a "NEW" JAPANEE DICE GAME TO CONSIDER

AN OLD BUDDY NAMED DANO TOLD ME ABOUT A DICE GAME HE JUST SAW IN JAPAN.. APPARENTLY IT'S A DICE GAME YOU PLAY WHEN DRINKING...
> -----------------------------------------
> From: "Dan Hyslop"
> To: "Wollner Marty"
> Cc:
> Sent: Sunday April 28 2024 9:17:03AM
> Subject: JAPANEZE DRINKING DICE GAME.
>
> Marty... I'm in this Tokyo joint and you can roll dice for drink specials.DOES THE HOUSE win on this?



(Response from Marty...

> On Apr 29, 2024, at 1:57 PM, Marty Wollner <MWOLLNER@mi.rr.com> wrote:
>
> 
> YOU CAME TO THE RIGHT PLACE. SPIKER SYSTEMS LOVES THESE PROBLEMS....

THESE PROBLEMS CAN BE SOLVED BY
adding up the component wins from which
we derive the overall probability of winning/losing..
Note however, the component wins may have varying payoffs.
So the odds of an overall win / loss might not even matter.

Here are the components...
>
> #1) Trow 2 dice, win even money if two match...

> The ways table for two dice is:
>
> Point = 2 Ways = 1 1-1
> Point = 3 Ways = 2 1-2, 2-1
> Point = 4 Ways = 3 1-3, 2-2, 3-1
> Point = 5 Ways = 4 1-4, 2-3, 3-2, 4-1
> Point = 6 Ways = 5 1-5, 2-4, 3-3, 4-2, 5-1
> Point = 7 Ways = 6 1-6, 2-5, 3-4, 4-3, 5-2, 6-1
> Point = 8 Ways = 5 2-6, 3-5, 4-4, 5-3, 6-2
> Point = 9 Ways = 4 3-6, 4-5, 5-4, 6-3
> Point = 10 Ways = 3 4-6, 5-5, 6-4
> Point = 11 Ways = 2 5-6, 6-5
> Point = 12 Ways = 1 6-6
>
> You can clearly count 6 total "pairs" among the 36 possible throws,
> so, the overall winning odds for this component is 1 / 6.
>
> 2) Even number with a total of 2 wins, and let us equate winning
> prizes in order to equalize the value of the wins.
>
> These are called "point numbers. They're saying certain points can be
> used for wins, so let's select all of the even point numbers first...
>
> Point = 2 Ways = 1 1-1
>
> Point = 4 Ways = 3 1-3, 2-2, 3-1
>
> Point = 6 Ways = 5 1-5, 2-4, 3-3, 4-2, 5-1
>
> Point = 8 Ways = 5 2-6, 3-5, 4-4, 5-3, 6-2
>
> Point = 10 Ways = 3 4-6, 5-5, 6-4
>
> Point = 12 Ways = 1 6-6
>
> Now, from what's left we select an even number with a total of 2?
> This is stupid, there's only one number that fits the description,
> which is the point number 2 (which can only be thrown as a 1-1).
> So the overall odds of this win is 1 / 36.
>
> 3) Odd number with a total of 2 wins, again, let us equate winning
> prizes in order to equalize the value of the wins. We then select
an Odd numbered point...
>
> Point = 3 Ways = 2 1-2, 2-1
>
> This is two wins, same as two winning units, which is 2 / 36 overall wins.
>
> Now, FOR THE OVERALL BET, totaling winning rolls, the odds = 6 / 6 + 1/36 + 2 / 36 = 9 /36.
>
WIN OR LOSE ON THE BET?

DANO, IF THE VALUE OF THE PAYOFF IS GREATER THAN OR EQUAL TO 1/4 THE BET, THEN IT IS AN ADVANTAGE BET FOR THE SHOOTER AND ITS WORTH IT.
> QED.
>




( FROM DANO)...
Bottom line, house always wins! Thanks for that.
Sent from my iPhone

HEY, WOV.. IS MY MATH CORRECT? (Marty)

There's a similar dice game in Taiwan where you gamble for sausages from street vendors.

I'm not sure why Discflicker mentions craps or points. That just confuses things.


OK. I have a big issue with the rules. I.E. Whoever translated them doesn't know English and/or doesn't understand this promo.

Rule 1: "Dices" ? Not a good start. Whatever. Same number on each die? 6 combinations out of 36.

Rule 2: "Even number with a total of two." Two what? Dice, or a total of 2?

If it's the latter, a total of 2 as I believe Discflicker thinks, then it's 1 out of 36. But I gotta assume you're not gonna get one free AND pay half (even if you get two at once). Plus I can't think you're gonna get a free drink more often than a half price drink, so although it's 18 out of 36, 6 are the results from rule 1, so it's 12 of 36.

Rule 3: "Odd number with a total of two." Again, two what? Well, since 2 is not an odd number, gotta think it's two dice. That's 18 out of 36. And it kinda confirms that we're talking two dice for Rule 2.

