ANSWER - Balls In A Jar !!!

[DanStokes]

Re: Balls In A Jar !!!

Ask Keith about "Balls in a Jar". He says most of us with women should know about this.

Dan

[Monster]

Re: Balls In A Jar !!!

OK, ok you guy's are a riot !!
Time for the answer ...
;D

Answer
The number of combinations to take 10 balls out of 20 is 20!/(10!)^2=184,756.

The number of combinations to take 5 balls of each color is (10!/(5!)^2)^2=63,504.

This means in 184,756 games you lose 63,504 games and win 121,252 games in average,
totaling a loss of $5,756.
In other words, you lose about $0.03116 per game in average.

Therefore, the gamble is not fair.

[pdub]

Re: ANSWER - Balls In A Jar !!!

Wow, that's some pretty scientifc stuff there. I started to follow it but the LED's on by calculator first turned to all 8's and then it briefly spelled out an obscene four-letter word and blew smoke.

Now that you've solved that one, here's another one that has puzzled folks for a long time. You've got a standard coin and are just flipping it. By some quirk of luck and nature it comes up tails 12 times in a row. What are odds it'll come up tails again on the 13th flip?

[truckpuller]

Re: ANSWER - Balls In A Jar !!!

Wow, that's some pretty scientifc stuff there. I started to follow it but the LED's on by calculator first turned to all 8's and then it briefly spelled out an obscene four-letter word and blew smoke.

Now that you've solved that one, here's another one that has puzzled folks for a long time. You've got a standard coin and are just flipping it. By some quirk of luck and nature it comes up tails 12 times in a row. What are odds it'll come up tails again on the 13th flip?

50% that it comes up tails

[pdub]

Re: ANSWER - Balls In A Jar !!!

Okay, that was too easy.

[DanStokes]

Re: ANSWER - Balls In A Jar !!!

Actually, isn't there some weight bias on any coin toward one side or the other? Seems unlikely that the US or whichever) mint would design coins to be perfectly symmetrical in regard to weight. I don't know this - I'm just wondering. Or maybe I'm overthinking this.

Dan

[Hemi Joel]

Re: ANSWER - Balls In A Jar !!!

OK, I have no idea how you figure that out, it is way over this roofers head. But since you know how, I have another one for you to figure out. This is about beating the roulette wheel.
The wheel has 38 spaces, 2 are green, 18 are red, 18 are black, The wheel is spun and a ball is spun and randomly drops into a space. This strategy involves betting on either red or black. If the ball lands in the color that you bet on, the house pays you double your bet. If it lands on green, or on the other color, you lose your bet. Let's assume that the minimum allowable bet is $5, and the maximum is $4000.
So to start out, you bet $5 on black. If you win, you bet $5 on red. If you win, you bet $5 on black, and so on. Anytime you lose, you bet on the same color again, but double the bet plus add $5. So loosing bets will go as follows:

5
15
35
75
155
315
635
1275
2555
So if the ball never goes against your color more than 10 times in a row, you will always be guaranteed a profit of $5 per spin.

So what is the probability the the ball will go against a color more than 10 times in a row and you run up against the table limit and loose? Is there a point before the table limit that you could go back to your $5 bet and have a probability of recovering your losses?
How many spins on average would it take for it to go against you 11 times? How much would you profit before it went against you?
I find this interesting, but have no way of figuring it out. Thanks, Joel

Edit: OK, I did some math. If the ball went against you 11 times in a row and you lost everything bet so far, that would be a $5065 loss. So if you win $5 per spin on average when you win, it would take 1013 winning spins to make up for each loosing streak. So can your formula tell us if it takes more than 1013 spins or less than 1013 spins on average to get to the point where you lose 11 in a row?

[pdub]

Re: ANSWER - Balls In A Jar !!!

I found a way to at least take a LONG time to lose all of your money on roulette, if you brought some to start with.

Place a bet of one (whatever your bet is) on either black or red, and one on either even or odd, and also one on one of the three columns. It's a slow burn, and you can even come out ahead (I did the only time I ever played in a real casino). It aggravates the table master something awful cause he can't get rid of you and you ain't placing or losing any high bets.

