best chance of win

malc

New member
Joined
Jul 24, 2012
Messages
4
if i throw 2 dice and am rewarded if the total amount scored is an odd figure,
and not rewarded if the total amount scored is an even figure then;
i assume as 2,4,6,8,10 and 12 can be scored or
3,5,7,9 and 11 then you have less chance of being rewarded.
other people suggest the odds are the same.
please can you clarify ?
many thanks
 
if i throw 2 dice and am rewarded if the total amount scored is an odd figure,
and not rewarded if the total amount scored is an even figure then;
i assume as 2,4,6,8,10 and 12 can be scored or
3,5,7,9 and 11 then you have less chance of being rewarded.
other people suggest the odds are the same.
please can you clarify ?
many thanks

You can throw a pair of dice in how many ways - 36

You can throw 2 in only one way 1 + 1

However you can throw "7" several ways: (1+6, 2+5, 3+4, 4+3, 5+2 and 6+1) - 6 ways

similarly you throw a "6" in - (1+5, 2+4, 3+3, 4+2, 5+1) - 5 ways

So the probability getting a 2 is 1/36, probability of getting total "6" is 5/36 and probability of getting total "7" is 1/6.

Those are not all the same!!

So now what .....
 
thanks for the reply,
i agree that a pair of dice can be thrown in 36 combinations,
i work out 21 of those outcomes result in an even total ie 58.33 %
and 15 of the outcomes result in an odd total ie 41.66%
so can i safely argue , down the pub that the odds/evens chances
are not the same ?







You can throw a pair of dice in how many ways - 36




You can throw 2 in only one way 1 + 1

However you can throw "7" several ways: (1+6, 2+5, 3+4, 4+3, 5+2 and 6+1) - 6 ways

similarly you throw a "6" in - (1+5, 2+4, 3+3, 4+2, 5+1) - 5 ways

So the probability getting a 2 is 1/36, probability of getting total "6" is 5/36 and probability of getting total "7" is 1/6.

Those are not all the same!!

So now what .....
 
thanks for the reply,
i agree that a pair of dice can be thrown in 36 combinations,
i work out 21 of those outcomes result in an even total ie 58.33 %
and 15 of the outcomes result in an odd total ie 41.66%
so can i safely argue , down the pub that the odds/evens chances
are not the same ?
How did you get that? There are nine pairs of even numbers and nine pairs of odd numbers.
 
thanks to all who have answered, let me explain the actual reason for the question ;
whilst in a pub at weekend they had a little quirk whereby after buying a round of drinks
you were offered the opportunity of rolling two dice, a free round if you throw an odd total,
nothing if you threw an even total. the bar staff weren`t bothered the order the two dice landed
and only the total. i reasoned that although a bit of fun the odds were slightly in favour of the pub
and a long argument ensued with arguments both ways.
the combinations for a win, to me were 1-2,1-4,1-6,2-3,2-5,4-3,4-5,5-6,6-3 = 9ways
and the combinations for a loss, to me were 1-1,2-2,3-1,3-3,4-2,4-4,5-1,5-3,5-5,6-2,6-4,6-6= 12ways
are there any other combinations ?
sadly i`m not as clever as you guys and only wondered whether my reasoning was correct,
thanks for the replies






Think about a red die and green die.

Calculate probability that sum is even. That means both dice are even or both are odd.

