WizardofEngland
WizardofEngland
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November 11th, 2010 at 6:51:30 AM permalink
A recent carnival was offering a tic tac toe style game. For £1 a go you throw three incredibly bouncy balls towards a large wooden box with 9 pockets in the bottom. The front face had clear perspex on the front so you can see where the balls landed.

I had a go and to be able to aim is nigh on impossible, so I figured the chances are purely random. The prize was either a massive cuddly toy, or £20. I was obviously going for the £20 ;-)

I spent £5 before giving up.

I have now attempted to figure the odds, assuming random placement.

I believe there are three distinct patterns.

1) The first ball lands in the middle, the second ball can land anywhere, the third has one slot to hit. So i figure 1/9 * 8/8 * 1/7 = 1/63
2) The first ball lands on a corner, the second ball has 4 possible pockets, the third again has one. So 1/9 * 4/8 * 1/7 = 1/126
3) The ball lands on the outer middle, but has exactly the same odds as above.

I think I now combine these odds, and get a probability of 1 in 31.5

Is this right?
http://wizardofvegas.com/forum/off-topic/general/10042-woes-black-sheep-game-ii/#post151727
cardshark
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November 11th, 2010 at 7:02:05 AM permalink
How many different ways can 3 the balls land? There are 9 slots, so we get 9*8*7 = 504
How many different ways cause a win? You have to count them by hand:

3 of this style
x.x.x
o.o.o
o.o.o

3 of this style

x.o.o
x.o.o
x.o.o

and 2 of this style

x.o.o
o.x.o
o.o.x

So 8 ways, and the ball can form one of these 8 patterns in 6 ways (123,132,213,231,312,321) 8*6 = 48

Answer 48/504 = 1 in 10.5

Of course, this assumes there isn't some sticky tape in a hole, or some other dishonest trick!
WizardofEngland
WizardofEngland
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November 11th, 2010 at 7:06:00 AM permalink
Pretty sure I was able to get a ball in every hole, but I am sure it was £20, so it that respect it was a good bet, but I watched for ages before having a go and no-one won. Hmmm
http://wizardofvegas.com/forum/off-topic/general/10042-woes-black-sheep-game-ii/#post151727
DJTeddyBear
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November 11th, 2010 at 7:11:01 AM permalink
Quote: WizardofEngland

2) The first ball lands on a corner, the second ball has 4 possible pockets, the third again has one. So 1/9 * 4/8 * 1/7 = 1/126

If the first ball lands in a corner, there are SIX good pockets for the second ball.


But I agree with CardShark: The overall odds really depend on there being equal odds for any available pocket on any throw. Carnival games, almost by definition, can be counted on to NOT be played fairly.
I invented a few casino games. Info: http://www.DaveMillerGaming.com/ ————————————————————————————————————— Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
WizardofEngland
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November 11th, 2010 at 7:16:35 AM permalink
Quote: DJTeddyBear

If the first ball lands in a corner, there are SIX good pockets for the second ball.


But I agree with CardShark: The overall odds really depend on there being equal odds for any available pocket on any throw. Carnival games, almost by definition, can be counted on to NOT be played fairly.



Your very right, don't know how I missed that
http://wizardofvegas.com/forum/off-topic/general/10042-woes-black-sheep-game-ii/#post151727
FleaStiff
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November 11th, 2010 at 7:27:04 AM permalink
Usually with tossing balls into a container its rigged to bounce out. Often if the game uses smaller balls the Carney will place one of them inside the container and then toss another ball inside to prove that it will stay there, if tossed. But its that first ball that is the one likely to bounce out.

If they use larger balls then its often that from the angle you can see all containers are the same but in actuality the valuable container will have a bottom that encourages a bounce.
Wizard
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Wizard
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November 11th, 2010 at 8:04:01 AM permalink
Here is how I would solve it. There are 8 ways to win (three rows, three columns, and two diagonals). There are combin(9,3)=84 possible combinations of 3 squares out of 9. 8/84 = 1 in 10.5.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
SOOPOO
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November 11th, 2010 at 8:11:38 AM permalink
So since wiz and card shark figured out if it is fair then the carnie will be losing money hand over fist. Most likely is the center hole is then least likely to accept a ball and the 4 non corners the most likely. This will skew the odds in the carnies favor significantly. Good for you that you only spent 5 pounds for the lesson.
miplet
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November 11th, 2010 at 8:19:29 AM permalink
Quote: SOOPOO

So since wiz and card shark figured out if it is fair then the carnie will be losing money hand over fist. Most likely is the center hole is then least likely to accept a ball and the 4 non corners the most likely. This will skew the odds in the carnies favor significantly. Good for you that you only spent 5 pounds for the lesson.


