find the odds in favor of rolling the sum of 6 when rolling a pair of dice

Answer 1

Ann,

We assume that the "game" or "trial" is: roll 2 d6 dice, and add up the values appearing on the tops.

2 important questions then: how large is the "sample space" of outcomes? How many distinct outcomes do we expect? Since values can be repeated (a 6 on the 1st die doesn't disqualify a 6 on the 2nd die), the multiplication rule for independent events gives us : 6x6 = 36 possible distinct outcomes.

Question 2: how many of those outcomes are sums = 6? For example, 2+4 = 6. Count them up. Let's say there are "K" different outcomes that =6.

For PROBABILITY, we would write a ratio, K/36 (which may be read, "K out of 36 outcomes" if you wish),

but to quote ODDS IN FAVOR, we would write (# outcomes = 6) : (# outcomes NOT = 6), or

something like K : (36-K) ....the two values sum up to 36 -- we are splitting the outcomes into two sets -- things that = 6, and things that do not.

For example, if there are TWO ways to make 6, then K = 2 and the ODDS IN FAVOR are 2:34.

Regards,

Phil C.