One in 50 probability

American Roulette with 0 and 00.. Probability of any random number winning is 1-38; Pays 35-1.

I am asking, what is the probability that any random number will hit at least once within 50 spins.
------
I have a :sigh: betting system for roulette.

Total Bankroll: 450$
Play the same number, each spin, until
1. it hits once... leave
2. 50 losses in a row and lost 450$
------
Spins 1-20 bet 5$ per spin.
Spins 21-40 bet 10$ per spin.
Spins 41-50 bet 15$ per spin.

If the number hits at any time, you will have a profit.. the amount that you profit depends on which spin it hits on. I made a chart earlier, I am pretty sure that "ideally" you win on Spin 21... but I would always take Spin 1, personally.
----------------
I understand that on each and every spin.. I have a 1-38 probability of winning.
I understand that the piece of ivory does not have a memory.
I understand that the past results have zero effect on the future.
Roulette tables at casinos have the tree that shows the previous numbers that hit. These numbers have 0 effect on the future bets.

I suggest picking a number that has not hit for the entire tree. Then going with that for the 50 spins. If the tree shows the previous 20 winning numbers... and willing to bet 50 spins.... The history has zero effect on the future... So now I would like to know.. what is the probability that a number will hit at least once during 70 spins....

You want the probability it hits at least once in 50 spins (which was the original question), not 70 spins (the wrong question).

The answer is 1-(37/38)^50 = 73.64%

Good luck!

Not too difficult to calculate Joe.

The probability that you will NOT hit after 1 spin is 37/38.

So, the probability that you will NOT hit after 50 spins is (37/38)^50 (to the power of 50).

This equals 0.2635763 or roughly 1 in every 3.8 times.

So, you will have a 73.6% success rate of hitting your number within 50 spins but you will lose your complete bankroll every 3.8 goes.



edit: beaten by 3 seconds :-)

Quote: minnesotajoe

what is the probability that any random number will hit at least once within 50 spins.

1.0. There is a 1.0 probability that a random number will hit each and every spin.

" Not too difficult to calculate Joe. "

I respectfully beg to differ. Not all of us are math ***** !

Quote: buzzpaff

" Not too difficult to calculate Joe. "

I respectfully beg to differ. Not all of us are math ***** !

Math Stars?

Close , very, very, close !

removed
silly

Quote: s2dbaker

Math Stars?

Chocolate stars.

No It is MATH ***** Do not report me to the authorities !
.

If you know that the tree doesn't affect the future, why are you betting a number not on it? Rather, the probability of a number coming up in seventy spins and that of it coming up in seventy spins, given it hasn't come up in twenty, are two very different questions - the latter is simply the probability of it coming up in fifty. A number has a 15.46% chance of not coming up in seventy spins, sight unseen, and 26.36% if you already know that it hasn't come up in the first twenty. The reason for this is that you've already shut out the 41.34% of worlds where a number comes up in the first twenty spins, which are all in the multiverse where it comes up in seventy. Just put it on double zero, or your father/grandfather's birth year, or whatever, and have fun.

I can never determine what is more ridiculous, someone implementing a betting strategy on Roulette, or actually playing American Roulette when the more player friendly European Roulette is widely available.

Anyway, if your going to go with a 50 spin regiment, might as well go all 50 and hope for a multiple hit, because if you do hit your number on 50 spins, you're slightly more apt to hit the number more than once than once only. (this changes nothing in the house edge, but since all roulette "systems" are ridiculous anyway, might as well go for a little gusto)

Odds of hitting # on 50 spins of American Roulette wheel (rounded to hundreths)

0- 26.38 %
1 -35.63%
2 - 23.58%
3- 10.19%
4 - 3.23%
5 - 0.80%
6- 0.16%
7+ - 0.03%

Wouldnt gambling be so much more fun if you could buy and sell contracts on 50 spin blocks and negotiate terms to bring the house edge to 0.50%(or whatever 2 parties agree to)? Like buying a 50 spin block, minimum of 2 hits on a number for +162 .

I give anyone 20x back on their bet (including the stake) on rolling 4 dice all the same with 10 rolls to try it. A little 9 % edge for me, which I need to make it worth my while to bet 1/19 , and a fun way for someone at a bar to pick up 19-1 on a $10 or $20 play just for entertainment. Interestingly enough, if I give someone 11 rolls instead of 10, there is almost no edge at all (less that 1%), and if I give 12 rolls, it becomes an 8% edge for the player.

your numbers do not support the claim that you are more likely to hit more than once.
it looks like you are more likely to not hit than you are to hit twice.

Quote: WongBo

your numbers do not support the claim that you are more likely to hit more than once.
it looks like you are more likely to not hit than you are to hit twice.

2+3+4+etc hits is greater than 1 hit or 0 hits.

i wouldn't bet on it

Quote: WongBo

i wouldn't bet on it

You think there's something wrong with the Maths?

Given the average number of hits of one number is about 1.3, and it's a poisson distribution, I'm not surprised that 1 or 2+ hits is -roughly- the same same frequency.

Quote: WongBo

your numbers do not support the claim that you are more likely to hit more than once.
it looks like you are more likely to not hit than you are to hit twice.

Try reading the original post, only S L O W E R . What I said is, that since the premise of his attack was to try and hit once in a 50 spin cycle and quit, i said to spin all 50 since IF you hit your number, you are slightly more apt to hit 2 or more than only once. Obviously 1 and 0 together are much more likely. I just did a bi-nomial equation to figure individual hit probability, and you think I cant do simple addition?