esedes666
Joined: May 1, 2016
• Posts: 6
May 1st, 2016 at 2:22:01 AM permalink
Good day!
We have a casino here in the Philippines and there's a game that is simply called the Red and White ball game. There are 3 ping pong balls that you'll shoot in a funnel. At the bottom, there are boxes that the balls would fall in. Those boxes are either red, white or yellow. There are about 90 or 100 boxes, only 2 are yellow. You will either bet on red or white for a 1 to 1 payout. Yellow would be 15 is to 1 payout. The side bets are, you get either triple white or triple red, that's 7 to 1 payout. The minimum bet during weekdays is 100 pesos and during weekends the minimum bet is 200 pesos. How would you best play this game?
Last edited by: esedes666 on May 1, 2016
OnceDear
Joined: Jun 1, 2014
• Posts: 6725
May 1st, 2016 at 5:01:19 AM permalink
Quote: esedes666

Good day!
We have a casino here in the Philippines and there's a game that is simply called the Red and White ball game. There are 3 ping pong balls that you'll shoot in a funnel. At the bottom, there are boxes that the balls would fall in. Those boxes are either red, white or yellow. There are about 90 or 100 boxes, only 2 are yellow. You will either bet on red or white for a 1 to 1 payout. Yellow would be 15 is to 1 payout. The side bets are, you get either triple white or triple red, that's 7 to 1 payout. The minimum bet during weekdays is 100 pesos and during weekends the minimum bet is 200 pesos. How would you best play this game?

By walking away.

Unless you mean that the side bet is just one side bet where all three are red or all three are white and you don't have to predict which colour.

Assuming 100 boxes: 49 White: 49 Red: 2 Yellow. All the same shape and size and randomly dispersed.
Assuming that the side bets are separately 'All Red pays 7:1' and 'All White pays 7:1'

House edge betting White = 1-2*(49/100)=2%
House edge betting Red = 1-2*(49/100)=2%
House edge betting Yellow = 1-16*(2/100)=68%
House edge betting 'All Red' Side bet = 1- 8 * (49/100) * (49/100) * (49/100)=5.88%
House edge betting 'All White' Side bet = 1- 8 * (49/100) * (49/100) * (49/100)=5.88%

If there is just one side bet 'All the same colour' paying 7:1 then fill your boots: You'd have a 88% advantage on that.

1- 8 * 2 * (49/100) * (49/100) * (49/100)=-88%

Sounds a pretty naff game. Possibly not totally random and subject to physical biases, e.g yellow balls over to one edge or directly under the funnel.

Are you sure I didn't just do your maths homework question?

e&oe
\$:o)
Beware. The earth is NOT flat. Hit and run is not a winning strategy: Pressing into trends IS not a winning strategy: Progressives are not a winning strategy: Don't Buy It! .Don't even take it for free.
esedes666
Joined: May 1, 2016
• Posts: 6
May 1st, 2016 at 6:00:47 AM permalink
Thanks!
This is not my Math homework. Hahahaha!
BTW, you can bet on all side bets. So even if i place a 100 peso bet on red, white or yellow, I could still place bets on either or both triple red or triple white side bet. Actually I could place my bets on all of them.
One more condition that I forgot to mention is, if one ball lands on atleast 1 yellow, all other color loses.

I've tried this, I placed 500 peso on red, 100 peso on all red, 100 peso on all white and 100 peso on yellow. If 2 balls landed on 2 red and 1 on white, I'll win 500 pesos, lose 300 pesos. I earn 200 pesos.

Best case scenario, If all 3 balls landed in red then I'll get 500 on the 2 red balls and get 700 on all red. I'd the earn 1000, that's already deducting the 100 each on all white and yellow.

Second best case scenario, if one ball landed on yellow, I'll earn 800 pesos, already deducting 700 that I've lost from 2 red, all red and all white.

If all white wins, I'll get my 800 pesos back.

The only time I'll lose is if only 2 balls landed on white.

What do you think?!
Last edited by: esedes666 on May 1, 2016
OnceDear
Joined: Jun 1, 2014
• Posts: 6725
May 1st, 2016 at 6:04:42 AM permalink
Quote: esedes666

Thanks!
This is not my Math homework. Hahahaha!
BTW, you can bet on all side bets. So even if i place a 100 peso bet on red, white or yellow, I could still place bets on either or both triple red or triple white side bet. Actually I could place my bets on all of them.
One more condition that I forgot to mention is, if one ball lands on atleast 1 yellow, all other color loses.

