one die question

Quote: seven

did you use any code key to come up with the >pays (to 1)< compared to the fair odds?

no
i just plugged in a few values to see what the he (ev per 1 unit bet) could be

one could use a simple formula depending on the house edge requirements
but that should also come from the player experience of playing the game

the balance between the casino's take and the variance the players have to win over time
can make or break a game

take Roulette, Blackjack and Craps
all those popular games
evolved
some bets came and went, some stayed, some got adjusted many times
rules changed, and on and on it went


i like playing this simple horse race dice game
but the bets get in the way for me

good luck to you with your game

i love dice games (not all) and some math so this was fun for me
thank you for sharing
Sally

Quote: mustangsally

no
i just plugged in a few values to see what the he (ev per 1 unit bet) could be

one could use a simple formula depending on the house edge requirements
but that should also come from the player experience of playing the game

the balance between the casino's take and the variance the players have to win over time
can make or break a game

take Roulette, Blackjack and Craps
all those popular games
evolved
some bets came and went, some stayed, some got adjusted many times
rules changed, and on and on it went


i like playing this simple horse race dice game
but the bets get in the way for me

good luck to you with your game

i love dice games (not all) and some math so this was fun for me
thank you for sharing
Sally

ok now I understand why there were some big house edge % and some not.

>>>i like playing this simple horse race dice game
but the bets get in the way for me<<<

thanks for giving your view and feeling. very interesting for me. what you are saying is place the bets, let the horses run :)
and the winner takes all.
will think about it and ask some friends.

@All

Sally wrote in his last posting
>>>i like playing this simple horse race dice game
but the bets get in the way for me<<<

I really would appreciate opinions of other users. Should I forget about the bets after each die throw and stay with the simple version of a race from start to finish?

simple race from start to finish would actually be a click on start button and the die will roll and checkers will move accordingly until one checker will cross the finish line.

thanks

Quote: seven

thanks for giving your view and feeling. very interesting for me. what you are saying is place the bets, let the horses run :)
and the winner takes all.
will think about it and ask some friends.

i do like the in-race betting, just not every roll
it takes away from the race on the 6hit version, for me it does

every 3 rolls seems to be ok to a point and then i want the race to finish
and some races go down to 2 or 3 at square 5
that is fun and MAYbe a great place to bet too
Sally

Quote: mustangsally

i do like the in-race betting, just not every roll
it takes away from the race on the 6hit version, for me it does

every 3 rolls seems to be ok to a point and then i want the race to finish
and some races go down to 2 or 3 at square 5
that is fun and MAYbe a great place to bet too
Sally

I understand your point and thanks for that. IMO it will be difficult for me as operator to decide when to stop the game and let people make some new bets.
some one coded me a small program so I could test the feeling how it is if I do it (as a user) roll by roll or with one click let the race run to the end.
simple way is to click once on start and see the rolls of the die and see the race. there are 6 possible speed options for testing

@all

I asked this question before and PeeMcGee was so kind to answer it. now I have the feeling that I asked the question in a wrong way.

as I am looking for a jackpot idea I have to ask what are chances that 111111 will happen? 6 times one of the 6 numbers in a row?

The probability that a specific number occurs six times in a row is simply (1/6)6.
The probability that any one of the numbers occurs six times in a row is (1/6)5.

let me ask the question again but a bit different and it could well be that the answer still applies not sure though.

what are the chances that 111111 (6x1) will show from the start of the game (1st roll) to the end of the game (6th roll)without any other number will be rolled in between?
as I phrased my last first question one could understand that 111111 (6x1) could show also in the middle of the game so that after some rolls a number will show 6 times in a row and win the game.

does the answer of PeeMcGee still apply?

thanks

hi guys

please help me to get the right answer to my last question. I really would like to know if it matters that 6 times the number 1 is rolled in a row at the start of the game to the finish as seen here in the example Example

or at the start and in the game till finish? is it in both events 1/6^6

thanks in advance

Quote: seven

does the answer of PeeMcGee still apply?

thanks

i say yes

there are 6 values each die can take on one roll
so 6 rolls would make 6*6*6*6*6*6 = 46,656 sequences (where order matters)
that all have the same probability of showing
examples:
111111
222222
333333
444444
555555
666666
153562
222333
123456
654321
all the same chance of 1 in 46,656

so 111111 = 1 in 46,656
111111 and 222222 are mutually exclusive (meaning they can not happen at the same time)
so we can add the individual probabilities to get the total for the 6
111111
222222
333333
444444
555555
666666
6 in 46,656


and it must be easier when more rolls are attempted one should think
to still get a run of 6 (111111)
it could be 2111111
or 3111111
or 4111111
5111111
6111111
one can add up those probabilities too

more ways to a run of 6 with 7 rolls
(1/6)^6 + (5/(6^7)) = 11/279936
1 in 25,448.7

yes, easier

have you come up with the betting ideas other than the long shots?

