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re: Tell me if my thoughts on roulette are incorrect
Posted on 6/13/18 at 12:53 pm to donRANDOMnumbers
Posted on 6/13/18 at 12:53 pm to donRANDOMnumbers
quote:
There is just as much an opportunity as 1 hitting 100 times in a row as your number hitting once.
Best way to gamble is to be consistent and know the rules.
What?
The probability of hitting 1 - 100x in a row is basically 0%.
The probability of hitting your number at least once in 100 throws is 93%. So, no there is not the same opportunity between those two scenarios.
However, if a 1 had been rolled 99 times in a row, the probability of hitting a 1 again or hitting your number on the 100th throw are the same or ~2.6%.
0
Posted on 6/13/18 at 1:03 pm to donRANDOMnumbers
quote:
There is just as much an opportunity as 1 hitting 100 times in a row as your number hitting once.
Jesus Christ. This is also not true
These are two separate math problems being conflated:
1. The probability of an event occurring given x number of potential outcomes
2. The probability of an event occurring given x number of potential outcomes and n number of attempts.
The probability of OP (at least) winning his money back betting one number for 36 spins has already been stated: 61.7%
The probability of winning on an individual spin has also been stated: 2.6%
The probability of winning 100 times in a row is so small that I can’t calculate it on my iPhone without getting an error.
ETA: Excel says it's 1.0511E-156 %
This post was edited on 6/13/18 at 1:09 pm
Posted on 6/13/18 at 1:03 pm to SoFunnyItsNot
You have a 1/38 chance every spin regardless of the previous spin. you have a ~2.6% to hit your number which only pays 35/1 not 38/1 so even if you hit once every 38 spins, you still lost $15.
Posted on 6/13/18 at 1:14 pm to iamAG
Let's get all Monte Carlo up in this bitch.
import random
def spin():
. outcome = [True] + 37 * [False]
. return random.choice(outcome)
def try36():
. money = 180
. for i in range(36):
.. money = money - 5
.. if spin():
... money = money + 35*5 + 5
. return money >= 180
wins = 0
for trials in range(100000):
. if try36():
.. wins = wins + 1
wins / 100000
'spin' just picks True one out of 38 times - that tells you if you just 'won' your spin of the wheel.
'try36' starts with $180, bets $5 36 times, and tells you if you came out with at least $180.
The for loop runs 'try36' 100,000 times, and tells you how many times your strategy would have at least broken even.
'wins / 100000' just converts that to the fraction.
It comes out fairly close to 61.8 % -- that is, if you start with $180, bet $5 36 times, then quit, you have a 61.8 % chance of at least winning your money back.
import random
def spin():
. outcome = [True] + 37 * [False]
. return random.choice(outcome)
def try36():
. money = 180
. for i in range(36):
.. money = money - 5
.. if spin():
... money = money + 35*5 + 5
. return money >= 180
wins = 0
for trials in range(100000):
. if try36():
.. wins = wins + 1
wins / 100000
'spin' just picks True one out of 38 times - that tells you if you just 'won' your spin of the wheel.
'try36' starts with $180, bets $5 36 times, and tells you if you came out with at least $180.
The for loop runs 'try36' 100,000 times, and tells you how many times your strategy would have at least broken even.
'wins / 100000' just converts that to the fraction.
It comes out fairly close to 61.8 % -- that is, if you start with $180, bet $5 36 times, then quit, you have a 61.8 % chance of at least winning your money back.
This post was edited on 6/13/18 at 1:17 pm
Posted on 6/13/18 at 1:14 pm to BaylorTiger
quote:
So if you flip a coin a trillion billion times, the split should be very close to 50/50.
Yeah. If you are allowed to flip a coin 10 times 4 different times.. and the results are as followed.
flip 1: 4 heads
flip 2: 3 heads
flip 3: 7 heads
flip 4: 5 heads
Then 48% of the time it landed on tails, but if you do it again, you could have completely different results.
flip 1: 2 heads
flip 2: 4 heads
flip 3: 3 heads
flip 4: 7 heads
This would be heads 40 percent of the time.
I think the most likely outcome would be anywhere between 40%-60%.