\(\displaystyle \dfrac{1}{36}\ for\ red\ 1\ and\ green\ 1.\) Sum is 2

\(\displaystyle \dfrac{1}{36}\ for\ red\ 1\ and\ green\ 3.\) Sum is 4

\(\displaystyle \dfrac{1}{36}\ for\ red\ 1\ and\ green\ 5.\) Sum is 6

\(\displaystyle \dfrac{1}{36}\ for\ red\ 2\ and\ green\ 2.\) Sum is 4

\(\displaystyle \dfrac{1}{36}\ for\ red\ 2\ and\ green\ 4.\) Sum is 6

\(\displaystyle \dfrac{1}{36}\ for\ red\ 2\ and\ green\ 6.\) Sum is 8

\(\displaystyle \dfrac{1}{36}\ for\ red\ 3\ and\ green\ 1.\) Sum is 4

\(\displaystyle \dfrac{1}{36}\ for\ red\ 3\ and\ green\ 3.\) Sum is 6

\(\displaystyle \dfrac{1}{36}\ for\ red\ 3\ and\ green\ 5.\) Sum is 8

\(\displaystyle \dfrac{1}{36}\ for\ red\ 4\ and\ green\ 2.\) Sum is 6

\(\displaystyle \dfrac{1}{36}\ for\ red\ 4\ and\ green\ 4.\) Sum is 8

\(\displaystyle \dfrac{1}{36}\ for\ red\ 4\ and\ green\ 6.\) Sum is 10

\(\displaystyle \dfrac{1}{36}\ for\ red\ 5\ and\ green\ 1.\) Sum is 6

\(\displaystyle \dfrac{1}{36}\ for\ red\ 5\ and\ green\ 3.\) Sum is 8

\(\displaystyle \dfrac{1}{36}\ for\ red\ 5\ and\ green\ 5.\) Sum is 10

\(\displaystyle \dfrac{1}{36}\ for\ red\ 6\ and\ green\ 2.\) Sum is 8

\(\displaystyle \dfrac{1}{36}\ for\ red\ 6\ and\ green\ 4.\) Sum is 10

\(\displaystyle \dfrac{1}{36}\ for\ red\ 6\ and\ green\ 6.\) Sum is 12

Which adds up (at least until almost closing time) to \(\displaystyle \dfrac{18}{36} = 50\%.\)

You can use those lines to achieve the same result a different way, but it will answer the true but irrelevant objection that the odds of rolling every sum are NOT equal.

Probability of the sum being two = \(\displaystyle \dfrac{1}{36}.\)

Probability of the sum being four = \(\displaystyle \dfrac{3}{36}.\)

Probability of the sum being six = \(\displaystyle \dfrac{5}{36}.\)

Probability of the sum being eight = \(\displaystyle \dfrac{5}{36}.\)

Probability of the sum being ten = \(\displaystyle \dfrac{3}{36}.\)

Probability of the sum being twelve = \(\displaystyle \dfrac{1}{36}.\)

And \(\displaystyle \dfrac{1}{36} + \dfrac{3}{36} + \dfrac{5}{36} + \dfrac{5}{36} +\dfrac{3}{36} + \dfrac{1}{36} =\dfrac{18}{36} = 50\%.\)
 
thanks for the info Jeff but in the eyes of the bar staff and in the absence of any colour on the dice, three can only be achieved one way ie; 2 and a 1
so in my situation am i more likely to win or lose a free drink ?
cheers
You were miscounting because you did not "see" that rolling the sum of 3 can happen TWO ways whereas rolling the sum of 2 can happen only ONE way. That is why I suggested thinking about a red die and a green die. Once you think that way, it is obvious that there are TWO ways to roll the sum of 3, namely red 1 and green 2 or else red 2 and green 1. But there is only ONE way to roll the sum of 2, namely red 1 and green 1. It's not that we are so smart; it's just that we have learned the tricks of the trade. One of MY tricks is to solve puzzles BEFORE going to the bar.
 
thanks for the info Jeff but in the eyes of the bar staff and in the absence of any colour on the dice, three can only be achieved one way ie; 2 and a 1
so in my situation am i more likely to win or lose a free drink ?
cheers
I'm not sure what you are saying here. Are you saying that you got a "1 and a 2" the bar staff would not give you a free drink but if you got a "2 and a 1" they would? How do they distinguish between a "2 and a 1" and "1 and a 2"? Whether or not the two dice add to an even or odd number does not, in any case, depend upon "the eyes of the bar staff"! The odds of "even or odd" are exactly the same.
 
Top