Quote: mustangsally

i say yes

there are 6 values each die can take on one roll
so 6 rolls would make 6*6*6*6*6*6 = 46,656 sequences (where order matters)
that all have the same probability of showing
examples:
111111
222222
333333
444444
555555
666666
153562
222333
123456
654321
all the same chance of 1 in 46,656

Thank You very much again for trying to help. where order matters chance is 1/6^6 Gotcha

Quote: mustangsally

so 111111 = 1 in 46,656
111111 and 222222 are mutually exclusive (meaning they can not happen at the same time)
so we can add the individual probabilities to get the total for the 6
111111
222222
333333
444444
555555
666666
6 in 46,656

Gotcha :)

Quote: mustangsally

and it must be easier when more rolls are attempted one should think
to still get a run of 6 (111111)
it could be 2111111
or 3111111
or 4111111
5111111
6111111
one can add up those probabilities too

I understand what you are saying but this kind of betting proposal I cant give because I want to propose 111111 from start to finish without any other number showing up.

Quote: mustangsally

more ways to a run of 6 with 7 rolls
(1/6)^6 + (5/(6^7)) = 11/279936
1 in 25,448.7

yes, easier

maybe this is what I was looking for when I asked if chances are the same if I propose 111111 from start to finish so it must show up in 6 die throws. lets say a player starts the game and the winner arrives at the finish after 15 die throws. if we take my proposal the moment he gets to the 7th die roll he couldn't have hit the 111111 because if he would hit 111111 the game would be over.
during a game with 15 die throws the player could hit the 111111 also if he rolls for example 432464642111111 and if I understood you then chances would be much less than 1/6^6 or what did I miss here?
I thought that if a player will play the game 100 times then he has 100 times the possibility to hit the 111111 in 6 die throws. but if I would propose it different lets say he can hit 111111 from start to finish even the game will last 31 die throws then the 111111 has a chance to come more often because of more die throws during the game. sorry for my english and if didn't explain it understandable I will try again :)

Quote: mustangsally

have you come up with the betting ideas other than the long shots?

I decided to listen to your opinion and will leave out all side bets. but I will give the 111111 Jackpot proposal. the one who hits 111111 while betting on whatever number of the 6 checkers will get xxx $

Quote: seven

I decided to listen to your opinion and will leave out all side bets. but I will give the 111111 Jackpot proposal. the one who hits 111111 while betting on whatever number of the 6 checkers will get xxx $

ok then,
you could make the jackpots at any point during the race as long as a checker is at 0, meaning it has not yet moved
and it would still be for the next 6 rolls at the same prob = (1/6)^6
and it would not have to be for 111111
it could be for less too
say
1111 wins a smaller prize

i would think
the bet(s) offered does not have to be based on the end of the race

you could offer a bet, after the 1st roll or 2nd or 3rd roll, for the 1st checker to reach the 3rd square

lots of choices
have fun

Quote: mustangsally

ok then,
you could make the jackpots at any point during the race as long as a checker is at 0, meaning it has not yet moved
and it would still be for the next 6 rolls at the same prob = (1/6)^6
and it would not have to be for 111111
it could be for less too
say
1111 wins a smaller prize

i would think
the bet(s) offered does not have to be based on the end of the race

you could offer a bet, after the 1st roll or 2nd or 3rd roll, for the 1st checker to reach the 3rd square

lots of choices
have fun

Yes lots of choices :) I will start regular game with Jackpot 111111 or any other of the 6 numbers. side bets during the game could delay the game and could be boring.
agree actually to your opinion of one of your earlier postings.

still looking for a Jackpot option. lets see if this makes sense. any help is very much appreciated.

Example

Quote: seven

still looking for a Jackpot option. lets see if this makes sense. any help is very much appreciated.

Example

If it must be in that specific order—then yes, the probability of that is 1/6⁶.

