Re: Probability of x within n attempts - two answers
You only did the first four rows on your tree, out of eight. I did all eight and found the answer to be equivalent to the first answer, as expected. You can apply the geometric series' closed form to prove this analytically if you're interested.
Even though after 8 bets you would expect to make $1 with 99.411% probability, you have might lose $1+2+4+8+16+32+64+128=$255 with 0.589% chance. You would then have an expected profit of $1*0.99411=$0.99411, and an expected loss of $255*0.00589 = $1.50195. Overall, you'd expect to lose $1.50 - $0.99 = $0.51 with this betting technique.
If you had infinite money and a casino willing to take arbitrarily large bets, you could make money with this scheme eventually with arbitrarily high probability (keep doubling indefinitely). Too bad casinos don't extend infinitely large lines of credit or I'd be rich tomorrow....
The time you enjoy wasting is not wasted time. Bertrand Russell
Re: Probability of x within n attempts - two answers
I would like to give my contribution to this thread with a simulation model. I am not familiar with games at all. Therefore I am not absolutely sure if the model replicates the roulette game precisely. I leave it up to you to judge and let me know.
Re: Probability of x within n attempts - two answers
First, wow - thanks for all the feed back.
Yes I did indeed mean to type:
P(Red) = 18/38
P(Not RED) = 20/38
and was obviously having a stupid moment. A big thanks to assis and NickThissen for spotting and working that one out
NickThissen you hit the nail on the head which is the doubling (in a very exponential style) soon crosses the last hope of reasonableness event-horizon and takes us into a world of silly loss. The other problem is that the earnings if played with a sanity check to stop inflation to stupidity is too slim even at the sixth or so iteration to be worth anything (from my numbers I can see that at the sixth attempt if you win or give up and play for long enough (forever) you are fairly sure not too loose much or to have a slight gain) and again as jemidiah points out to be utterly sure you need almost infinite credit.
Huge thanks to jemidiah for completing the expectation calculations - I just could not remember how to do that (it has been a very long time).
Big props to assis for building a model to test things out - I will have to run that later.
I've attached my working (which I lazily did with an Open Office spread sheet) which gave me the two answers. jemidiah's working conforms to my first answer (where I multiplied using powers) but I still have a nagging voice that says the loss expectation is bigger care of the second set of figures... but as I have no reasonable reason I'm going to go with the group consensus which is plenty good enough for me.
The original question of which method moot point in light of the larger answer because between us we have shown that whatever the most accurate/correct calculation might be even the more "generous" one gave an overall loss (which is what I suspected).
This has pretty much satisfied my curiosity and silenced a few silly notions I could not rationally address that had gotten into my head (I hate it when that happens).
Which only leaves me to enquire if anyone else has heard of other "systems" of this nature? (The subject has got me interested now and I can see me wanting to examine the numbers on more such ideas).
Add up all these chances and you'll get about 0.9941 = 1 - (20/38)^8, which is your first answer accounting for the typo. You only did the first 4 rows corresponding to 4 spins instead of 8, which is why you came up with a lower chance of winning than your first (and correct) calculation showed. I'm not really sure why you stopped at 4 rows... I guess it was just a brain fart
Of course, 1-(20/38)^n is always at least slightly less than 1, and with that, you can use an expected loss analysis like I did above to prove that you'll never expect to make money with this system.
Last edited by jemidiah; Jul 26th, 2008 at 10:46 PM.
The time you enjoy wasting is not wasted time. Bertrand Russell
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Re: Probability of x within n attempts - two answers
