Roulette 18®
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March 13th, 2017 at 4:49:50 AM
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Roulette-18® http://buff.ly/2m2tRqu (I welcome comments and suggestions).
Exciting, Space-Saving alternative to traditional Roulette!
Roulette-18® uses a 20-sided die (a Wheel or a deck of Roulette-18 Cards) and a unique table layout that does not require a roulette-sized table.
Like traditional Roulette, Players can place straight-up, split, street, corner, line or even-chance bets.
Patented and Registered Design. ©Stephen Au-Yeung, All right reserved.
Math Analysed at wizardofodds.com: http://buff.ly/2mxbb5V
Also see:
Roulette 18® Wheel-Deal™ http://buff.ly/2ncSrXF
Roulette 18® Rollette® http://buff.ly/2mMzihm
Roulette 18® CardWin® http://buff.ly/2nid6tl
Exciting, Space-Saving alternative to traditional Roulette!
Roulette-18® uses a 20-sided die (a Wheel or a deck of Roulette-18 Cards) and a unique table layout that does not require a roulette-sized table.
Like traditional Roulette, Players can place straight-up, split, street, corner, line or even-chance bets.
Patented and Registered Design. ©Stephen Au-Yeung, All right reserved.
Math Analysed at wizardofodds.com: http://buff.ly/2mxbb5V
Also see:
Roulette 18® Wheel-Deal™ http://buff.ly/2ncSrXF
Roulette 18® Rollette® http://buff.ly/2mMzihm
Roulette 18® CardWin® http://buff.ly/2nid6tl
Stephen Au-Yeung (Legend of New Table Games®) NewTableGames.com
March 13th, 2017 at 10:16:32 AM
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I think this would be good for the California market where roulette can't be played with a ball. I'm sure it has lots of game speed and protection advantages over regular roulette. However, I think player will prefer the traditional game with a wheel. Something about spinning wheels is very pleasing to people. It must explain why I like unicycles so much.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
March 13th, 2017 at 11:36:05 AM
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I'd rather play this than that stupid version of roulette we have in Oklahoma that uses cards stuck on the wheel.
The dice version would make a great home game. Hope they market it for that someday, cause I would totally buy one.
The dice version would make a great home game. Hope they market it for that someday, cause I would totally buy one.
March 13th, 2017 at 12:48:56 PM
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Another DI opportunity! Dice school to follow!
When a rock is thrown into a pack of dogs, the one that yells the loudest is the one who got hit.
March 13th, 2017 at 1:42:39 PM
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Math time!
One version of this game mentioned two icosahedrons. Assuming at least one of them landed on a 2, what is the probability both did?
One version of this game mentioned two icosahedrons. Assuming at least one of them landed on a 2, what is the probability both did?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
March 13th, 2017 at 2:11:48 PM
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1 in 400?
Edited for spoiler tag
Last edited by: TigerWu on Mar 13, 2017
March 13th, 2017 at 3:02:06 PM
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Quote: WizardMath time!
It's the same trick as with the normal dice. Let the die be N-sided and assuming there's a black die and a white die.
There were originally N^2 possible outcomes. For the specific outcome being on the black die, there is 1 way for the black die to show it and there are N ways (as we don't care at this stage) for the white die: i.e. 1xN ways. Similarly the other way round is 1xN ways. However 1 way is where both are the desired result.
So in total there are 2N-1 ways to get at least one die showing the required outcome, but only 1 of these has both dice.
The information you have could be any of the 2N-1 scenarios, however only 1 of them would have the other die also winning.
Hence the probability is 1 / (2N-1). In this case N=20, Pr=1/39.
There were originally N^2 possible outcomes. For the specific outcome being on the black die, there is 1 way for the black die to show it and there are N ways (as we don't care at this stage) for the white die: i.e. 1xN ways. Similarly the other way round is 1xN ways. However 1 way is where both are the desired result.
So in total there are 2N-1 ways to get at least one die showing the required outcome, but only 1 of these has both dice.
The information you have could be any of the 2N-1 scenarios, however only 1 of them would have the other die also winning.