[Monster]

Re: ANSWER - Balls In A Jar !!!

How many spins on average would it take for it to go against you 11 times?

Hemi Joel, believe it or not, I had a system figured out very similar to this, where I would bet either red or black for $1.
If you win, you get $2.
If you lose, you double the original bet and keep doubling until you win and then you get all your money back plus winning the original bet.

Showboat Casino, Las Vegas, 1985, I bet on Red.
Black hit 11 times in a row and I lost all the cash I had on me and could not continue.
What a bitch !!
So much for my system ....
:P

[mlcraven]

Re: Balls In A Jar !!!

OK, ok you guy's are a riot !!
Time for the answer ...
;D

Answer
The number of combinations to take 10 balls out of 20 is 20!/(10!)^2=184,756.

The number of combinations to take 5 balls of each color is (10!/(5!)^2)^2=63,504.

This means in 184,756 games you lose 63,504 games and win 121,252 games in average,
totaling a loss of $5,756.
In other words, you lose about $0.03116 per game in average.

Therefore, the gamble is not fair.

But if I only play the very first game, seems to me that it's almost 2-1 that I will win at it. And if I do win, I stop.

Big risk taker, huh?

[joe_rocket45]

Re: ANSWER - Balls In A Jar !!!

How many spins on average would it take for it to go against you 11 times?

Hemi Joel, believe it or not, I had a system figured out very similar to this, where I would bet either red or black for $1.
If you win, you get $2.
If you lose, you double the original bet and keep doubling until you win and then you get all your money back plus winning the original bet.

Showboat Casino, Las Vegas, 1985, I bet on Red.
Black hit 11 times in a row and I lost all the cash I had on me and could not continue.
What a bitch !!
So much for my system ....
:P

lots of us have had this idea, on both roulette and on craps (craps has a similar system that should be even more favorable than roulette)...

at anyrate I am 100% convinced that the tables are rigged. If on roulette you wait until there are 3 reds or blacks ina row your odds are increased greatly, yet you will still loose. don't think the casinos do not know these games/tricks. it's their house, and they are not legit.

I came up with a way on craps to win, there are free craps games you can download offline and I played for 2 weeks before showing my brother. He couldn't believe it worked and said he would have to see it with real dice, so we got real dice and tried. it worked and I also showed my friend. He was so convinced he said we were going to the casino "tonight". and we did.
and we lost. bigtime.

cheatin' mofo's I tell ya.

[std]

Re: Balls In A Jar !!!

OK, ok you guy's are a riot !!
Time for the answer ...
;D

Answer
The number of combinations to take 10 balls out of 20 is 20!/(10!)^2=184,756.

The number of combinations to take 5 balls of each color is (10!/(5!)^2)^2=63,504.

This means in 184,756 games you lose 63,504 games and win 121,252 games in average,
totaling a loss of $5,756.
In other words, you lose about $0.03116 per game in average.

Therefore, the gamble is not fair.

I don't think that your math works in real life. Life isn't math. It's a game of chance and I think every game is it's own game. I think if you play184,756 games, you will win.

Only 1 way to prove it. And its your game Steve.Let us know how it turns out

[BigBlockMopar]

Re: ANSWER - Balls In A Jar !!!

std, that's mean. Everyone knows the homeless usually don't have internet... ;D

[joe_rocket45]

Re: Balls In A Jar !!!

[i]Answer
The number of combinations to take 10 balls out of 20 is 20!/(10!)^2=184,756.

The number of combinations to take 5 balls of each color is (10!/(5!)^2)^2=63,504.

can you elaborate on that math please?
20! / (10!) ^ 2 = 184,756

[SuperBuickGuy]

Re: ANSWER - Balls In A Jar !!!

I used to live in Reno, and would laugh at the posters on the walls saying you have a 97% chance of winning.... what they mean is for every dollar you bet, you get 97 cents back (now consider that billions are gambled every year, and you realize how they built those nice buildings)..... there's a guy (too lazy to google his name) who did the math, upon which, all casinos are built - and he provided the formula you have WM.... it's interesting stuff to me, but I'm weird that way. ;)