and the answer to

"Somebody bet he that he could roll a total of 6 and 8, with two dice, before I could roll a total of seven twice."

https://wizardofodds.com/ask-the-wizard/179/

*****

07-17-2013

1. "What is the probability of Rolling a Sum of 6 and 8 before an opponent rolls two sums of 7s?"

"2. Roll two dice to show that the experimental probability of rolling a sum of 8 and a sum of 6 before rolling two sums of 7 is approximately 55 percent. If you’re familiar with probability theory, prove that the theoretical probability is 54.6 percent."

Quote:Q#1

This looks to be about two players each rolling the dice

and only their roll counts towards their event(s)

in other words, A Race game.

Q#2

The book question is a popular one all over the net and in school.

To me, one or more persons rolls the dice to see which event happens first for that series of dice rolls.

https://forumserver.twoplustwo.com/25/probability/roll-six-eight-before-two-sevens-1352937/

*****

Math 361, Problem Set 2

September 17, 2010

University of Denver

3. (1.4.7) A pair of 6-sided dice is cast until either the sum of seven or eight

appears.

(a) Show that the probability of a seven before an eight is 6/11.

(b) Next, this pair of dice is cast until a seven appears twice (as a sum)

or until each of a six and an eight have appeared at least once. Show

that the probability of the six and eight occurring before two sevens

is 0.546.

http://cs.du.edu/~paulhorn/361/assn2-solns.pdf

two different questions

and

the Wizard, imo, answered Q2 (not the question asked in #179)

and answered 3b in Math 361, Problem Set 2

(b) Next, this pair of dice is cast until a seven appears twice (as a sum)

or until each of a six and an eight have appeared at least once.

Show that the probability of the six and eight occurring before two sevens

is 0.546.

'a well written question I must agree'

my answer using pari/gp

can do online too

https://pari.math.u-bordeaux.fr/gp.html

(08:42) gp > b=10/16 * 6/11* 5/11 ;\\6or8,a 7, then other

(08:42) gp > c=6/16 * 10/16 * 5/11 ;\\a 7, then 6or8, then other

(08:42) gp > a+b+c

%4 = 4225/7744

(08:42) gp > 4225/7744.

%5 = 0.54558367768595041322314049586776859504

there should be an easy way to solve this

(IF the question was, for example,

player A rolls 2d6 first and wants one sum of 6 B4 player B rolls 2nd and wants one sum of 7, one could use the sum of a infinite geometric series to solve and that is easy and shown all over the internet. answer for this example is Player A wins with a prob of 30/61,Player B wins with a prob of 31/60 and IF Player B rolls 1st, well, that winning prob looks to be 36/61)

"Somebody bet he that he could roll a total of 6 and 8, with two dice, before I could roll a total of seven twice." being that Player A rolls 1st

and

is trying for at least one 6 and one 8

the answer must be higher than 0.546

or is it?

Sally

Quote:mustangsallyhow to solve the below question (or bet)

(I used a Markov chain and a recursion and ran a simulation to get an answer)

I think your handle should be Markov Sally. Yes, I agree that would be the elegant way to solve it. However, I was probably afraid at the time I posted the question that just mentioned in Markov Chain would scare off the audience. I tried to show an answer using as simple math as possible.

you answered a different question than the one asked in #179, imoQuote:WizardHowever, I was probably afraid at the time I posted the question that just mentioned in Markov Chain would scare off the audience. I tried to show an answer using as simple math as possible.

you showed an answer to this

(b) Next, this pair of dice is cast until a seven appears twice (as a sum)

or until each of a six and an eight have appeared at least once.

Show that the probability of the six and eight occurring before two sevens

is 0.546.

the answer to

"Somebody bet he that he could roll a total of 6 and 8, with two dice, before I could roll a total of seven twice."

this has yet to have an answer shown (I keep mine to myself until at least another shows their answer and solution)

I think this can be solved using the sum of an infinite geometric series

I may be wrong

Sally

for at least one 6 and eight is rolled by Player A who rolls 1st

B4

two 7s are rolld by Player B rolling after Player A rolls

[1] 0.5750781

> print(end.time - start.time)

Time difference of 5.997747 mins

> p = count68/sims

> error = 3.29*sqrt(p*(1-p)/sims)

> error

[1] 0.0009897647

> p

[1] 0.5750781

> p-error

[1] 0.5740884

> p+error

[1] 0.5760679

> 1-p

[1] 0.4249219

> ##########################

> sims

[1] 2700000

> count68

[1] 1552711

> count7

[1] 1147289

> count68.1

[1] 563953

> count68.1/count68

[1] 0.3632054<<< at least 1 of either 6 or 8 was rolled

ok for now

Sally

Quote:mustangsally"Somebody bet he that he could roll a total of 6 and 8, with two dice, before I could roll a total of seven twice."

I think that if we are to be adequately nerdy, we would have to ask some questions like "Who gets to start rolling first?" and "Who can roll more times the fastest?" I mean, if I can make twenty rolls before the other guy can make his second roll, I think I have a much better chance of reaching my goal before he reaches his.

Yes, I'm just having some fun here.

;-)

nowQuote:WizardSally, please check if you approve of the new wording.

that is like (I use the word 'like' a lot)

the question that can be found in a text book and the math paper I linked too.

I say you answered that question in #179

How about the actual question from

ANTHONY FROM INDIANA

it is a different type of question, very close to the text book question, but still different

as the 'odd number rolls' are for the 6 and 8 event (at least 1 of each)

and the even rolls are for the two 7s

I am looking for an answer to that (I have one but want to see how other(s) would do it)

Sally

and I couldn't figure out any way. I don't think there is any way to use it to your advantage. if there was it would be well known. but on the rare chance that one of you guys or gals knows one please let it fly.

exactly. the question should leave no room for interpretation.Quote:DocI think that if we are to be adequately nerdy, we would have to ask some questions like "Who gets to start rolling first?"

The text book question, I think, makes it very clear.

the bar bet version that player A rolls first followed by Player B seems to be asked more and that is how I understood it. they alternate the rolls until a winner is found (and payment collected)

of course, bar bets are that way

Sally

btw, I am NOT saying you are wrong in your 1st answer in #179, just that you answered a different question and the new wording now fits your answerQuote:WizardSally, please check if you approve of the new wording.

Sally