Hello Mathsters, I have what I think is a fairly straightforward question. Let's say I am offered a gambling game in which I have a supposed Two % edge. How many trials (instances) of the game would I need to play in order to prove that this two % supposed edge is real with a "very" high degree of confidence ?? Thanks, Brooklyn
If you had a 2% edge in a gambling game, how many trials would you need to perform in order to prove out the edge with a high level of certainty
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BigBeachBanana
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Based on your choice of confidence level, you can figure out the sample size based on the margin of error formula for a binomial proportion.
[math]\text{Margin of Error} = z_{\gamma} \times \sqrt{\dfrac{p(1-p)}{n}}[/math]
Solve for [imath]n:[/imath]
[math]n = \dfrac{z_{\gamma}^2 \times p(1-p)}{\text{margin of error}^2}[/math]
where:
- [imath]\gamma [/imath] is the confidence level
- z is the z-score corresponding to the desired level of confidence.
- p is the estimated probability
- n is the sample size
Big Beach, Thanks so much for the formula. So if I understand correctly, I need to first solve for the margin of error because I need that for the second part to solve for N trials. I have a few more questions: If we say that I want to go to a 95% confidence level then how do I find the Zscore for that? And lastly when I look at the margin of error formula above, I see Zy at the front, does that mean multiply Z times Y then do the square root operation. Thanks, AL
BigBeachBanana
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No. If your confidence interval is 95% then your margin of error is 5%. Just wanted to show you where the formula comes from but you only need the second equation.
Standard Normal Table or Excel, Online Calculator, etc...
[imath]\gamma[/imath] is a subscript to indicate the z-score corresponding with the chosen confidence level. So no you don't multiply the z with [imath]\gamma[/imath]
This online calculator might be your interest.
Hey, I was so focused on getting a response I didn't realize what you said... I'll be more specific The Flatbush and Maple St. Bklynkid !!! Long Live Brighton Beach Long Live Prospect Park... I am on the West Coast now, but the New York state of mind never goes away...
OK Big Beach, So I have the Z score which I believe is 1.65. My only hurdle now is understanding what is meant by estimated probability. My EV is assumed to be 2 %. In my thinking this means my estimated probability is 1.02 Can you clear this up for me.. Thanks again, Brooklyn
BigBeachBanana
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Should be 1.96 for a 95% confidence interval.
What's EV? The probability can't be over 1.
Going off of your original post when you say you have a 2% edge over the game. My interpretation is that you have a 51% chance of winning vs. the house is 49%. Thus, your probability estimate is 51% or 0.51
At the risk of having of a meta verse slide rule hurled at my cranium at great velocity, just to cap this thing off, I came up with the number of trials as 383.. with 95% confidence level, 1.96 z score, and .51 probability. I did refresh with the PEMDOS hack. Hopefully it didn't hack me.. Am I close ??
BigBeachBanana
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Close. 385.