THEREFORE: If you choose to throw the dice, you will never pay the regular price. And since any even number is a discount, let's focus on Rule 3.

How much booze is in the mega highball?
What's the price?
Would you otherwise drink multiple regular highballs to match the quantity of the mega?

That's too many unknowns to calculate.

Quote: DJTeddyBear

OK. I have a big issue with the rules. I.E. Whoever translated them doesn't know English and/or doesn't understand this promo.

Rule 1: "Dices" ? Not a good start. Whatever.
link to original post

"Dices" is an extremely common word in Asian cultures for people who speak English as a second language.

Quote: discflicker

AN OLD BUDDY NAMED DANO TOLD ME ABOUT A DICE GAME HE JUST SAW IN JAPAN.. APPARENTLY IT'S A DICE GAME YOU PLAY WHEN DRINKING...
> -----------------------------------------
> From: "Dan Hyslop"
> To: "Wollner Marty"
> Cc:
> Sent: Sunday April 28 2024 9:17:03AM
> Subject: JAPANEZE DRINKING DICE GAME.
>
> Marty... I'm in this Tokyo joint and you can roll dice for drink specials.DOES THE HOUSE win on this?

g]

(Response from Marty...

> On Apr 29, 2024, at 1:57 PM, Marty Wollner > wrote:
>
> 
> YOU CAME TO THE RIGHT PLACE. SPIKER SYSTEMS LOVES THESE PROBLEMS....

THESE PROBLEMS CAN BE SOLVED BY
adding up the component wins from which
we derive the overall probability of winning/losing..
Note however, the component wins may have varying payoffs.
So the odds of an overall win / loss might not even matter.

Here are the components...
>
> #1) Trow 2 dice, win even money if two match...

> The ways table for two dice is:
>
> Point = 2 Ways = 1 1-1
> Point = 3 Ways = 2 1-2, 2-1
> Point = 4 Ways = 3 1-3, 2-2, 3-1
> Point = 5 Ways = 4 1-4, 2-3, 3-2, 4-1
> Point = 6 Ways = 5 1-5, 2-4, 3-3, 4-2, 5-1
> Point = 7 Ways = 6 1-6, 2-5, 3-4, 4-3, 5-2, 6-1
> Point = 8 Ways = 5 2-6, 3-5, 4-4, 5-3, 6-2
> Point = 9 Ways = 4 3-6, 4-5, 5-4, 6-3
> Point = 10 Ways = 3 4-6, 5-5, 6-4
> Point = 11 Ways = 2 5-6, 6-5
> Point = 12 Ways = 1 6-6
>
> You can clearly count 6 total "pairs" among the 36 possible throws,
> so, the overall winning odds for this component is 1 / 6.
>
> 2) Even number with a total of 2 wins, and let us equate winning
> prizes in order to equalize the value of the wins.
>
> These are called "point numbers. They're saying certain points can be
> used for wins, so let's select all of the even point numbers first...
>
> Point = 2 Ways = 1 1-1
>
> Point = 4 Ways = 3 1-3, 2-2, 3-1
>
> Point = 6 Ways = 5 1-5, 2-4, 3-3, 4-2, 5-1
>
> Point = 8 Ways = 5 2-6, 3-5, 4-4, 5-3, 6-2
>
> Point = 10 Ways = 3 4-6, 5-5, 6-4
>
> Point = 12 Ways = 1 6-6
>
> Now, from what's left we select an even number with a total of 2?
> This is stupid, there's only one number that fits the description,
> which is the point number 2 (which can only be thrown as a 1-1).
> So the overall odds of this win is 1 / 36.
>
> 3) Odd number with a total of 2 wins, again, let us equate winning
> prizes in order to equalize the value of the wins. We then select
an Odd numbered point...
>
> Point = 3 Ways = 2 1-2, 2-1
>
> This is two wins, same as two winning units, which is 2 / 36 overall wins.
>
> Now, FOR THE OVERALL BET, totaling winning rolls, the odds = 6 / 6 + 1/36 + 2 / 36 = 9 /36.
>
WIN OR LOSE ON THE BET?

DANO, IF THE VALUE OF THE PAYOFF IS GREATER THAN OR EQUAL TO 1/4 THE BET, THEN IT IS AN ADVANTAGE BET FOR THE SHOOTER AND ITS WORTH IT.
> QED.
>


g]

( FROM DANO)...
Bottom line, house always wins! Thanks for that.
Sent from my iPhone
link to original post

This is a fascinating post! The findings from Stanford Wong and Little Joe definitely raise some interesting questions about the nature of randomness in dice rolling. The fact that they rolled significantly fewer sevens than expected—74 instead of the 83.33 average—suggests there might be more at play here than pure chance. It would be great to hear more from the Wizard on this; his insights are always invaluable.