Die probability question

Say that the bank rolls one die and the player one die. The player wins if her score is higher. What is the probability for the player to win (the bank wins if the scores are the same)?

What if the bank and the player may roll two times each and use the highest score?

(Say that the bank can roll his die n times and use that highest score and that the player can roll her die m times and use the higher score. Is there a general equation that can be used to find the odds for the player winning?)

This sounds suspiciously like a homework assignment.

Did you work out the math for the first question? That's the easiest.

For the second (and third), I'm inclined to think that the odds are the same as the first question.


Then again, I know the math for the first question, and didn't get out my pencil for the second one yet.....

We're all gonna die!!!

For the second and third, I would think it favors the house more since there is a better chance of a tie.

For multiple rolls, the house edge is lower, as there is less chance for a tie score the more rolls are used; the final score range is wider. 10 rolls would average 35, and tying on 35 or any number would be harder. The final score range would be 10 to 60.

Look at the chances of tyoing on a coin toss, where heads equal 1 and tails equal 0.
In one roll, your score can be only 1 or 0, and there's a 50% chance of a tie. In ten coin tosses, the possible range of scores is 0 to 10, average of 5. In a million coin tosses, the range is 0 to 1,000,000, average 500,000 and the chances of tying on 500,000 or any number would be miniscule.

Quote: Paigowdan

For multiple rolls, the house edge is lower, as there is less chance for a tie score the more rolls are used; the final score range is wider. 10 rolls would average 35, and tying on 35 or any number would be harder. The final score range would be 10 to 60.

To me, the question sounds like they will only take their highest individual roll, so the final score range would still be 1 to 6 with 10 rolls. If this is the case, it is very likely that the player and the house will tie with a 6.

Yes, if they use the single highest score and not an additive score. House edge would be huge with one roll, forget about multiple rolls.

Hmmm....

Yeah, I get it now.

The more rolls, the greater the chance for a tie.

However, with ties excluded, there always remains an equal chance of either winning.

But when ties are considered a win for the bank, the more rolls, the greater the house edge.


But that brings me back to my original question (and my feeling that this is homework).

Did you do the math for one roll each? That's the easiest to calculate.

I recall that this is an actual game in a casino. I think the Wizard did an analysis.

Edit: Yes it is. See the article on "Beat the Dealer" at Turning Stone, here.

The first two questions are quite simple and I guess they would be on homework level. What I can't figure out (nor my collegues) is the probability when the bank gets more rolls than one (well two is rather straight forward since you can use a chart). That the bank has higher chance of winning if it has the same amount of dice is also intutive; but how do you calculate the probability if the bank has for example 3 rolls of the die and the player 4 rolls of the die?

That game is based upon a two-die TOTAL.

The original poster proposed a one-die game, as well as multiple rolls taking only the highest roll.

Sorry my bad - I meant 3 rolls of the die and 4 rolls respectivly.

so the reply should be:

The first two questions are quite simple and I guess they would be on homework level. What I can't figure out (nor my collegues) is the probability when the bank gets more rolls than one (well two is rather straight forward since you can use a chart). That the bank has higher chance of winning if it has the same amount of dice is also intutive; but how do you calculate the probability if the bank has for example 3 rolls of the die and the player 4 rolls of the die?

First, figure out the odds of the dealer getting each number given his three rolls (easy, right?). Then, figure out the odds of the player getting each number given his 4 rolls (also easy). Then, manipulate the player's odds to be in a "less than n" format, and put it in a table against the dealer's values. For example, the dealer gets a 5 x% of the time and the player gets a 5 or less y% of the time. So a dealer's 5 wins z% of the time (x*y). Repeat for each dealer's total, then sum to get the dealer's overall win %.

Quote: rdw4potus

First, figure out the odds of the dealer getting each number given his three rolls (easy, right?). Then, figure out the odds of the player getting each number given his 4 rolls (also easy). Then, manipulate the player's odds to be in a "less than n" format, and put it in a table against the dealer's values. For example, the dealer gets a 5 x% of the time and the player gets a 5 or less y% of the time. So a dealer's 5 wins z% of the time (x*y). Repeat for each dealer's total, then sum to get the dealer's overall win %.

Remember that the game is over if the bank rolls a six on any of their rolls. The player need not attempt any rolls if this occurs. The odds of this occuring are (n*(1/6), where n = number of rolls by the bank (or number of dice rolled by the bank).

Quote: Ayecarumba

Remember that the game is over if the bank rolls a six on any of their rolls. The player need not attempt any rolls if this occurs. The odds of this occuring are (n*(1/6), where n = number of rolls by the bank (or number of dice rolled by the bank).

You're not wrong, but the odds of a dealer's win on a 6 are 1:1 under my method above. The dealer has 6, and the odds of the player having 6 or less are 100%, so the odds of the dealer winning on a 6 are 100%. Multiply that by the odds of the dealer's 6, add that outcome to the odds of the dealer winning on his other numbers, and you've got the whole game plotted.

Quote: Ayecarumba

The odds of this occuring are (n*(1/6), where n = number of rolls by the bank (or number of dice rolled by the bank).

Consider n=60 :)

Quote: Ayecarumba

Remember that the game is over if the bank rolls a six on any of their rolls. The player need not attempt any rolls if this occurs. The odds of this occuring are (n*(1/6), where n = number of rolls by the bank (or number of dice rolled by the bank).

Not exactly. Even with 18 rolls, it is not guaranteed that the dealer will roll a 6. Sure, he will likely roll 3 sixes, but it's not guaranteed.

