Any mathematicians or probabilists out there?

36 possible combinations on the 2 dice

10 combinations to make 6

10 cambinations to make 8


20 winning combos divided by 36 possible combos gives you 0.55555555555555 using typical round up you get 0.56 he's right your wrong.

0.56 and i dont play dice

Gold Coast,and the 7's?
From the 36 throws there are 5 that make 6
5 1
4 2
3 3
1 5
2 4
........
and 5 that make 8
6 2
5 3
4 4
3 5
2 6
But thank you for trying to help.

Hmmm, I make it 58.14%, different again I think.

1 way to get 2
2 ways to get 3
3 ways to get 4
4 ways to get 5
5 ways to get 6
6 ways to get 7
5 ways to get 8
4 ways to get 9
3 ways to get 10
2 ways to get 11
1 way to get 12

Scrap that, found a mistake, will try again ...

Yeah, I also make it 0,5455, or well, actually 0,5456 :P

10/16*5/11+10/16*6/11*5/11+6/16*10/16*5/11=0,5456

I can't even work out what the question is

How many times does he roll the pair of dice?

Depends, as I understand it at least he stops whenever he has rolled both a 6 and an 8, or 7 two times, whatever happens first.

Yep, Mr Anderson and OP have it right. Couldn't see it for a while, but calc above explains it nicely.

Thank you Mr Anderson,I can sit back annd relax now.

I agree too but did it as

1-(3/8*3/8+3/8*5/8*6/11+5/8*6/11*6/11)

Well done Reverend!

Mr Anderson and Reverend Bayes,
It may amuse you to know that I have received an answer from another source.
The method used by the other person involves,wait for it,
diagonalized eigenvalues,matrices,Markov Chains and a couple of pages of figures.
And blow me down they very nearly got it right!

Sounds like Eldrick. Although he would get it right.

0.5455 rounds to 0.546
0.546 rounds to 0.55

Surely the answer is an infinite series with ever decreasing values the further it goes on. Because theres always the probability that each roll of the dice doesn't make a 6,7 or 8. I think thats where all the complicated maths comes into it?

Scores in the range 1-5, 9-12 don't affect the game so, the game can be won in 6 ways:
6,8
8,6
6,7,8
8,7,6
7,6,8
7,8,6
Probability of a 7 is 6/16ths
Probability of a 6 or an 8 are 5/16ths
Total of the above probabilities is .54558

I think Mr Anderson explained it best though.

agree with mr a et al

bongo, your method doesn't explicitly seem to deal with {6,6,8} and the like - i take it you know how to get around that bit?

Maybe it's Alzheimers but I just don't understand this at all.

There are 5 ways of throwing a total of 6 with 2 dice, 5 1, 4 2, 3 3, 2 4, 1 5.
6 ways of total 7, 6 1, 5 2, 4 3, 3 4, 2 5, 1 6.
5 ways of total 8, 6 2, 5 3, 4 4, 3 5, 2 6.

A 6/11 chance of throwing 7 before 6, that's 54.55%.

A 6/11 chance of throwing 7 before 8, that's 54.55%.

Chance of winning with 6 and 8 before losing with two 7's looks like 45.45%.

The lanterne rouge

Remember that the numbers other than 6,8 or 7 have no possible bearing on the result.
There can be no draw,the shooter or the bank wins,plain and simple.

Then work it from 1-P(bank wins) or P(shooter wins)
The tricky bit is that when a 6 or an 8 has shown the probabilities change.
At the start we have 6/36 for 7
5/36 for 6
5/36 for 8 but since only 6,7 or 8 count we only have the 5+5+6 to worry about therefore respective probabiities are
7 6/16
6 5/16
8 5/16
but when a 6 has shown it is no longer valid hence P(8)=5/11 and P(7)=6/11 at that time.
I hope that helps.

Thanks for your reply dunlaying.   I don't know the rules of the game and thought I answered the original question OK.   The "no longer valid" part has me puzzled.

Regards, The Lanterne Rouge.

The lantern Rouge
When a 6 or an 8 is thrown it would count as ,shall we say,a non runner.
Now ,remembering that there is only one shooter,the play is,to win for shooter,
68
86
768
786
678
876
all other numbers invalid.
So ,since 6 with 2 dice is 5/36
          8 with 2 dice is 5/36
          7 with 2 dice is 6/36
and the other numbers are "non runners"

the starting probabilities are for 6 5/16
                                   8 5/16
                                   7 6/16
but if we take the case where a 6 is the first valid number to show and we are working on P(shooter wins) then there are only 7's and 8's left that count.
therefore our total tally of valid throws would be 5/36 for the 8 and 6/36 for the 7 hence 5/11 for an 8
and 6/11 for a 7.
If I still have not made myself clear then ,please,let me know and I will try again.

Thanks dunlaying, now the penny has dropped.   My ignorance of the rules was the problem.   Regards, The Lanterne Rouge.