I wonder how skewed they differant holes have to be? Excel here I come.
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WizardofEngland
WizardofEngland
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November 11th, 2010 at 8:20:23 AM permalink
Quote: SOOPOO

So since wiz and card shark figured out if it is fair then the carnie will be losing money hand over fist. Most likely is the center hole is then least likely to accept a ball and the 4 non corners the most likely. This will skew the odds in the carnies favor significantly. Good for you that you only spent 5 pounds for the lesson.



You have to be in the spirit for a carnival though, and then there is the macho factor, trying to look good by winning, but ending up a loser.
http://wizardofvegas.com/forum/off-topic/general/10042-woes-black-sheep-game-ii/#post151727
MathExtremist
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November 11th, 2010 at 8:34:55 AM permalink
Quote: WizardofEngland

A recent carnival was offering a tic tac toe style game. For £1 a go you throw three incredibly bouncy balls towards a large wooden box with 9 pockets in the bottom. The front face had clear perspex on the front so you can see where the balls landed.

I had a go and to be able to aim is nigh on impossible, so I figured the chances are purely random.



I'd bet very strongly against equally-likely spots. If there is 100% chance of the first ball landing in one of the 9 spots, that means there's some sort of guard rail around it. The chances of bouncing off the guard rail and into an edge spot would seem to be different than into the center. I'm not sure how much different, but to me it's very unlikely that each spot is equiprobable.

However, under the premise of equiprobable pockets:

Quote:

1) The first ball lands in the middle, the second ball can land anywhere, the third has one slot to hit. So i figure 1/9 * 8/8 * 1/7 = 1/63
2) The first ball lands on a corner, the second ball has 4 possible pockets, the third again has one. So 1/9 * 4/8 * 1/7 = 1/126
3) The ball lands on the outer middle, but has exactly the same odds as above.



1 is right.
2 - there are 4 corners, so 4/9. The 2nd ball has 6 possible pockets: 6/8. The 3rd ball has 1: 1/7. = 3/63
3 - there are 4 middles, so 4/9. The 2nd ball has 4 possible pockets: 4/8. The 3rd ball has 1: 1/7. = 2/63
Total is 6/63 which has already been reported. The several other ways of arriving at the same answer are instructive as to solving such problems.
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
Doc
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November 11th, 2010 at 9:06:02 AM permalink
Quote: WizardofEngland

... For £1 a go you throw three incredibly bouncy balls towards a large wooden box ....

Maybe I didn't understand the game completely. This portion of the description does not convince me that 100% of the "incredibly bouncy" balls thrown "towards" the box actually end up in any of the nine holes. They could as well be somewhere on the floor behind or in front of the box or somewhere else when they stop bouncing. If that were the case, then all of the calculated probabilities that have been posted are inflated, and the game is partially one of skill.

So, WoEngland, do all of the balls necessarily end up in one of the holes or not?
Wizard
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November 11th, 2010 at 9:57:02 AM permalink
Quote: Doc

So, WoEngland, do all of the balls necessarily end up in one of the holes or not?



I can't speak for England, but I've seen that game here, and there are walls around the board at an angle so it would be hard to miss the board completely. It is pretty much a game of luck, as it would be difficult achieve an accuracy to get the ball in a specific square. Maybe a good dice setter could it. ;-).
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
DJTeddyBear
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November 11th, 2010 at 10:09:03 AM permalink
Quote: Wizard

Maybe a good dice setter could do it. ;-).

Uh, oh.

Do I hear the sound of a can of worms being opened? :B
I invented a few casino games. Info: http://www.DaveMillerGaming.com/ ————————————————————————————————————— Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
Doc
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November 11th, 2010 at 10:31:53 AM permalink
Quote: DJTeddyBear

Uh, oh.

Do I hear the sound of a can of worms being opened? :B

Aha! Where are our bouncy-ball influencers?

And thanks for the info, WoV. Long ago, I saw some toss-the-ball-in-the-hole games (or maybe it was toss the ring around the neck of a bottle) in which most of the tosses were complete misses.

Last night I happened upon an interesting Wiki article on the "illusion of control."
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