For the simple red and white bets, I'd considered that you were betting on individual balls not "At least one White and no Yellows"
Beware. The earth is NOT flat. Hit and run is not a winning strategy: Pressing into trends IS not a winning strategy: Progressives are not a winning strategy: Don't Buy It! .Don't even take it for free.
esedes666
Joined: May 1, 2016
• Posts: 6
May 1st, 2016 at 6:20:56 AM permalink
Nope, not individual. This is how it looks like on the betting board.
2 red balls = 1:1
2 white balls = 1:1
All red = 7:1
All white = 7:1
Atleast 1 yellow = 15:1 (all other color loses, even if 2 balls landed on the same color)

Under the funnel, it looks like a chessboard.
Wizard
Joined: Oct 14, 2009
• Posts: 25129
May 1st, 2016 at 6:32:41 AM permalink
Resorts World in Manila has the Red and White game. From the video we can see there are 100 squares as follows:

49 Red
49 White
2 Logo

I could only confirm the 7 and 15 to one pays in the post above. There was a bet for 2 Logos but the video doesn't say what it pays.

Update: This forum says two Logos pays 250 to 1.
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
esedes666
Joined: May 1, 2016
• Posts: 6
May 1st, 2016 at 6:46:43 AM permalink
That is exactly the game I'm talking about. LOL!
Wizard
Joined: Oct 14, 2009
• Posts: 25129
May 1st, 2016 at 6:50:38 AM permalink
Here is my analysis of the game.

Under these rules, the bets for two red/white lose if the third ball lands in yellow. I assume that the bet on one yellow will win on two yellows.

Bet Pays Combinations Probability Return
Two+ red 1 76,048 0.470303 -0.059394
Two+ white 1 76,048 0.470303 -0.059394
Three red 7 18,424 0.113939 -0.088485
Three white 7 18,424 0.113939 -0.088485
One yellow 15 9,604 0.059394 -0.049697
Two yellow 250 98 0.000606 -0.847879
Last edited by: Wizard on May 1, 2016
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
OnceDear
Joined: Jun 1, 2014
• Posts: 6725
May 1st, 2016 at 6:56:54 AM permalink
Quote: esedes666

Nope, not individual. This is how it looks like on the betting board.
2 red balls = 1:1
2 white balls = 1:1
All red = 7:1
All white = 7:1
Atleast 1 yellow = 15:1 (all other color loses, even if 2 balls landed on the same color)

Under the funnel, it looks like a chessboard.

Ah....
2red and one white or 3 reds pays 1:1..... lets see....
Probability of 3 reds =(49/100)^3
Probability of 2 reds and one white ( not yellow )= 3 x (49/100)^2 x (49/100)
Therefore probability of 'at least 2 reds and no yellow = (49/100)^3 + 3 x (49/100)^2 x (49/100)=0.352947
but it only pays 1:1
So the house edge on that is 1-2x0.352947=29.41%

If I have that right. Don't walk away. Run away screaming.

hmmmm My numbers don't match Mikes...... reviewing...
Quote: Wizard

Here is my analysis of the game.

Bet Pays Combinations Probability Return
One red 1 62,475 0.386364 -0.227273
One white 1 62,475 0.386364 -0.227273
Two red 1 59,976 0.370909 -0.258182
Two white 1 59,976 0.370909 -0.258182
Three red 7 18,424 0.113939 -0.088485
Three white 7 18,424 0.113939 -0.088485
One Logo 15 9,506 0.058788 -0.059394
Two logo 250 98 0.000606 -0.847879

Ah.... Because only one ball can go in one pocket....
P 3 reds=49/100 x 48/99 x 47/98 = .1139394 not the 0.117 that I gave... But then Mike had seen the game video.

Run away!!!! Don't bet on 2 yellow. That's the ultimate sucker bet.

Pretty Dealer: Nice action.
Last edited by: OnceDear on May 1, 2016
Beware. The earth is NOT flat. Hit and run is not a winning strategy: Pressing into trends IS not a winning strategy: Progressives are not a winning strategy: Don't Buy It! .Don't even take it for free.
esedes666
Joined: May 1, 2016
• Posts: 6
May 1st, 2016 at 7:12:12 AM permalink
So since I've tried this:
I placed 500 peso on red, 100 peso on all red, 100 peso on all white and 100 peso on yellow. If 2 balls landed on 2 red and 1 on white, I'll win 500 pesos, lose 300 pesos. I earn 200 pesos.

Best case scenario, If all 3 balls landed in red then I'll get 500 on the 2 red balls and get 700 on all red. I'd the earn 1000, that's already deducting the 100 each on all white and yellow.

Second best case scenario, if one ball landed on yellow, I'll earn 800 pesos, already deducting 700 that I've lost from 2 red, all red and all white.

If all white wins, I'll get my 800 pesos back.

The only time I'll lose is if only 2 balls landed on white.

Even if i distrubuted the 800 pesos that way, does it mean I'll only have a little less than 50% probability of winning?