Re: Just-Dice.com : Play or Invest : 1% House Edge : Banter++
I got the same as you, dooglus - except that you've included all 512 rolls, when it's only on the 9th that the first time 9 losses in a row can occur - so you can only include rolls 9 to 512. That changes the result slightly: (1 - 0.505**9) ** 504 = 0.3403612.
A couple of extras: Since this is a series of trials with the probability of 9 losses in a row occurring being 0.505**9, the number of rolls until 9 in a row occurs is a geometrically distributed variable for X = 1, 2, 3... - which makes expected values and confidence intervals easy to calculate.
Expected number of rolls until 9 losses in a row = 1/ 0.505**9 = 468.142
95% confidence interval for number of rolls until 9 losses in a row = 12 to 1726 rolls.
If the attempt is made multiple times, 50% of the time 9 losses in a row will occur before 325 rolls.
I wouldn't use this as an investment vehicle
A couple of extras: Since this is a series of trials with the probability of 9 losses in a row occurring being 0.505**9, the number of rolls until 9 in a row occurs is a geometrically distributed variable for X = 1, 2, 3... - which makes expected values and confidence intervals easy to calculate.
Expected number of rolls until 9 losses in a row = 1/ 0.505**9 = 468.142
95% confidence interval for number of rolls until 9 losses in a row = 12 to 1726 rolls.
If the attempt is made multiple times, 50% of the time 9 losses in a row will occur before 325 rolls.
I wouldn't use this as an investment vehicle
I used your formula, and I put it into a spreadsheet, and came up with this:
Quote
if p = the probability to win a game (the game percentage / 100 ) and n=number of losses in a row, then:
* The average number of loss runs until the next expected number of losses in a row: (1-p)^(1-1/p)/p
* The average number of loss runs until the next n losses in a row: ((1 - p)^-n)/p
* The average number of loss runs until the next greater than n run of losses in a row: (1 - p)^-(n+1)
* The average number of loss runs until the next expected number of losses in a row: (1-p)^(1-1/p)/p
* The average number of loss runs until the next n losses in a row: ((1 - p)^-n)/p
* The average number of loss runs until the next greater than n run of losses in a row: (1 - p)^-(n+1)
Code:
p n next next n greater n
0.877779 5 1.5266 41,772.0366 300,002.5896
0.877779 6 1.5266 341,774.6261 2,454,591.1878
0.877779 5 1.5266 41,772.0366 300,002.5896
0.877779 6 1.5266 341,774.6261 2,454,591.1878
Would you be so kind as to tell me what this means? I don't quite get "The average number of loss runs until the next ..."
The numbers are in order of your formulas. p, n, next expected, next n, next greater.
It's an excel exploit, and now all your coin belong to us!!1! Bwahahahhaa!!! etc.
All the information you need is in these three posts:
http://organofcorti.blogspot.com/2013/07/142-some-notes-about-just-dicecom.html
http://organofcorti.blogspot.com/2013/07/143-primer-on-advanced-martingale.html
http://organofcorti.blogspot.com/2013/07/144-some-notes-about-just-dicecom.html
To answer your last question, "loss runs" are series of losses in a row before a win, for 0, 1, 2, 3 ... For example, if you're tossing a coin, a head run of HHHT is 3, and a head run of T is zero. A "loss run" is just the same as a martingale sequence, and an analysis based on loss runs was how I intuitively understood martingale sequences, and how I thought everyone else would.
However I'm not a gambler, and the first quote shows me that you're interested in probabilities based on rolls, not loss runs. So the spreadsheet you made and the functions I posted are of no use to you.