Quote: PeeMcGee

If it must be in that specific order—then yes, the probability of that is 1/6⁶.

only 1 of the 6 roll sequences could be 1,2,3,4,5,6
out of possible 6^6 (46,656) sequences

now if the order does not matter
then we can find the number of ways that could happen
example:
1,2,3,4,6,5 would still result in a 6 way tie after the 1st 6 rolls
cool!

now
the 1st die could be any of the 6 values
the 2nd die any remaining 5
and the 3rd die, any of remaining 4
and so on
6*5*4*3*2*1 (this is written as 6!)

that is 720 ways to have a tie after the 1st 6 rolls
that could be a side bet
fair odds would be 63.8 to 1

FYI
after the 1st 3 rolls, the most common position of the race
would be that 3 checkers have moved one space each (5/9 probability or 120/216)

that would i thinks add some excitement to a 6-way tie bet

Quote: seven

still looking for a Jackpot option. lets see if this makes sense. any help is very much appreciated.

Example

yes it is meant in that specific order. ok so it is 1/6^6 but how about the player's bet? because if the player has a bet on checker number 4
he would have 1 in 5 chance to win the JP if this specific sequence showed up. does this change the chance for the player from 1/46656 to 1/xxxxxx ?
because if checker 4 (player's bet) will not win then he doesn't win the JP. therefore I thought this would change the odds.

thanks for helping

edit
he would have 1 in 5 chance to win the JP if this specific sequence showed up
sorry it would be 1/6 chance

Quote: mustangsally

Quote: PeeMcGee

If it must be in that specific order—then yes, the probability of that is 1/6⁶.

only 1 of the 6 roll sequences could be 1,2,3,4,5,6
out of possible 6^6 (46,656) sequences

now if the order does not matter
then we can find the number of ways that could happen
example:
1,2,3,4,6,5 would still result in a 6 way tie after the 1st 6 rolls
cool!

now
the 1st die could be any of the 6 values
the 2nd die any remaining 5
and the 3rd die, any of remaining 4
and so on
6*5*4*3*2*1 (this is written as 6!)

that is 720 ways to have a tie after the 1st 6 rolls
that could be a side bet
fair odds would be 63.8 to 1

FYI
after the 1st 3 rolls, the most common position of the race
would be that 3 checkers have moved one space each (5/9 probability or 120/216)

that would i thinks add some excitement to a 6-way tie bet

yes there would be again some nice side bets options. and I canceled (in my head) already all side bets and wanted to give Jackpot only

thanks again

so please let me try to put the question a bit different and maybe it will be more obvious if I am right or wrong.

as we already agreed chances to get the specific sequence of 123456 after the start so all checkers will be sitting on first row is 1/6^6 or 46656
if a player gets this situation he will be eligible for the Jackpot but only if he wins his original bet and lets say his original bet was checker 4. now lets say 123456 showed up after the start of the game the player has still a 1/6 chance to win. now if I am taking these facts in consideration I would think that the chance to hit the Jackpot is now 46656 * 6= 1/279936 or what did I miss here?

thanks

Hi guys

I am back with an important question. the screenshot will show you a simulation for the regular 1/6 chance for each checker.
Example

the range of
0 >>> 715,827,882 for checker 1
715,827,882 >>> 1,431,655,764 for checker 2
1,431,655,765 >>> 2,147,483,647 for checker 3
2,147,483,648 >>> 2,863,311,530 for checker 4
2,863,311,531 >>> 3,579,139,413 for checker 5
3,579,139,414 >>> 4,294,967,296 for checker 6

now I want to add the 1 in 279936 chance for the jackpot. would this kind of option make sense?
Example

I will explain for checker 1 what it takes to win Jackpot.
0 715,827,882 if this range is showing up it means that checker 1 wins with 1/6 = 16.66666% chance the regular bet
0 15,342 if this range is showing up it means that the bettor on checker 1 wins regular bet on checker 1 and Jackpot. = 1 in 279936

does this make sense?

thanks in advance ( I personally doubt that this is the right formula)

Quote: BruceZ

There is a remarkable way to solve this problem which is almost like magic. Perfect for the Wizard. The trick is to imagine that the die is rolled at points in time which are determined by a continuous Poisson points process. Clearly it makes no difference to the final answer when in time the die is rolled. We are free to imagine that these moments in time are chosen any way we want. So we choose to imagine them occurring at random points in time determined by the Poisson points process with rate 1. This means that the number of rolls that occur within any time interval depends only on the length of the time interval, independent of the number in any other interval, and it is distributed as a Poisson distribution with rate 1 so there is on average 1 roll per unit time.