Hence the probability is 1 / (2N-1). In this case N=20, Pr=1/39.
March 13th, 2017 at 4:04:57 PM
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These two products are equal:
1) (probability of two 2s) * (probability of at least one 2 given that there are two 2s)
2) (probability of at least one 2) * (probability of two 2s given that there is at least one 2)
The first product is ( 1 / 400 ) * ( 1 ).
And the second product is ( 39 / 400 ) * P.
Therefore P = 1 / 39.
1) (probability of two 2s) * (probability of at least one 2 given that there are two 2s)
2) (probability of at least one 2) * (probability of two 2s given that there is at least one 2)
The first product is ( 1 / 400 ) * ( 1 ).
And the second product is ( 39 / 400 ) * P.
Therefore P = 1 / 39.
March 14th, 2017 at 12:24:23 PM
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Quote: charliepatrickIt's the same trick as with the normal dice. Let the die be N-sided and assuming there's a black die and a white die.
There were originally N^2 possible outcomes. For the specific outcome being on the black die, there is 1 way for the black die to show it and there are N ways (as we don't care at this stage) for the white die: i.e. 1xN ways. Similarly the other way round is 1xN ways. However 1 way is where both are the desired result.
So in total there are 2N-1 ways to get at least one die showing the required outcome, but only 1 of these has both dice.
The information you have could be any of the 2N-1 scenarios, however only 1 of them would have the other die also winning.
Hence the probability is 1 / (2N-1). In this case N=20, Pr=1/39.
Correct! I owe you a pint of ale.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
March 15th, 2017 at 5:47:33 AM
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I must be missing something. The problem assumes that one die has rolled a 2 and the only issue is if the 2nd has as well. Isn't this therefore an example of conditional probability? If so, would not the answer be
20
My goal of being well informed conflicts with my goal of remaining sane.
March 15th, 2017 at 6:21:13 AM
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No - it's a difficult question! I'm guessing your answer is rather like saying what's the chance of a horse #2 winning the 2:30 race given horse #2 won the 2:00 one. That's not the question. The question is if you know that sometime in the afternoon a horse numbered 2 won a race, what are the chances both races were won by #2.Quote: TumblingBonesI must be missing something..
If you look at a similar problem using coins (where you can actually list all the outcomes, consider you know there's a H, what's the chances of both being H), and then perhaps 6-sixed dice, it may give a clue.
March 15th, 2017 at 9:36:03 AM
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Quote: TumblingBonesI must be missing something.
I think I am, too...haha... it sounds to me like the question is "what are the odds of two d20's both landing on 2," in which case shouldn't the answer be:
(1/20)(1/20) = 1/400
Guess I'm wrong, though, which explains why I'm neither a math whiz nor an advantage player!
March 15th, 2017 at 10:59:50 AM
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Nope, I'm still not seeing it. If I have a fair coin, the odds of a single toss being heads is 0.5 and the odds of 2 consecutive tosses both coming up heads is (0.5 x 0.5) = 0.25. But if we know that the 1st toss is a heads and you ask "what is the odds of the 2nd toss also being heads?" then the answer is 0.5, right? So how is this problem different?
My goal of being well informed conflicts with my goal of remaining sane.
March 15th, 2017 at 12:07:41 PM
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See my horse racing example consider two races the 2pm and 2.30pm.Quote: TumblingBones...if we know that the 1st toss is a heads and you ask "what is the odds of the 2nd toss also being heads...
The question IS NOT "horse #2 won the 2pm what are the chances #2 wins the 2:30."
The question IS "there were two races today 2pm and 2:30pm; and sometime during the afternoon #2 won a race - what are the chances #2 won both races".
If that doesn't help try the following problem...
There are a large number of tables in your restaurant and at each table there is a couple (say man and woman) having a meal. Each person has (independently) chosen red or white wine.
You are then told that, as you're running out of steak, only tables where both are drinking red will receive the meat dish. All the rest will receive fish.
(Q1) What are the chances of any specific table being served meat?