But either way, it really come down to each flip. There is no special power telling that coin "I flipped heads 10 times in a row, it's time for me to start hitting on tales".
This post was edited on 6/13/18 at 1:18 pm
Posted on 6/13/18 at 1:19 pm to iamAG
quote:
iamAG
Do you like to play cards by any chance? I'd be happy to get a game set up if you are willing to come play.
Posted on 6/13/18 at 1:23 pm to iamAG
A 1 in 5 chance doesn’t mean you are guaranteed to hit 1 in 5. Each roll is an independent roll. Just because you’ve rolled 4 times and haven’t hit, doesn’t mean that 5th will be a win. It means it still has a 1 in 5 chance.
Or simpler explanation would be a coin toss. You could flip it 100 times and it probably won’t be 50 heads and 50 tails just because it’s a 1 in 2 chance.
Or simpler explanation would be a coin toss. You could flip it 100 times and it probably won’t be 50 heads and 50 tails just because it’s a 1 in 2 chance.
Posted on 6/13/18 at 1:49 pm to Sao
quote:
Why not just play a single color with hedges on 0/00
Only way to play roulette
Posted on 6/13/18 at 1:54 pm to iamAG
Let's say you are going to bet on "2" every time. In your hypothetical, playing $5 a spin, and bringing $180, you have enough money for 36 spins.
On any given spin, there is a 37/38 chance that you lose. That's a 97.368% chance you lose $5 on your first spin. No problem! Now, you spin again. The chances of losing on two spins in a row is (37/38) (the first spin losing) multiplied by (37/38) the second spin losing. That's a 94.806% chance you lose both of the first two spins. It's (37/38) raised to the number n where n is the number of consecutive spins.
Your hypothetical is better when you're bringing enough money for 35 spins, because the payout for any one number is 35:1. So if we calculate the probability of losing 35 times in a row, it's (37/38) raised to the 35th power. That works out to a 39.32% chance that if you bet the same number 35 times in a row, it won't hit. Conversely, based on probability alone, there's a 60.68% chance you'll break even on 35 spins.
If we calculate the odds of *making* money, you would need to hit your number within 34 spins, so that you'd make $5. That would be (37/38) raised to the 34th. That's a 59.62% chance you make $5 when you're risking $170. The risk is just too high for the payout.
If you're wondering, the "50/50" is at 26 spins. If you put all bets on the same number, spin after spin, you have a 50/50 chance of "breaking even" after 26 spins. But at that point, you've risked $130 ($5/spin at 26 spins) just hoping to not lose. It's a lot of heartburn for the chance to win a very small amount.
On any given spin, there is a 37/38 chance that you lose. That's a 97.368% chance you lose $5 on your first spin. No problem! Now, you spin again. The chances of losing on two spins in a row is (37/38) (the first spin losing) multiplied by (37/38) the second spin losing. That's a 94.806% chance you lose both of the first two spins. It's (37/38) raised to the number n where n is the number of consecutive spins.
Your hypothetical is better when you're bringing enough money for 35 spins, because the payout for any one number is 35:1. So if we calculate the probability of losing 35 times in a row, it's (37/38) raised to the 35th power. That works out to a 39.32% chance that if you bet the same number 35 times in a row, it won't hit. Conversely, based on probability alone, there's a 60.68% chance you'll break even on 35 spins.
If we calculate the odds of *making* money, you would need to hit your number within 34 spins, so that you'd make $5. That would be (37/38) raised to the 34th. That's a 59.62% chance you make $5 when you're risking $170. The risk is just too high for the payout.
If you're wondering, the "50/50" is at 26 spins. If you put all bets on the same number, spin after spin, you have a 50/50 chance of "breaking even" after 26 spins. But at that point, you've risked $130 ($5/spin at 26 spins) just hoping to not lose. It's a lot of heartburn for the chance to win a very small amount.