Here's where the magic occurs. The rolls of the 6 numbers form 6 independent Poisson points processes each with rate 1/6. This is a basic property of Poisson points. Before we introduced the Poisson points, the players were not independent. If player 1 beat player 2, that would affect the probability that he beat the other players, so we could not simply raise the probability that he beats player 2 to the 5th power to get the probability that he beats all 5 opponents. But when we introduce the element of time in this manner with Poisson points, the probability that all of the opponents have moved 5 squares or less at a given time becomes simply the product of the probabilities for each opponent because the Poisson points for different numbers form independent Poisson points processes. Before we introduced time, if our player had moved say 4 squares in only 6 rolls, then no other player could have moved more than 2 squares. But now a player having moved 4 squares in some amount of time does not constrain in any way the number of squares the other opponents could have moved in that same amount of time because the waiting times between Poisson points are distributed as an exponential distribution which is memoryless. This allows us to compute the exact probability that player1 with the head starts wins like this:



= 223480979813228411 / 789730223053602816

I have been studying this problem as it relates to another problem.

This Poisson method shown here is remarkable indeed. I understand the concept and I will probably use it some time in the future.

However, when I enter the above integral (sum expanded, factorials evaluated) into any online calculator it fails. Maybe I'm entering the formula wrong but it appears to just be timing out. My calculus is basic and not used very often. Can someone help with a link or the actual solution with steps?

Quote: Ace2

This Poisson method shown here is remarkable indeed. I understand the concept and I will probably use it some time in the future.

However, when I enter the above integral (sum expanded, factorials evaluated) into any online calculator it fails. Maybe I'm entering the formula wrong but it appears to just be timing out. My calculus is basic and not used very often. Can someone help with a link or the actual solution with steps?

I did it in a spreadsheet, and I get something very close to that answer, although it's not a rational number by any means

First, move e^-t/6 from the right side of the sigma to the left side, as there is no k in it

Next, expand the summation to 1 + t/6 + t²/(2 x 36) + t³/(6 x 216) + t⁴/(24 x 1296) + t⁵/(120 x 7776)

Inside the square brackets, you now have (e^-t/6)⁵ x (1 + t/6 + t²/(2 x 36) + t³/(6 x 216) + t⁴/(24 x 1296) + t⁵/(120 x 7776))⁵
= e^{-(5/6) t} x (1 + t/6 + t²/(2 x 36) + t³/(6 x 216) + t⁴/(24 x 1296) + t⁵/(120 x 7776))⁵

(t / 6)⁴ can be rewritten as t⁴ / 1296; multiply this by the 1/4! term and the 1/6 term to get t⁴ / (24 x 7776)

e^-t/6 can be multiplied by e^{-(5/6) t} to get e^-t

The term of the integral is (1 + t/6 + t²/(2 x 36) + t³/(6 x 216) + t⁴/(24 x 1296) + t⁵/(120 x 7776))⁵ t⁴ / ((24 x 7776) e^t) dt;
move the 1 / (24 x 7776) to the left of the integral sign

At this point, I used Excel (well, LibreOffice Calc) on the integral expression with t = 0.5, 1.5, 2.5, and so on, and summed the terms, then divided by (24 x 7776) to get the solution. It only takes about 100 values of t to reach a stable result.

Quote: ThatDonGuy

I did it in a spreadsheet, and I get something very close to that answer, although it's not a rational number by any means

First, move e^-t/6 from the right side of the sigma to the left side, as there is no k in it

Next, expand the summation to 1 + t/6 + t²/(2 x 36) + t³/(6 x 216) + t⁴/(24 x 1296) + t⁵/(120 x 7776)

Inside the square brackets, you now have (e^-t/6)⁵ x (1 + t/6 + t²/(2 x 36) + t³/(6 x 216) + t⁴/(24 x 1296) + t⁵/(120 x 7776))⁵
= e^{-(5/6) t} x (1 + t/6 + t²/(2 x 36) + t³/(6 x 216) + t⁴/(24 x 1296) + t⁵/(120 x 7776))⁵

(t / 6)⁴ can be rewritten as t⁴ / 1296; multiply this by the 1/4! term and the 1/6 term to get t⁴ / (24 x 7776)

e^-t/6 can be multiplied by e^{-(5/6) t} to get e^-t

The term of the integral is (1 + t/6 + t²/(2 x 36) + t³/(6 x 216) + t⁴/(24 x 1296) + t⁵/(120 x 7776))⁵ t⁴ / ((24 x 7776) e^t) dt;
move the 1 / (24 x 7776) to the left of the integral sign

At this point, I used Excel (well, LibreOffice Calc) on the integral expression with t = 0.5, 1.5, 2.5, and so on, and summed the terms, then divided by (24 x 7776) to get the solution. It only takes about 100 values of t to reach a stable result.

