson distribution. The number of throws to result in one snake eye follows a binomial distribution.

I'm confused here.  I think the number of snake eyes in N throws follows a binomial distribution and the number of throws to result in one snake eye follows a geometric distribution.  Both of these distributions are built on the underlying base of independent identically distributed Bernoulli trials.  The Poisson distribution provides an approximation for the binomial distribution when n is large and p is small (and hence is an excellent choice in practice for handling bitcoin mining).

Yes, from what I understand atm, you couldn't show the start/finish boundary problem without summing that way. I was assuming, as you say, as many games in the past and the future as necessary to make the start and finish of a series of games irrelevant and tbh hadn't I considered the problem at all.

Ok.  I only brought it up because the assumption of many games in the past and future was not made in the question and did affect the answer.

Quote
The interesting bit is in seeing why the expected value of a throw is \$1.  If someone cannot see that it really is \$1 then I doubt using words like "binomial" and "distribution" won't help.

Well, no, but the conversation might interest people enough to do their own reading. I find that unless very well crafted, intuitive explanations can lead readers astray since the writer has to assume *some* level of knowledge on the part of the reader. If you assume to too little knowledge the reader gets bored and assume too much knowledge and the reader makes incorrect intuitive leaps. I can't find that happy medium so I try to make the math as simple as possible to follow.

I think you are wiser than I.

Well, that was great! Thanks for persevering teukon. I'm sure I wasn't the only one to find that interesting.

Comp ends in only 3 days. Any other entrants?

You're welcome.  I was briefly interested in spending some time (a full day) looking at all the reward systems mathematically, perhaps inventing one or two of my own in the process.  I discovered however that a decent body of work on this subject already exists.

organofcorti
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 October 19, 2011, 08:13:33 AM

Expected values for dicecoin - competitions answers

Question 1: is there a winning strategy for this gambling game? What is it?

Both competition entrants iopq and 3phase put a lot more work into this than I expected anyone to make! Good job both of you and I'll make some comments about your results after I give the answer.

I was actually assuming that entrants would simply try the game and see what happens, as I mentioned in one of the hints. If you had tried that you would have obtained something like the following 100 games, which took quite a while using the online dice generator I linked to:

Code:
Number of throws per game:4, 3, 5, 13, 52, 26, 23, 2, 5, 5, 3, 4, 6, 3, 14, 8, 12, 15, 6, 2, 14, 39, 5, 1, 13, 5, 1, 12, 3, 6, 5, 18, 17, 8, 3, 4, 3, 2, 1, 9, 4, 3, 5, 3, 2, 3, 7, 1, 2, 4, 2, 2, 8, 20 ,17 ,2 ,2, 9, 9, 1, 1, 12, 6, 5, 9, 17, 24, 7, 2, 2, 2, 8, 11, 9, 29, 12, 14, 27, 17, 3, 4, 7, 4, 8, 14, 8, 6, 4, 24, 15, 6, 6, 3, 28, 5, 40, 14, 12, 1, 25

If you weren't going to use probability theory to solve this, the first intuitive leap you had to make was that assessing one throw per game before the costs and prizes were distributed amongst the players seems much easier to assess than trying to work out a per player result (which would depend on how many players were present). If you accept this for now, I'll derive the result mathematically later.

To work out the value of the first throw, we have to find all of the games which contain one or more throws - i.e. all of them. The profit for the first throw of a game before costs and winnings are split amongst players is:
Winnings for the throw minus cost of the throw = \$100/(number of throws in the game)-\$10

Code:
Average profit for the first throw of each game:
= mean(\$100/4-\$10, \$100/3-\$10, \$100/5-\$10, \$100/13-\$10, …. , \$100/25-\$10)
= mean(\$100/(all throws per game) - \$10)
= mean(25.24 - \$10) = \$15.24

So our average profit (when we define profit as winnings minus cost) for playing only the first throw in a game for this series of games is \$15.24

To estimate the value of the second throw, we have to find all of the games which contain two or more throws -
Code:
Number of throws per game, for games>=2 throws: 4, 3, 5, 13, 52, 26, 23, 2, 5, 5, 3, 4, 6, 3, 14, 8, 12, 15, 6, 2, 14, 39, 5, 13, 5, 12, 3, 6, 5, 18, 17, 8, 3, 4, 3, 2, 9, 4, 3, 5, 3, 2, 3, 7, 2, 4, 2, 2, 8, 20, 17, 2, 2, 9, 9, 12, 6, 5, 9, 17, 24, 7, 2, 2, 2, 8, 11, 9, 29, 12, 14, 27, 17, 3, 4, 7, 4, 8, 14 ,8, 6, 4, 24, 15, 6, 6, 3, 28, 5, 40, 14, 12, 25

Calculating the average profit as above results in \$9.62 per game for the second throw of each game.