Posted on 6/13/18 at 1:56 pm to TH03
Look, do what I did. Took my daughter out to eat, no pics, after she got off work. I was playing golf in the area and called her at work to set it up. Met her at her apartment and showered and off we went. It was at the casino so after our dinner we strolled over to the casino. She was only 21 at the time and never ever gambled at a casino. So we strolled around the tables, slots, craps etc., and gave her my limited knowledge of the games. We get to the roulette table and she asks how does this work? I say it's simple, you just put money on a number, color, first, second or third twelves blah, blah, blah. I said you wanna play and she says yea sure. So I throw 40 on the table and get forty chips of a color at dollar each. I tell her she has to bet the minimum at least or bet it all at one time. So she randomly puts chips on different numbers and they roll the ball. She hits. We high five, laugh, enjoy the moment and they pay her 35.00 smackers. She says do we quit now, I said yea if you want but I threw 40.00 win or lose. So she does it again with different numbers. She hits. Woo hoo, high fiving yea we got this. I said that's good, but this doesn't often only betting the minimum. Loses the next spin, wins the next after the losing spin. Cash out a little later, get my forty and she keeps the rest. 50.00 dollars or more. And leave. Now, do understand the probability of that?
Posted on 6/13/18 at 1:59 pm to iamAG
You're completely wrong. Past results don't affect future outcomes in a game like roulette. Every single spin has a 1/38 chance of landing on a specific number.
There is a weird thing with probability that sticking with 1 number every time will give you a slight advantage over changing but that's it; this is why casinos have that board that shows the last ~20 spins: they want you to constantly move your bet depending on what's happening.
Remember, everything in a casino is designed to take your money.
There is a weird thing with probability that sticking with 1 number every time will give you a slight advantage over changing but that's it; this is why casinos have that board that shows the last ~20 spins: they want you to constantly move your bet depending on what's happening.
Remember, everything in a casino is designed to take your money.
Posted on 6/13/18 at 2:02 pm to iamAG
You’re math is fine. You win 36x multiplier 1 of 38 times so if you’re betting a dollar, you’re down 2 net dollars for 38 times played which means every hand you’re losing $2/38 so about 5 cents a turn
Posted on 6/13/18 at 2:14 pm to Big_Slim
quote:
You’re math is fine. You win 36x multiplier 1 of 38 times so if you’re betting a dollar, you’re down 2 net dollars for 38 times played which means every hand you’re losing $2/38 so about 5 cents a turn
No, you have a 63.7% chance of winning (at least) 1 of 38 times. The math is not fine.
BUT you did hit on a point I haven’t seen in this thread - the OP’s math wasn’t calculating his odds to break even. It was actually calculating the house edge (which is about 5%).
Posted on 6/13/18 at 2:17 pm to Big_Slim
OK now that we're all in agreement about roulette, if Monty Hall opens a door with a goat behind it and asks if you want to change your pick, you say "yes".
Posted on 6/13/18 at 3:01 pm to iamAG
It doesn't matter how many hands you play. You have the same 1 in 38 chances on each spin.
Posted on 6/13/18 at 3:34 pm to iamAG
The payments for picking one number are not equal to the odds of hitting.
Also, each instance you have the same odds. Its not like if you lose 90 times, your odds get better that you will win on the next one. You can lose every bet for a year.
Also, each instance you have the same odds. Its not like if you lose 90 times, your odds get better that you will win on the next one. You can lose every bet for a year.
Posted on 6/13/18 at 3:41 pm to iamAG
Always switch doors.
Posted on 6/13/18 at 3:45 pm to LSU999
quote:
Although not the smartest it works a good bit of the time, betting 1st and 2nd 12. You will only clear half of what you bet. You end up covering 24 numbers with 14 (including greens) to bust you. Again, not the smartest play, but makes for fun entertainment. I've done good a few times and done bad a few times. Entertainment is the idea.
EX. If you bet $5 on two sets of 12 and win, you get $15 but lose $5. So in essence you are betting $10 to clear $5.
used to do this all the time.... we would always team up and do it. had some pretty decent hauls like this. always wondered if i was the only dumb frick that tried it this way
Posted on 6/13/18 at 3:54 pm to iamAG
The odds of your number hitting the first spin is the same as the last spin.
Posted on 6/13/18 at 4:19 pm to iamAG
quote:
With each spin that does not hit the number you choose, the “probability” that the number you selected hits will go up. This is what probability is.
No, no it is not. the probability of A number coming up is exactly the same spin to spin.
They dont take one out of play.
Somewhere out there is a quantitative methods professor jumping off a bridge.
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