(Q2) What are the chances of any specific table being served fish?
(Q3) What are the chances of any specific table having both people drinking white wine?
(Q4) You are told that someone at the table nearest the window is drinking white wine. What is the chance that both of them are?
(Q5) Instead of red or white wine, you have 19 types of Gins and 1 type of Whisky. Only those who are both drinking Gin get meat. Use similar logic to ask the chances of both drinking Whisky.
I'm sure there's a better idea that having Whisky going well with fish. Anyone remember the James Bond film, Airplane etc. For the record I prefer a nice Gin and Tonic!!
March 15th, 2017 at 12:14:44 PM
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In this case, you don't know if it's the first coin or the second that is heads (it could also be both).Quote: TumblingBonesNope, I'm still not seeing it. If I have a fair coin, the odds of a single toss being heads is 0.5 and the odds of 2 consecutive tosses both coming up heads is (0.5 x 0.5) = 0.25. But if we know that the 1st toss is a heads and you ask "what is the odds of the 2nd toss also being heads?" then the answer is 0.5, right? So how is this problem different?
Say you flip two coins simultaneously, and a friend catches both so that you can't see the result of either. He then peeks at them and states, "At least one result is H." Now, what are the odds that both results are H? Well, let's take a look at the (all equally likely) possibilities:
1) H-H
2) H-T
3) T-H
4) T-T
Now, for results #1-3, your friend can say, "At least one result is H." But he can't make this statement if it is result #4. So, of the 3 possible results where at least one coin is heads, only 1 is H-H. Therefore, the answer is 1/3.
Extrapolating to Wizard's two 20-sided dice question, you can see that there are 20 distinct outcomes where die #1 is "2" (e.g, 2-1, 2-2, 2-3... 2-20), and there are 19 more distinct outcomes where die #2 is "2" (e.g, 1-2, 3-2, 4-2... 20-2). Note that we already counted the "2-2" result when totaling the die #1 outcomes.
Now, of these 39 outcomes where "2" is a result of at least one of the dice, only one of these results is "2-2." Thus, the odds of this happening are 1/39.
"Dealer has 'rock'... Pay 'paper!'"
March 15th, 2017 at 12:28:47 PM
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charliepatrick, Joeman:
Thanks! Now I get it. This probably gives you a good idea why Probability and Statistics was my least favorite math course.
Thanks! Now I get it. This probably gives you a good idea why Probability and Statistics was my least favorite math course.
My goal of being well informed conflicts with my goal of remaining sane.
March 20th, 2017 at 8:10:55 AM
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Quote: TigerWuI'd rather play this than that stupid version of roulette we have in Oklahoma that uses cards stuck on the wheel.
The dice version would make a great home game. Hope they market it for that someday, cause I would totally buy one.
Hi TigerWu,
Thanks for liking Roulette 18®.
I will be giving away Roulette 18® Dice for FREE at this year’s G2E-2017.
If all goes well, I will be show 6 New Table Games at G2E-2017:
1. Roulette 18®. http://buff.ly/2mlgsPp
2. Open 21® Blackjack. http://buff.ly/2nVY2Bl
3. Common 21® Blackjack (Unlimited-Players). http://buff.ly/2nWbm8H
4. Casino Hold'em®. http://buff.ly/2nW3rrU
5. Casino Hold'em® Poker-Plus (Poker-Progressive). http://buff.ly/2nVZfZ5
6. Lucky Draw Baccarat®. http://buff.ly/2nWi1ji
Last edited by: MrCasinoGames on Mar 20, 2017
Stephen Au-Yeung (Legend of New Table Games®) NewTableGames.com
March 20th, 2017 at 9:56:01 AM
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Quote: Wizard
One version of this game mentioned two icosahedrons. Assuming at least one of them landed on a 2, what is the probability both did?
Oh Mike. You b*gger. Years of therapy undone in one post.
Psalm 25:16
Turn to me and be gracious to me, for I am lonely and afflicted.
Proverbs 18:2
A fool finds no satisfaction in trying to understand, for he would rather express his own opinion.
