sleepingvery Joined: May 14, 2017
• Posts: 16
May 14th, 2017 at 6:18:34 AM permalink
for the life of me i have a simple question that i am sure must be wrong, yet i don't know where i calculate it wrong.

it is about sic bo, the double bet i.e. any two of the three dices have same number 1,1 or 2,2 etc.

i place bets on all 6 double bets (1,1 to 6,6) costing 6 units.

the probability of me not getting paid at all, is ofcourse the three dices having 3 different numbers, otherwise i would get paid on 1 of my 6 bets, which even in worst case macau pays 8:1.

so i calculate the probability of 3 different dice numbers as 5 x 4 / 36 = 0.555555556

on the other hand me getting paid on at least 1 of the 6 bets is 1-0.555555556
= 0.444444444

therefore the expectation value of my bet (costing 6 units on 6 bet areas) =
0.555555556 x (-6)
plus
0.444444444 x (+8)

= 0.222222222

which is positive! how come?

surely there cannot be positive EV bets in the casino? at least not on such a simple and visible level?

any help in my calculations would be much appreciated! thanks!
beachbumbabs Joined: May 21, 2013
• Posts: 14214
May 14th, 2017 at 6:32:05 AM permalink
There are 4 ways to get at least a pair of any particular number (from a related thread running concurrently ):

For 6's

6x6 x66 66x 666

multiply by 6, for each die face.

4 x 6 = 24 total ways of getting at least a pair.

216 total rolls @ 6x6x6 possibilities

24/216 =.1111, or 11%, or 1 in 9.

The casino pays 8:1, when true odds a win would pay 9:1.
If the House lost every hand, they wouldn't deal the game.
sleepingvery Joined: May 14, 2017
• Posts: 16
May 14th, 2017 at 6:43:13 AM permalink
Quote: beachbumbabs

There are 4 ways to get at least a pair of any particular number (from a related thread running concurrently ):

first of all, thank you very much for your help. :)

but i still don't know what went wrong in my calculations...

besides, from mathproblems . info i know that the odds of double for any number is:

"There are 6 to the power of 3=216 total permutations, so the probability of winning is 16/216 = 0.0740741"

now if i place bets on all 6 double bets, does that mean the probability of winning is (16 x 6) / 216 ?

which is still 0.444444444

which still gives the positive EV in my first post above.

if anyone can help it would be wonderful!

i am positive something is not right with my calculations yet i don't know what...
beachbumbabs Joined: May 21, 2013
• Posts: 14214
May 14th, 2017 at 6:58:43 AM permalink
Quote: sleepingvery

first of all, thank you very much for your help. :)

but i still don't know what went wrong in my calculations...

besides, from mathproblems . info i know that the odds of double for any number is:

"There are 6 to the power of 3=216 total permutations, so the probability of winning is 16/216 = 0.0740741"

now if i place bets on all 6 double bets, does that mean the probability of winning is (16 x 6) / 216 ?

which is still 0.444444444

which still gives the positive EV in my first post above.

if anyone can help it would be wonderful!

i am positive something is not right with my calculations yet i don't know what...

If you have to get EXACTLY 2 numbers, and 3 of a kind does not pay on this bet, there are only 3 solutions per number, x 6 numbers, for 18/216, or 8.33%, which is 1 in 12. Even worse than 9:1 paying 8:1, 12:1 only paying 8:1 is truly horrendous.

For some reason, you are thinking there are 16 non-winning solutions per number, when there are only 3 wins. Calculating similarly to your method, there are 33 ways to NOT get that exact pair, not 16. For all numbers, that means198/216 rolls lose, or 33/36 when the fraction is reduced, or 11/12.