To calculate the probability that the dealer ends with a total score of 6 with 3 rolls, sum up the following:

dealer rolls 6, 3 times: (1/6)^3 * (5/6)^0 * combin(3,3)
dealer rolls 6, 2 times: (1/6)^2 * (5/6)^1 * combin(3,2)
dealer rolls 6, 1 time: (1/6)^1 * (5/6)^2* combin(3,1)

The 1/6 is the likelihood of rolling the 6 and the 5/6 is the likelihood of rolling a 5 or less on the other rolls.

Similarly, calculate the probability that the dealer ends with a total score of 5 with 3 rolls:

dealer rolls 5, 3 times: (1/6)^3 * (4/6)^0 * combin(3,3)
dealer rolls 5, 2 times: (1/6)^2 * (4/6)^1 * combin(3,2)
dealer rolls 5, 1 time: (1/6)^1 * (4/6)^2* combin(3,1)

Continue this for all possible totals. In the end, you should find this distribution:


Probability of Ending with Final Score

Score	dealer (3 rolls)	player (4 rolls)
1	0.00462963	0.000771605
2	0.032407407	0.011574074
3	0.087962963	0.050154321
4	0.171296296	0.135030864
5	0.282407407	0.284722222
6	0.421296296	0.517746914

Quote: CrystalMath

To calculate the probability that the dealer ends with a total score of 6 with 3 rolls, sum up the following:

Or just 1-(5/6)^3 = 0.421296296 :)

Thank you all for the (gentle) correction.

Quote: weaselman

Or just 1-(5/6)^3 = 0.421296296 :)

Yes.

Collaboration usually leads to better answers. I normally start with a brute force, then sometimes realize easier ways.

To expand on weaselman:

p(6) = 1-(5/6)^3
p(5+) = 1-(4/6)^3
p(4+) = 1-(3/6)^3
p(3+) = 1-(2/6)^3
p(2+) = 1-(1/6)^3
p(1+) = 1-(0/6)^3

To get the probability of getting a score of 5, take p(5+) - p(6) = (5/6)^3-(4/6)^3
p(total score of 4) = p(4+)-p(5+) = (4/6)^3-(3/6)^3

Easy!

Quote: CrystalMath

Yes.

Collaboration usually leads to better answers. I normally start with a brute force, then sometimes realize easier ways.

To expand on weaselman:

p(6) = 1-(5/6)^3
p(5+) = 1-(4/6)^3
p(4+) = 1-(3/6)^3
p(3+) = 1-(2/6)^3
p(2+) = 1-(1/6)^3
p(1+) = 1-(0/6)^3

To get the probability of getting a score of 5, take p(5+) - p(6) = (5/6)^3-(4/6)^3
p(total score of 4) = p(4+)-p(5+) = (4/6)^3-(3/6)^3

Easy!

EASY !!!!!!!!!! I have a better chance of figuring out Who's on first. LOL

Let f = the face of the die
Let d = the number of dice
The probability of the highest number being f is:
(f^d-(f-1)^d)/6^d

Quote: CrystalMath

To expand on weaselman:

p(6) = 1-(5/6)^3
p(5+) = 1-(4/6)^3
p(4+) = 1-(3/6)^3
p(3+) = 1-(2/6)^3
p(2+) = 1-(1/6)^3
p(1+) = 1-(0/6)^3

To get the probability of getting a score of 5, take p(5+) - p(6) = (5/6)^3-(4/6)^3
p(total score of 4) = p(4+)-p(5+) = (4/6)^3-(3/6)^3

My first thought was to do it the other way around:

p(1) = (1/6)³
p(2) = p(2 or less) - p(1) = (2/6)³ - (1/6)³
p(3) = p(3 or less) - p(2 or less) = (3/6)³ - (2/6)³
p(4) = p(4 or less) - p(3 or less) = (4/6)³ - (3/6)³
p(5) = p(5 or less) - p(4 or less) = (5/6)³ - (4/6)³
p(6) = p(6 or less) - p(5 or less) = (6/6)³ - (5/6)³

Note (N/6)³ - ((N-1)/6)³ = (N³ - (N³ - 3N² + 3N - 1)) / 6³ = (3N²- 3N + 1)/216

This gets the same numbers as weaselman, but in a slightly different way.

I have a Matrix solution (Markov chain?)

a is an initial state vector, after 1 roll:
a =

0.1667 0.1667 0.1667 0.1667 0.1667 0.1667

D is the matrix that lets you roll another die:
D =

0.1667 0.1667 0.1667 0.1667 0.1667 0.1667
0.0000 0.3333 0.1667 0.1667 0.1667 0.1667
0.0000 0.0000 0.5000 0.1667 0.1667 0.1667
0.0000 0.0000 0.0000 0.6667 0.1667 0.1667
0.0000 0.0000 0.0000 0.0000 0.8333 0.1667
0.0000 0.0000 0.0000 0.0000 0.0000 1.0000

So that a*D^(n-1) gives you the probability that your highest die will be any of the 6 states.

Then you need a win matrix W:

W =

0 1 1 1 1 1
0 0 1 1 1 1
0 0 0 1 1 1
0 0 0 0 1 1
0 0 0 0 0 1
0 0 0 0 0 0

that will compare dealer totals to player totals.

Given the dealer rolls n dice and the player rolls m dice, they compare the highest, dealer winning ties, the general formula to answer the original question is:

a * D^(n-1) * W * (a * D^(m-1))'

For n = 3, m = 4, I get 0.4028 player win.

If you are trying to make a nice gambling game, n=2,m=3 gives a more even 0.4716.

Or, if you have enough dice, n=3, m=7 gives 0.4892