[resolved] Probability of x within n attempts - two answers
I've been looking at a "system" someone suggested for a particular roulette table and so wanted to test their claims a little more closely. The "system" says that if you bet red with $1 and then double until you win you should increase the pot by $1 a time and always be better off.
The table in questions has 18 reds, 18 blacks and two greens (0 and 00).
That gives us 38 possible outcomes.
(opps)
P(Red) = 18/32
P(Not RED) = 20/32
(/opps)
P(Red) = 18/38
P(Not RED) = 20/38
So far so good.
What I am looking for is
P(Red in 8 or less tries)
Now if I take the probability of NOT RED 8 times I have (20/32)^8 and could assert that 1 - (20/32)^8 = P(Red With 8 tries) = 0.9941
which makes the system look good.
However, something bugged me about that it seemed to be TGTBT (too good to be true) so I ran a different set of numbers. In my head I read off all the ways a win could happen probability tree style.
Answer one gives a positive expectation over a large number of rounds of the system but answer two gives a negative expectation over a large number of rounds.
My rusty memory of probability says that the second calculations are correct and the house still wins but I am not sufficiently full of myself these days to state this with 100% certainty.
Would someone else check my working and comment?
Last edited by Matt_T_hat; Jul 26th, 2008 at 07:28 AM.
Reason: silly error
Re: Probability of x within n attempts - two answers
Originally Posted by Matt_T_hat
The table in questions has 18 reds, 18 blacks and two greens (0 and 00).
That gives us 38 possible outcomes.
P(Red) = 18/32
P(Not RED) = 20/32
I would like you to clear what you mean with "two greens (0 and 00)". If you say there are 38 possible outcomes, how is it possible that you neglect the two greens in the calculus that follows? How did it come (the number 32)? There must be an easy explanation, I am sure.
Re: Probability of x within n attempts - two answers
Originally Posted by assis
I would like you to clear what you mean with "two greens (0 and 00)". If you say there are 38 possible outcomes, how is it possible that you neglect the two greens in the calculus that follows? How did it come (the number 32)? There must be an easy explanation, I am sure.
He probably meant 18/38 and 20/38 (18 + 20 = 38 )
This is ofcourse about a game of roulette? There are 18 red, 18 black and 2 green possibilities. That indeed gives a possibility of 18/38 red and 20/38 non-red (black or green).
I haven't read your post in detail (and I'm not that good at chances and the like), but is this about the 'famous' system where you just bet on one color every time, and if you lose you double up?
If so, I have followed a long thread on another forum where they ran simulations etc, and the conclusion was that:
- If you stop after your last win, you always win back your first bet.
For example, if you start with a $20 bet, and you lose 4 times in a row:
You lose:
$20 - $40 - $80 - $160
Then you win and you get paid out double, so you win 160*2 - (160 + 80 + 40 + 20) = $20.
Alright, so far so good. However, there are a few major buts here!
First of all, look how quickly the numbers grow:
20, 40, 80, 160, 320, 640, 1280, 2560, 5120, 10240...
Well, no problem you say, since you will always win 20 at the end?
Yeah, but what if you keep losing until you are broke? What if you don't have more than 10240$ and lose again? You just lost everything!
Ofcourse the chances of losing 10 times in a row are very minute, but still... 4 or 5 times in a row is not uncommon and then you still lose 640$!
To be honest, risking 10240$ just to win 20$ is a bit insane, lol!
Last edited by NickThissen; Jul 25th, 2008 at 01:14 PM.
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
[resolved] Probability of x within n attempts - two answers
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Originally Posted by Matt_T_hat
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Re: Probability of x within n attempts - two answers