Earlier, you were using 36 as your denominator, when all calculations should use 216 because of the square values; this threw off your numbers.
If the House lost every hand, they wouldn't deal the game.
sleepingvery Joined: May 14, 2017
• Posts: 16
May 14th, 2017 at 7:03:29 AM permalink
Quote: beachbumbabs

If you have to get EXACTLY 2 numbers, and 3 of a kind does not pay on this bet, there are only 3 solutions per number, x 6 numbers, for 18/216, or 8.33%, which is 1 in 12. Even worse than 9:1 paying 8:1, 12:1 only paying 8:1 is truly horrendous.

so what i understand is, a triple does NOT pay for the double?? even if it is the same number??

if i bet on a double for 1,1 and the dice outcome is 1,1,1 - i lose??
beachbumbabs Joined: May 21, 2013
• Posts: 14214
May 14th, 2017 at 7:04:32 AM permalink
I think, as far as plugging in numbers to mathproblems site, somehow the numerator is being squared before the calculation, and it should not be. 4 per number, x 6 numbers, /216 total rolls, gives you a correct answer for AT LEAST a pair on 3 dice.
If the House lost every hand, they wouldn't deal the game.
beachbumbabs Joined: May 21, 2013
• Posts: 14214
May 14th, 2017 at 7:05:38 AM permalink
Quote: sleepingvery

so what i understand is, a triple does NOT pay for the double?? even if it is the same number??

if i bet on a double for 1,1 and the dice outcome is 1,1,1 - i lose??

I will check WoO website. Stand.by. Will edit and link.

Edit: the specific instance you're trying to calculate appears to be represented as "double" on a sic Bo table. Wizard says this pays between 8:1 and 11:1 on tables, and may or may not pay the Triple as a Double win, depending on the house layout.

I was addressing the math in your specific example, and am not an expert on Sic Bo payouts for the game itself (it is weird and complex by design). I suggest you read thoroughly on the linked page, and especially the index at the bottom of that page where the Wizard explains how he calculated individual bets there. Hope.all this helps.
If the House lost every hand, they wouldn't deal the game.
sleepingvery Joined: May 14, 2017
• Posts: 16
May 14th, 2017 at 7:13:05 AM permalink
thanks very much beachbumbabs, i appreciate your help

i guess the cruz of the matter is - what is the probability of getting 3 different numbers in any one roll?

the reasoning that led me to my answer of 0.555555556 is this:

you can have any number for the first dice i.e. N1

now to have 3 different numbers the second dice N2 cannot be the same as N1, therefore there are (6-1)=5 possibilities

similarly, for the third dice N3 it can neither be the same as N1 or N2 (which be definition must also be different from each other), therefore for N3 there are (6-2)=4 possibilities

therefore the possibility of getting 3 different numbers is 6 x 5 x 4 / 216

which is 0.555555556

which still gives the positive EV in first post above, even based on lousy macau 8:1 payout

i know it cannot be this good so something must be wrong...
ChesterDog Joined: Jul 26, 2010
• Posts: 836
May 14th, 2017 at 7:13:39 AM permalink
Quote: sleepingvery

...on the other hand me getting paid on at least 1 of the 6 bets is 1-0.555555556
= 0.444444444

therefore the expectation value of my bet (costing 6 units on 6 bet areas) =
0.555555556 x (-6)
plus
0.444444444 x (+8)

= 0.222222222

which is positive! how come?...

When one of your doubles bet win, your other five doubles bets lose. So, your calculation can be amended to: E = 0.555555556 x (-6) + 0.444444444 x (+8 - 5) = -2, which you divide by 6 units to get -0.333333333. And this is the same as the Wizard's answer.
sleepingvery Joined: May 14, 2017
• Posts: 16
May 14th, 2017 at 7:19:41 AM permalink
Quote: ChesterDog

When one of your doubles bet win, your other five doubles bets lose. So, your calculation can be amended to: E = 0.555555556 x (-6) + 0.444444444 x (+8 - 5) = -2, which you divide by 6 units to get -0.333333333. And this is the same as the Wizard's answer

ar yes! ofcourse! there is my error, i now get it.

i have done this kind of error before :facepalm: