one die question

Quote: seven

Hi all

please remember that I am no a math or formula expert and therfore I am back with a question. thanks

please look here Example
this game is played with one die and after each throw the checker with the thrown number is moved one panel ahead. here the first throw was the number 1 and the checker was moved one panel ahead. the winner is the first who will end in the finish panel.

at start the chances to win is 5:1 and winner will end with 6

if you see this starting pic. what are now the chances for 1 or the others to cross the finish line first? actually chances will change after every move. what formula do I need to use to get always the right odds after each move till finish?

and what is the chance that until the end of the game the winner ended in finish and one of the other 5 checkers didn't move at all and is still sitting on the starting panel? that means that the number of the die was not thrown even once during the game.

General Solution

Let,
x = the number of spaces needed to win for the ‘checker’ that you want to win
p,q,r,s,t = the number of spaces needed to win for each of the other checkers

Then, the probability for the checker you want to win is:

Quote: PeeMcGee

Quote: seven

Hi all

please remember that I am no a math or formula expert and therfore I am back with a question. thanks

please look here Example
this game is played with one die and after each throw the checker with the thrown number is moved one panel ahead. here the first throw was the number 1 and the checker was moved one panel ahead. the winner is the first who will end in the finish panel.

at start the chances to win is 5:1 and winner will end with 6

if you see this starting pic. what are now the chances for 1 or the others to cross the finish line first? actually chances will change after every move. what formula do I need to use to get always the right odds after each move till finish?

and what is the chance that until the end of the game the winner ended in finish and one of the other 5 checkers didn't move at all and is still sitting on the starting panel? that means that the number of the die was not thrown even once during the game.

This would be a fairly difficult homework problem—unless I’m overlooking a really simple way to do this (which I have a habit of doing).

I’ll keep the explanation at a minimal for now. I basically just used the negative binomial distribution to derive the following solution:

General Solution

Let,
x = the number of spaces needed to win for the ‘checker’ that you want to win
a,b,c,d,e = the number of spaces needed to win for each of the other checkers

Then, the probability for the checker you want to win is:
Summation {from a=0 to x=(a-1)}
Summation {from b=0 to x=(b-1)}
Summation {from c=0 to x=(c-1)}
Summation {from d=0 to x=(d-1)}
Summation {from e=0 to x=(e-1)}
((a+b+c+d+e+x-1)! / (a!b!c!d!e!(x-1)!)) * (1/6)^a+b+c+d+e+x

Solution for Starting Picture

In your picture, the 1 checker needs 5 spaces while the rest each needs 6. So, plug in x=5 and plug in 6 for the rest of the variables. Compute to get 0.28298. So, the 1 checker now has a 28.3% chance to win. Therefore, the rest of the field has a 71.7% chance to win or 14.3% each.

Quote:

I don't know about the math, but I think these questions are germane:

Do you have to hit the goal exactly, or can you go over?

Do the players go in the same order each round? Does the game end when one checker reaches the goal, or do the others still have an opportunity?

Quote: Ayecarumba

Quote: PeeMcGee

Quote: seven

Hi all

please remember that I am no a math or formula expert and therfore I am back with a question. thanks

please look here Example
this game is played with one die and after each throw the checker with the thrown number is moved one panel ahead. here the first throw was the number 1 and the checker was moved one panel ahead. the winner is the first who will end in the finish panel.

at start the chances to win is 5:1 and winner will end with 6

if you see this starting pic. what are now the chances for 1 or the others to cross the finish line first? actually chances will change after every move. what formula do I need to use to get always the right odds after each move till finish?

and what is the chance that until the end of the game the winner ended in finish and one of the other 5 checkers didn't move at all and is still sitting on the starting panel? that means that the number of the die was not thrown even once during the game.

This would be a fairly difficult homework problem—unless I’m overlooking a really simple way to do this (which I have a habit of doing).

I’ll keep the explanation at a minimal for now. I basically just used the negative binomial distribution to derive the following solution:

General Solution

Let,
x = the number of spaces needed to win for the ‘checker’ that you want to win
a,b,c,d,e = the number of spaces needed to win for each of the other checkers

Then, the probability for the checker you want to win is:
Summation {from a=0 to x=(a-1)}
Summation {from b=0 to x=(b-1)}
Summation {from c=0 to x=(c-1)}
Summation {from d=0 to x=(d-1)}
Summation {from e=0 to x=(e-1)}
((a+b+c+d+e+x-1)! / (a!b!c!d!e!(x-1)!)) * (1/6)^a+b+c+d+e+x

Solution for Starting Picture

In your picture, the 1 checker needs 5 spaces while the rest each needs 6. So, plug in x=5 and plug in 6 for the rest of the variables. Compute to get 0.28298. So, the 1 checker now has a 28.3% chance to win. Therefore, the rest of the field has a 71.7% chance to win or 14.3% each.

I don't know about the math, but I think these questions are germane:

Do you have to hit the goal exactly, or can you go over?

Do the players go in the same order each round? Does the game end when one checker reaches the goal, or do the others still have an opportunity?

Quote: PeeMcGee

Quote:

I don't know about the math, but I think these questions are germane:

Do you have to hit the goal exactly, or can you go over?

Do the players go in the same order each round? Does the game end when one checker reaches the goal, or do the others still have an opportunity?

My understanding of the game is there are no "players". You simply roll a six sided die and move the corresponding number one space towards the finish. As soon as one of the checkers reaches the final space the game is over.

Quote: seven

Quote: PeeMcGee

Quote:

I don't know about the math, but I think these questions are germane:

Do you have to hit the goal exactly, or can you go over?

Do the players go in the same order each round? Does the game end when one checker reaches the goal, or do the others still have an opportunity?

My understanding of the game is there are no "players". You simply roll a six sided die and move the corresponding number one space towards the finish. As soon as one of the checkers reaches the final space the game is over.

there could be 1 player and there could be 100 players. the idea is that at the start of the game each player could chose one or more from 1-6 that he thinks will end the winner.
but after each throw and move of a checker the chances are changing and players could bet again with different odds. that was the reason I posted my question for a formula to offer the right odds.

Quote: PeeMcGee

Quote: seven

Quote: PeeMcGee

Quote:

I don't know about the math, but I think these questions are germane:

Do you have to hit the goal exactly, or can you go over?

Do the players go in the same order each round? Does the game end when one checker reaches the goal, or do the others still have an opportunity?

My understanding of the game is there are no "players". You simply roll a six sided die and move the corresponding number one space towards the finish. As soon as one of the checkers reaches the final space the game is over.

there could be 1 player and there could be 100 players. the idea is that at the start of the game each player could chose one or more from 1-6 that he thinks will end the winner.
but after each throw and move of a checker the chances are changing and players could bet again with different odds. that was the reason I posted my question for a formula to offer the right odds.

That was my understanding. The expression I gave in my post will provide what you are seeking.

Quote: PeeMcGee

I’ll keep the explanation at a minimal for now. I basically just used the negative binomial distribution to derive the following solution:

General Solution

Let,
x = the number of spaces needed to win for the ‘checker’ that you want to win
a,b,c,d,e = the number of spaces needed to win for each of the other checkers

Then, the probability for the checker you want to win is:
Summation {from a=0 to x=(a-1)}
Summation {from b=0 to x=(b-1)}
Summation {from c=0 to x=(c-1)}
Summation {from d=0 to x=(d-1)}
Summation {from e=0 to x=(e-1)}
((a+b+c+d+e+x-1)! / (a!b!c!d!e!(x-1)!)) * (1/6)^a+b+c+d+e+x

Solution for Starting Picture

In your picture, the 1 checker needs 5 spaces while the rest each needs 6. So, plug in x=5 and plug in 6 for the rest of the variables. Compute to get 0.28298. So, the 1 checker now has a 28.3% chance to win. Therefore, the rest of the field has a 71.7% chance to win or 14.3% each.

Quote: DJTeddyBear

This smells like homework.

Quote: mustangsally

no way
looks like another game idea from seven

ok

so the checkers move one square to the right depending on the # that is rolled.

say the 1st roll = 1 (OP example)
so checker is now at square1 (one die roll)
the others still at 0

say we speed this up and roll 2 dice
next roll = 3,1

checker moves 1 space to right and is now at square 2
c2 is still at space 0
c3 now moves one space to the right and is at aquare 1
c4,c5 and c6 are just watching at square 0

next roll = 1,1
c1 moves 2 squares to the right and is now at 4
the others are where they were the last roll
1st to square #6 is the winner

lots of states this game can be in
and i love dice games
so i would solve this for each step with a Markov chain

also looks like no winner on the first 5 rolls of 1 die
and there has to be a winner on the 31st roll (6*5, all checkers at 5 and still having the same chance to be the winner as B4 the game started)

OP
is this your game idea and do I have the rules correct?
the number on the die moves that checker# one space or square to the right
next roll
Sally

Quote: gordonm888

Quote: PeeMcGee

I’ll keep the explanation at a minimal for now. I basically just used the negative binomial distribution to derive the following solution:

General Solution

Let,
x = the number of spaces needed to win for the ‘checker’ that you want to win
p,q,r,s,t = the number of spaces needed to win for each of the other checkers

Then, the probability for the checker you want to win is:
Summation {from a=0 to x=(a-1)}
Summation {from b=0 to x=(b-1)}
Summation {from c=0 to x=(c-1)}
Summation {from d=0 to x=(d-1)}
Summation {from e=0 to x=(e-1)}
((a+b+c+d+e+x-1)! / (a!b!c!d!e!(x-1)!)) * (1/6)^a+b+c+d+e+x

Solution for Starting Picture

In your picture, the 1 checker needs 5 spaces while the rest each needs 6. So, plug in x=5 and plug in 6 for the rest of the variables. Compute to get 0.28298. So, the 1 checker now has a 28.3% chance to win. Therefore, the rest of the field has a 71.7% chance to win or 14.3% each.

I'm not familiar with the negative binomial distribution, so I'm trying to learn. When I plug x=5 and a=b=c=d=e=6 into the equation
((a+b+c+d+e+x-1)! / (a!b!c!d!e!(x-1)!)) * (1/6)^a+b+c+d+e+x I calculate:

(34!/(6!⁵*4!)) * (1/6)³⁵ = 0.000739651.

Now, what are the values of those 5 summation terms and how do I apply them to arrive at an answer of 0.28198?

Quote: seven

hi Sally

Quote: seven

<snip>but could you please ELI5

Quote: mustangsally

Hi

ok, i try, i am not a very good teacher

lots of states this game can be in

kind of like water
it can be in 3 "states"
liquid, solid or a gas

so your game as it unfolds can be in different states after each roll
depending on the # of successes for each number rolled (your checkers)
1 thru 6 inclusive

after the 1st roll
we could be in 1 of 6 states
1.1 (checker 1 with 1 success or hit) 2.0, 3.0, 4.0, 5.0, 6.0 (all other checkers at 0 hits)
or
1.0, 2.1, 3.0, 4.0, 5.0, 6.0
or
1.0, 2.0, 3.1, 4.0, 5.0, 6.0
and on for 1 each

now from that first roll
each state can move to other states and there will be more of them (states)
like
1.0, 2.1, 3.0, 4.0, 5.0, 6.0
could move to
1.0, 2.2, 3.0, 4.0, 5.0, 6.0 with a probability = 1/6
or
1.1, 2.1, 3.0, 4.0, 5.0, 6.0
or
1.0, 2.1, 3.1, 4.0, 5.0, 6.0

and so one

it becomes a fun exercise on finding all the combinations that all states can be in (always a trick)
after each roll

this is how i would do this
other than a simulation
but you said you want to know the probability of winning for each checker
after each roll and that is exactly what a Markov chain method can do for you (and me too)

this one looks fun
once all the states and the transition probabilities are entered into the matrix, just simple matrix multiplication returs lots of answers
lots of them
even for the absorbing states (1.6 or 2.6 and so on)

no lesson on that part here
"and i love dice games" i do
"so i would solve this for each step with a Markov chain" yes i would do just that if I find the time

so this game is just a RACE
could have 1st and 2nd place prizes too

it could be even a party game using people instead of checkers
lots of "states" this game can be in ;)
Sally

Quote: PeeMcGee

My apologies, I wasn’t sure how to make the expression clear on the forums. Hopefully, this helps…



It’s very computational—you’ll want a program that can solve summations.

Quote: PeeMcGee

My apologies, I wasn’t sure how to make the expression clear on the forums. Hopefully, this helps…



It’s very computational—you’ll want a program that can solve summations.

Quote:

Question: each summation is for a variable from 0 to "that particular variable - 1" (e.g the first summation is the sum over all values of a from 0 to "a-1"). Are the variables a,b,c,d,e as the summation variables different from the a,b,c,d,e "target" values, in which case, you should change one of those sets of variables to, say, p,q,r,s,t?

Quote:

By the way, in your summations it is a little unconventional for a parameter such as "a" to be defined as ranging from 0 to "a-1" - but I do understand exactly what you mean, lol -and I thank you for the effort to post the equation in a format that is so clear.

Quote:

Wow. I agree that the equation is probably too complicated for Excel. Do you use Mathematica? or what computational software do you use, or would you recommend?

Quote:

I understand that the fraction is the number of permutations of (a + b + c + d + e + (x-1)) items where a are one value, b are another, c are a third, d are a fourth, e are a fifth, and the remaining (x-1) are a sixth.

Quote: seven

thanks for the patience and explanation very much appreciated. I will later talk to google about Markov :)

yes the game is a race idea. it could be horses, cars, people etc. but I thought it would be boring if there would be bets at the start and then
just to wait who will be first to cross the finish line. IMO it would be much more interesting to give new quotes/odds after each move.

Quote: PeeMcGee

I wrote a program in Python that solves it.
Excel could do it, but it would require a lot of work on the user. There might be some creative way to do it easy though.
Also, WolframAlpha might be an option.

Quote: mustangsally

i agree with you
i played a simple version by meself (a 2 hit version)
after i set up a 70x70 transition matrix in Excel of course

it was a blast and i showed it to me hubby B4 dinner and he said
this is fun
where did you find this game?

the Angles won and it was an under
so i won again and again
i am so lucky
Sally

thank you for sharing!

Quote: seven

what is advantage of the 2 hit version?

Quote: seven

tbecause it ends a bit quicker what would maybe good for testing?

Quote: seven

thanks for all the help to solve the problem. what is your opinion regarding Sally's idea to solve it with Markov? and lets say I want to bring this idea online
and I want to give the users after each move of a checker the the right odds so each user can bet on the winner at any time of the game as long no checker crossed the winner line . what program should I use? I understood Excel would not be good or not work. would the python program that you did solve this problem?

import math

x = 6
p = 6
q = 6
r = 6
s = 6
t = 6
total = 0
for a in range(p):
    for b in range(q):
        for c in range(r):
            for d in range(s):
                for e in range(t):
                    total += (math.factorial(a+b+c+d+e+x-1) / (math.factorial(a) * math.factorial(b) * math.factorial(c) * math.factorial(d) * math.factorial(e) * math.factorial(x-1)) * ((1/6)**(a+b+c+d+e+x)))

print(total)

Quote: PeeMcGee

Markov is a valid way to solve it. It’s just that there are so many states. I found a way to maybe reduce it down to like 400 or so states—but that’s still quite a bit.

Yes, Python can do what you are seeking. Do you know Python? Anyways, here is my program:

import math

x = 6
p = 6
q = 6
r = 6
s = 6
t = 6
total = 0
for a in range(p):
    for b in range(q):
        for c in range(r):
            for d in range(s):
                for e in range(t):
                    total += (math.factorial(a+b+c+d+e+x-1) / (math.factorial(a) * math.factorial(b) * math.factorial(c) * math.factorial(d) * math.factorial(e) * math.factorial(x-1)) * ((1/6)**(a+b+c+d+e+x)))

print(total)

The x,p,q,r,s,t are all the variables I discussed in my original post. So when one of the checkers moves, you can have the program reduce the corresponding variable by one and then calculate the new probability.

Quote: seven

I will look for a coder to code the game with a nice design. like horse trotters race, regular horse, cars, motorcycle etc so each user can choose the race he likes.

Quote: seven

could the coder implement your python program so it will show next to the checker (horse etc ) the new odds so they can bet on it?

Quote: mustangsally

that sounds wonderful
a good one could easily do that
but for the game of 6hits for example
you would probably not want each roll to offer a new bet
MAYbe every so many rolls a new bet(s) a futures bet so to speak

and the prorgram would already have the odds to offer for each bet in a table
not calculating every time

like after 3 rolls say 1,2,3 rolled, in any order (3 single hits)
prob for 1,2 or 3 to win = abouts 22.5% each
4,5,and 6 now at = abouts 10.8% each
odds can easily come from that


is this a
for fun play money game
or
for a house edge real money game to offer?

in other words
what are you after?
(other than fame or fortune)
Sally

Quote: seven

it looks that no one paid attention to my last sentence (question) in my first posting. or if yes maybe you guys thought that the answer is obvious. here is my question again

and what is the chance that until the end of the game the winner ended in finish panel and one or more of the other 5 checkers didn't move at all and is still sitting on the starting panel? that means that the number of the die was not thrown even once during the game.

thanks

edit:
the idea for this question is to give out something like a bad beat price in case one player didn't move at all during a game

Quote: PeeMcGee

Take my expression

Quote: seven

the idea for this question is to give out something like a bad beat price in case one player didn't move at all during a game

Quote: mustangsally

there could be at least 1 player (checker) not moving one space by the time 1 is a winner.

a simple simulation shows that 1 to 5 players (checkers) could move not one space during the race
wow
1,1,1,1,1,1 WINNER! (it happened in my sim)

"RIGGED RACE!"

and agree with 25.3% as that answer
feels about right too

(abouts 20.1% of races will have just 1 checker not moving at least 1 space
and 4.2% there were 2 checkers watching each other AND the others race)

Sally

Quote: seven

@ PeeMcGee thanks again

lets see if I understood it. taking the 25.3% would mean that every 4th game at least 1 checker
will not move from the start field? or am I wrong?

Quote: PeeMcGee

You would expect it to happen about 1 in 4 games. Correct.

Quote: mustangsally

i agree too
on average that is

abouts 1 in 4 games

not every 4th game - that would be a game design if needed

but a possibility that in the next 4 games (even 8 games)
there would be not 1 game (race) where at least 1 checker did not move

averages are 1 in 4
25% is the event success probability

so they are related
but are not the same thing

Quote: seven

with 1 in 4 the only thing that comes to my mind is an insurance side bet that it won't happen.

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?
yes Sally your simulation gave me this idea :)

Quote: PeeMcGee

Which...it should be pointed out that the 1 in 4 probability was for any one of the numbers to not move.

Quote: PeeMcGee

For a specific number to not move…well, it would be one sixth of that or about 1 in 24.

Quote: PeeMcGee

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.

Which...it should be pointed out that the 1 in 4 probability was for any one of the numbers to not move. For a specific number to not move…well, it would be one sixth of that or about 1 in 24.

Quote: mustangsally

yes, and at least 1
that seems too easy of an answer to me
so i looked at me sims
and they all showed between 1 in 19 to 1 in 19.4 for a specific checker not to move (i used #1 both times)
i could be wrong here too
that seems too easy too

but I have played this 6hit game many times in Excel
and it is still fun
(MAY not be fun when I be in Las Vegas next week starting tomorrow)

wow
look at this race


after roll 11

who bet on #3 to win?
not i
all liked #5 (me too) and one can see why

what a finish!!

and who came in second?
i did not take a photo of that

Sally

i be finished, won my Angel over, until after next week
have a fun weekend
WSOP Event #3

Quote:

now I am a bit confused. the math is telling us 1 in 24 for a specific number not to move and you are saying it is about 1 in 19.

Quote: seven

Hi all

please remember that I am no a math or formula expert and therfore I am back with a question. thanks

please look here Example
this game is played with one die and after each throw the checker with the thrown number is moved one panel ahead. here the first throw was the number 1 and the checker was moved one panel ahead. the winner is the first who will end in the finish panel.

at start the chances to win is 5:1 and winner will end with 6

if you see this starting pic. what are now the chances for 1 or the others to cross the finish line first? actually chances will change after every move. what formula do I need to use to get always the right odds after each move till finish?

SymPy session

P = 6   # Players
N = 6   # Race to N
H = 1   # Head start for player 1

dt = N*P/10000
k = 0:(N-1)
prob = 0
for (t in seq(0,10*N*P,dt)) {
  prob = prob + sum((t/P)^k/factorial(k)*exp(-t/P))^(P-1)*(t/P)^(N-H-1)/factorial(N-H-1)*exp(-t/P)*1/P
}
prob = prob*dt
prob

Quote: seven

and what is the chance that until the end of the game the winner ended in finish and one of the other 5 checkers didn't move at all and is still sitting on the starting panel? that means that the number of the die was not thrown even once during the game.

Quote: seven

Quote: mustangsally

wow
look at this race

thank you for the race sample. very interesting one for the house because it would attract many side bets.

Quote: seven

@PeeMcGee could you please check your program and tell us what would have been the odds for the player
after the the 5th #5 showed up? after roll 15.

Quote: BruceZ

There is a remarkable way to solve this problem which is almost like magic.

Quote: BruceZ

Perfect for the Wizard.

Quote: BruceZ

<snip>

Quote: seven

what do you mean with Head start? what or who is head start?

Quote: mustangsally

Hi Bruce!
in what way exactly?
(I ask that question a lot)

Quote: mustangsally

where have you been hiding?
you do not have to answer (love you!)

Quote: mustangsally

OK
the solution you showed
looks nice
but it took me almost forever to get PeeMcGee's method down
(i got it and was still going to work with R using the python code)

Quote: mustangsally

you do know this game is popular all over the world
it is a horse race game with dice
i have never seen anyone trying out math to solve for winning probs
so this IS fun

i checked it after a small sim (100,000 - should be 10M or 100x that for closer results)
my sim in Excel (no vba just a little add-in used)

Player probs to win the race after the 15th roll
current positions
1,2,2,2,5,3
1: 1100 >> 0.011
2: 3264 >> 0.03264
3: 3248 >> 0.03248
4: 3221 >> 0.03221
5: 79699 >> 0.79699
6: 9468 >> 0.09468
calculated with the python code
(have not yet converted this to R code
and it now will be a 2 player race!)

1: 0.010940937
2: 0.032622172
3: 0.032622172
4: 0.032622172
5: 0.796962623
6: 0.094229924

you can take the probability distribution and create a payout table

that looks like fun too!

WSOP Event #3 coming on the 28th!
Sally

Quote: seven

need yet to figure out what odds the bookmaker is giving (odds less his %).
need to see what a healthy % would be.

Quote: BruceZ

You can answer the OPs first question with 2 lines in R:

> t = seq(0,5000,.01)
> sum((1-pgamma(t/6,6))^5*(t/6)^4/factorial(4)*exp(-t/6)*1/6)*.01
[1] 0.282983952354144

Quote: BruceZ

We could have 100 players with a 100-sided die,

Quote: BruceZ

and you wouldn't need 99 nested sums. You would only need the one integral which is like 1 sum.

Quote: BruceZ

Then we can easily modify this to do the more general problem where different players start on different squares.

Quote: BruceZ

Also where the die is loaded.

Quote: mustangsally

yes of course

for the race i showed after the 15th roll
it could be something like this

player5 is such an odds on favorite to win at abouts 80%, the odds offered in this example are 1:5 (bet 5 to win 1)



fun
Sally