Keep in mind that for a constant player, the expected profit is \$0 per game.

If we continue calculating the expected profits for a throw as above, we can generate the following results:

When the expected or average value of the throw is below \$0, the throw is more likely to reduce your overall winnings than increase them. So for this dataset the strategy would be to stay on for the first 5 throws. I would have accepted this as an answer. It turns out (interestingly) that when the expected values are more accurately calculated the strategy would still be a winner if the 4th throw or the 5th throw is used as the final throw.

Both iopq and 3phase came up with similar results. Although 3phase was first with
Quote
The best number is given by the strategy of playing 5 throws and then leaving.
he later added that any consecutive 5 throws would be a winning strategy (although I can't find that quote in the post anymore, only in the discussion between iopq & 3phase - can you confirm that for me 3phase?)

So, contingent on 3phase confirming his "any consecutive 5 throws" statement, I'm awarding iopq the huge 1btc prize! Congratulations iopq, and well done to both iopq and 3phase.

So I think that's enough for the moment. A little later I'll post a proof for the above, then a discussion of both sets of results from iopq and 3phase, and finally I explain how to calculate the expected profit from the game if you use this strategy.

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3phase
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 October 19, 2011, 08:39:50 AM

Both iopq and 3phase came up with similar results. Although 3phase was first with
Quote
The best number is given by the strategy of playing 5 throws and then leaving.
he later added that any consecutive 5 throws would be a winning strategy (although I can't find that quote in the post anymore, only in the discussion between iopq & 3phase - can you confirm that for me 3phase?)

So, contingent on 3phase confirming his "any consecutive 5 throws" statement, I'm awarding iopq the huge 1btc prize! Congratulations iopq, and well done to both iopq and 3phase.

I admitted I was wrong in making that statement, and edited my initial post (post #9) just after iopq's comments. See "EDIT" notes in post #13.

So I missed the prize for trying to play clever  .

Anyway, I did enjoy thinking about it and reading the posts here. Thanks, organofcorti !!!

Fiat no more.
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organofcorti
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 October 19, 2011, 08:48:37 AM

Both iopq and 3phase came up with similar results. Although 3phase was first with
Quote
The best number is given by the strategy of playing 5 throws and then leaving.
he later added that any consecutive 5 throws would be a winning strategy (although I can't find that quote in the post anymore, only in the discussion between iopq & 3phase - can you confirm that for me 3phase?)

So, contingent on 3phase confirming his "any consecutive 5 throws" statement, I'm awarding iopq the huge 1btc prize! Congratulations iopq, and well done to both iopq and 3phase.

I admitted I was wrong in making that statement, and edited my initial post (post #9) just after iopq's comments. See "EDIT" notes in post #13.

So I missed the prize for trying to play clever  .

Anyway, I did enjoy thinking about it and reading the posts here. Thanks, organofcorti !!!

You're welcome - glad you enjoyed it. And I don't think you were trying to be clever - you were just interpreting what you that the results told you. A good attempt anyway.

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organofcorti
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 October 19, 2011, 11:41:29 AM

Expected values for dicecoin: Answers part 2

So here I'll derive the strategy mathematically. I name all variables except number of players and cost as per Analysis of Bitcoin pooled mining reward systems to make the derivation easier to follow. A read of How to hop 7: Expected values will also be helpful.

D = 10, the 'difficulty' of throwing a ten on a ten sided die.
probability of a winning throw, p = 1/D
probability of a losing throw, (1-p) = 1-1/D
I = throw previous to the one being assessed
Y = number of players
N = number of throws
B = total reward per game = 100
C = cost per throw = 10
winnings = 100/(N*Y)
costs = 10/Y per round

Expected profit of the first throw = probability * value

Probabilities:

Code:
Probability of game ending at current throw, p1 = (1-1/D)^(1-1) * 1/D
Probability of game ending at next throw, p2 = (1-1/D)^(2-1) * 1/D
Probability of game ending at the throw after next, p3 = (1-1/D)^(3-1) * 1/D
…etc

if last throw = I
probability Nth after Ith throw wins
1/D*(1-1/D)^(N-I-1)

Values:
Code:
winning value first throw:
if there is 1 other player, the strategic player gets B/1/2
2 other players, strategic player gets B/1/3
3 other players, strategic player gets B/1/4
n other players, strategic player gets B/1/(Y+1)
Code:
winning value second throw:
if there is 1 other player, the strategic player gets B/1/2
2 other player strategic player gets B/2/3
3 other player strategic player gets B/2/4
n other players strategic player gets B/2/(Y+1)

So winning value Nth throw = B/(N*(Y+1))

Costs:
Code:
cost value first throw:
if there is 1 other player, the strategic player pays C/1/2
2 other player strategic player pays C/3
3 other player strategic player pays C/4
n other players strategic player pays C/(Y+1)

Code:
cost value second throw:
if there is 1 other player, the strategic player pays C/1/2
2 other player strategic player pays C/3
3 other player strategic player pays C/4
n other players strategic player pays C/(Y+1)

So the cost value Nth throw = C/(Y+1)

Profit \$ value per throw (winnings minus cost)

B/(N*(Y+1)) - C/(Y+1)
= 1/(Y+1)*(B/N-C)

Profit can also be expressed as total income/total outgoings,
eg if it costs \$9 to earn \$10, the percentage profit is 10/9 = 111.11%, or an 11.11% increase. Expressed as a fraction instead, 10/9 = 1.1111

Code:
Fractional profit value per throw (winnings/cost) = B/(N*(Y+1))/(C/(Y+1))
= B/N * 1/(Y+1) * (Y+1)/C
= B/(C*N)

This allows us to make the number of players irrelevant when determining expected fractional profit of a throw.

Expected fractional profit first throw:

1/D*(1-1/D)^(1-1)*B/C/1 + 1/D*(1-1/D)^(2-1)*B/C/2 + …..

Expected fractional profit second throw:

= 1/D*(1-1/D)^(1-1)*B/C/2 + 1/D*(1-1/D)^(2-1)*B/C/3 + …..

Expected fractional profit Nth throw:

Code:
= sum from (N = I+1) to infinity {1/D*(1-1/D)^(N-I-1)/N} * B/C

Renaming I (the previous throw) to J for clarity and using wolfram alpha to calculate the first 5 throws:

2.55843, 1.73159, 1.36843, 1.15011, 1.00012

The first twenty throws in terms of fractional profit:

Since the fifth throw is approximately 1.000, a strategy of leaving the game at 4 or 5 throws is equally effective. If you got free drinks per throw, then you'd be better off staying for 5 throws. If instead of drinks there were lots of other tables playing the same game, then you'd be better off leaving at 4 throws - or earlier if another game starts sooner.

I'll post an answer about the expected profits of the game later on, unless someone beats me to it or points out a mistake in my reasoning

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iopq
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 October 20, 2011, 06:49:14 AM

the expected fractional profit does not actually tell you the correct strategy for the game with a lower number of players
this is because the fifth throw has a positive expected fractional profit, but it's a lesser EV than not throwing for the player who is playing strategically

the expected fractional profit ONLY shows the correct strategy as the number of players approaches infinity because of the effect of players gaining shares in later rounds that devaluate your winnings
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 October 20, 2011, 07:41:23 AM

the expected fractional profit does not actually tell you the correct strategy for the game with a lower number of players
this is because the fifth throw has a positive expected fractional profit, but it's a lesser EV than not throwing for the player who is playing strategically

the expected fractional profit ONLY shows the correct strategy as the number of players approaches infinity because of the effect of players gaining shares in later rounds that devaluate your winnings

Can you show me what the mistake in the derivation is?

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 October 20, 2011, 07:48:32 AM

Also, the expected fractional profit for the 5th throw is 1.0, or winnings=costs. Any value above 1.000 is profit, any below 1.000 is a loss. At 1.000 you have neither. The expected fractional profit for throw 5 is 1.00012, so it won't have a likelyhood of producing a profit unless you play thousands of games.

