# How to simulate a loaded dice in Igor Pro?

Wed, 03/10/2021 - 08:12 am

Hi. I am looking for a way to generate random states with non-equal probabilities. A simple example to illustrate this would be a loaded dice, where outcome of 1 is 2p/3, the probability of obtaining 2, 3, 4 or 5 is p each, and the probability of obtaining 6 is 3p/2. My approach so far has been to pick random numbers from a wave that would have 6x100 points with the first 66 points equal 1, next 100 equal 2, and so on, with the last 137 points equal to 6. This code shows how a loaded coin would work:

Variable Probability

Make/O/N=100/FREE ProbW=1*(x<Probability*100)+0*(x>=Probability*100)

Return ProbW[50+enoise(50)]

End

However this is computationally very expensive and takes too long to be practical, as I need to generate large sets of data. Also has low resolution, as sometimes the probabilities are at the 7th digit, and making a wave with Nx10^7 only makes things worse.

Is there a function in Igor Pro similar to the one in R that can do this faster/in one shot? This is how that code would look like:

x <- sample(1:6, size = 100, replace = TRUE, prob = myprobs)

how about

Variable Probability

Return (enoise(0.5)+0.5)<Probability

End

March 10, 2021 at 12:39 pm - Permalink

A roll of the die is floor(enoise(3) + 4), so for p(6) > 1/6, try this:

variable p6

return floor(min(6, enoise(3) + 3 + 6*p6))

end

March 10, 2021 at 11:10 pm - Permalink

for p(6) = 1/4, as per your example:

March 10, 2021 at 11:48 pm - Permalink

In your example, IVP, the probabilities of the loaded dice do not add up to 100%. Is that intentional? If the probability of rolling a 6 is 3/2*1/6 I would expect the probability of rolling a 1 to be 1/2*1/6. This is assuming that the only effect of the loaded dice is to convert a given fraction of 1s into 6s.

To make that clear, I would write the code like this:

// Rolls a loaded six-sided dice, where a roll of 1 is converted to a roll of 6 with a probability of LoadEfficiency

Variable LoadEfficiency

// Generates a random value between 0 and 6

Variable Roll=enoise(3)+3

// Converts a roll of 1 to a 6 with a probability of LoadEfficiency

if (Roll<LoadEfficiency)

Roll=6

// Rounds up the generated random number to the closest integer, producing rolls of 1, 2, 3, 4, 5 and 6

else

Roll=Ceil(Roll)

endif

// Returns the rolled value

Return Roll

end

March 11, 2021 at 03:04 am - Permalink

or do you need something more general where you have a list of events, each with different probabilities?

event 1 0.10

event 2 0.20

event 3 0.20

event 4 0.40

event 5 0.90

etc...

March 11, 2021 at 03:15 am - Permalink

This is how I would handle the more generalized case. I haven't tested the histogram. You'll want to make sure to do that before you use the code.

// Creates a not-normalised probability wave with 7 events, each with a given probability

Make/O root:MyProbabilityWave={0.01, 0.05, 0.05, 0.20, 0.30, 0.90, 0.90}

Wave MyProbabilityWave=root:MyProbabilityWave

// Creates a list of random events based on the probabilities in MyProbabilityWave

Make/O/N=100 root:ListOfRandomEvents=RandomEventGenerator(MyProbabilityWave)

Wave ListOfRandomEvents=root:ListOfRandomEvents

// Displays the probability wave and the list of random events

Edit MyProbabilityWave, ListOfRandomEvents

end

Function RandomEventGenerator(ProbabilityWave)

// Selects a random event, based on a list of probabilities for each event

Wave ProbabilityWave

// Calculates the sum off all probabilities in the wave. For a normalised wave this should be 1. This step is included just in case it is not

Variable SumOfAllProbabilities=sum(ProbabilityWave)

// Generates the random number used to select an event

Variable RandomNumber=enoise(SumOfAllProbabilities/2)+SumOfAllProbabilities/2

// Integrates the values in the probability wave. This is needed to select an event with the random number

Duplicate/O/FREE ProbabilityWave IntegratedProbabilityWave

Integrate ProbabilityWave /D=IntegratedProbabilityWave

// Counts through the events one at a time and selects the event where the combined probability have passed the random number

Variable NumberOfEvents=NumPnts(IntegratedProbabilityWave)

Variable Event=0

for (Event=0; (RandomNumber>IntegratedProbabilityWave[Event]) && (Event<NumberOfEvents); Event+=1)

endfor

// Returns the selected event

Return Event

end

March 11, 2021 at 04:02 am - Permalink

In reply to In your example, IVP, the… by olelytken

Hi Olelytken,

It was unintentional... I picked that example from the internet... Probabilities should add up to 1.

Thanks

March 11, 2021 at 06:16 am - Permalink