Given the term in the integral (from 0 to infinity) is (1 + t/6 + t²/(2 x 36) + t³/(6 x 216) + t⁴/(24 x 1296) + t⁵/(120 x 7776))⁵ * t⁴ / ((24 x 7776) e^t) dt

Of course, you could just bring e^t from the denominator into the numerator as e^-t and the integral becomes a sum of integrals (from 0 to infinity) each with a term of the general form c * tⁿ * e^-t. I haven't checked my calculus text but I think that integral form is solvable, so that you can reduce the integral into an algebraic expression? I think you solve it by using integration by parts. Or, is a numerical method required to solve the integral of xⁿ * e^-x dx from 0 to infinity?

Quote: gordonm888

Of course, you could just bring e^t from the denominator into the numerator as e^-t and the integral becomes a sum of integrals (from 0 to infinity) each with a term of the general form c * tⁿ * e^-t. I haven't checked my calculus text but I think that integral form is solvable, so that you can reduce the integral into an algebraic expression? I think you solve it by using integration by parts. Or, is a numerical method required to solve the integral of xⁿ * e^-x dx from 0 to infinity?

Er, isn't that the Gamma function for (n + 1) - which means the integral evaluates to n! ?

Quote: ThatDonGuy

Quote: gordonm888

Of course, you could just bring e^t from the denominator into the numerator as e^-t and the integral becomes a sum of integrals (from 0 to infinity) each with a term of the general form c * tⁿ * e^-t. I haven't checked my calculus text but I think that integral form is solvable, so that you can reduce the integral into an algebraic expression? I think you solve it by using integration by parts. Or, is a numerical method required to solve the integral of xⁿ * e^-x dx from 0 to infinity?

Er, isn't that the Gamma function for (n + 1) - which means the integral evaluates to n! ?

Good observation

The thing that I missed is that the polynomial string with 5 terms inside the integral is taken to the 5th power. So, it becomes a polynomial string with many (3125?) terms of the form xⁿ * e^-x and the solved integral will become a sum of 3125 terms, each term involving an n! divided by a different constant coefficient. I think that this solution to the integral can be then combined and written as an algebraic expression that sums about 25 terms - each term with a factorial ranging from 4! to 29!

Sounds like a good project for an undergrad or graduate math student.

Quote: ThatDonGuy

I did it in a spreadsheet, and I get something very close to that answer, although it's not a rational number by any means

First, move e^-t/6 from the right side of the sigma to the left side, as there is no k in it

Next, expand the summation to 1 + t/6 + t²/(2 x 36) + t³/(6 x 216) + t⁴/(24 x 1296) + t⁵/(120 x 7776)

Inside the square brackets, you now have (e^-t/6)⁵ x (1 + t/6 + t²/(2 x 36) + t³/(6 x 216) + t⁴/(24 x 1296) + t⁵/(120 x 7776))⁵
= e^{-(5/6) t} x (1 + t/6 + t²/(2 x 36) + t³/(6 x 216) + t⁴/(24 x 1296) + t⁵/(120 x 7776))⁵

(t / 6)⁴ can be rewritten as t⁴ / 1296; multiply this by the 1/4! term and the 1/6 term to get t⁴ / (24 x 7776)

e^-t/6 can be multiplied by e^{-(5/6) t} to get e^-t

The term of the integral is (1 + t/6 + t²/(2 x 36) + t³/(6 x 216) + t⁴/(24 x 1296) + t⁵/(120 x 7776))⁵ t⁴ / ((24 x 7776) e^t) dt;
move the 1 / (24 x 7776) to the left of the integral sign

At this point, I used Excel (well, LibreOffice Calc) on the integral expression with t = 0.5, 1.5, 2.5, and so on, and summed the terms, then divided by (24 x 7776) to get the solution. It only takes about 100 values of t to reach a stable result.

Thanks for your reply. This problem got me thinking more about this extremely useful Poisson approach and I've now solved several other problems with it. For instance, the problem of median/mean number of rolls to hit each number of a single die at least twice had been stumping me for a while. But with this method both can be easy calculated with the formula (1 - e^(-t/6) * (t/6 +1) )^6. This reminds me of when I first learned about Markov chains...all of the sudden a bunch of problems became easy to solve. I always thought the Poisson formula was special since it basically replaces time with expected value, I just hadn’t thought of applying it this way.