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 October 20, 2011, 08:35:11 AM

the expected fractional profit does not actually tell you the correct strategy for the game with a lower number of players
this is because the fifth throw has a positive expected fractional profit, but it's a lesser EV than not throwing for the player who is playing strategically

the expected fractional profit ONLY shows the correct strategy as the number of players approaches infinity because of the effect of players gaining shares in later rounds that devaluate your winnings

Can you show me what the mistake in the derivation is?
your non-participation changes the number of players in the game in later rounds

this affects the game differently based on how many players there are to begin with, since your shares vs. their shares scale differently

if there's two other players:

you throw first 5, they throw 10 in that time
then they throw for 5 more and get 10 shares in that time
we all have 25 shares

your share is 5/25 = 1/5 of the total prize pool for \$20
you paid 1/3 of \$10 five times, so \$50/3 = \$16.(6)

your profit is \$3.33, so 1/5 on top of your original investment

if there's three other players:

you throw the first 5, they throw 15 in that time
they throw 5 more and get 15 more in that time
so total shares is 35, you get 1/7 of the prize pool or almost \$14.3

you paid 1/4 of \$10 five times, or \$12.5
so you made about 14.3% profit or 1/7 on top of our original investment

how do you reconcile these differences under your fractional profit model? When there's a different amount of players, even when the game goes the same way, the profits are DISTRIBUTED differently
this is because the more other players there are, the more each player cuts into the strategic player's shares

so the strategic player could be compensated less than 1.0 even when the fractional is above 1.0
and in smaller games I've found this to be the case, tossing 4 is better than 5 because your fifth share is worth less than the cost

and this is for the player himself, the fractional for all players is still above 1.0, but the EV for one player != fractional for all players
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 October 20, 2011, 08:42:15 AM

thought experiment:

the game is getting really stupid now, 9 strategic players play it around the clock professionally
one player plays it fairly

he has to pay \$1 on the first toss and \$10 on the 6th+ toss
but he only takes ONE share per toss away from the strategic players
and they all pay \$1 per round as well until they all stop tossing at 5 tosses because they read the organofcorti guide

it's super cheap for them and super expensive for the poor fair player
in fact, he ends up paying nearly half of the cost but gets only twice as many shares each strategic player, getting about ~20% of the reward

my point is:
strategic players and "it's due for a 10" players change the distribution of the payouts BETWEEN the players, so by playing the game you're changing the distribution of the payout between the players since you're a strategic player

so that's why my comment was your post outlines "the correct strategy as the number of players approaches infinity" at which point your strategic play doesn't disturb the payouts since you're an infinitesimal part of the field

this was already pointed out earlier in 3phase's post where he wrote:
"The number of players participating in the game is relatively stable and big enough so that your participation or your non participation does not make a significant difference (say number of players N=100,000)"

I did not make this assumption and wrote the EXACT EV for the strategic player assuming he was the only strategic player in the game
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 October 20, 2011, 08:55:08 AM

because a strategic player actually messes with the distribution of the payouts by his own presense, he can get away with "paying less, and receiving more"

he does this by participating in cheap rounds (rounds with many players in them) and not participating in expensive rounds (rounds with few players in them)

who cares about the EV for all players when some players pay less to receive more?
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 October 20, 2011, 08:59:23 AM

Understood - multiple players affect expected winnings. But does it change the fractional profit per strategic player, ie winnings/cost? You might make less per round, but I think the fractional profit will remain the same.

Let me think a bit more on it, I'll do a post on expected profit and see if I can get similar results to you.

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 October 20, 2011, 09:03:35 AM

Understood - multiple players affect expected winnings. But does it change the fractional profit per strategic player, ie winnings/cost? You might make less per round, but I think the fractional profit will remain the same.

Let me think a bit more on it, I'll do a post on expected profit and see if I can get similar results to you.

the strategic player himself increases his own profits by being in the game
this effect is eliminated at high enough number of players and at sufficiently large number of players the results would be close to the fractional for all players even for the strategic player himself
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 October 26, 2011, 03:33:45 PM

New 'How to hop' post on a simple method to calculate 'c' for a 'Slush scored' pool.

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 October 26, 2011, 09:35:37 PM

lambert_w

How is that implemented in R?

organofcorti
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 October 27, 2011, 12:03:30 AM

I didn't realise it when I posted How to hop 8, but the Lambert W function in R is actually from the GNU scientific library. Anything in GNU-sl can be used on any platform that can use C and C++ since the source is available.

A good simple reference is here.

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 October 27, 2011, 05:29:51 AM

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