How to add odds? (or how I got the wrong sandwich)

Question

Jon and Frank ordered 2 footlong sandwiches, but instead got 4 6-inch sandwiches. Jon had ordered spicy italian, and Frank chicken teriyaki. The sandwiches aren't labeled and are wrap, so they can't tell which sandwiches are which.

What would be the odds of Jon picking the 2 sandwiches that are are his own? (the 2 spicy italian sandwiches)?

I know that there is a 50/50 chance of choosing a correct sandwich, since there are 2 terikaki and 2 spicy, but once 1 is chosen, there is only 1/3 chances to get the second one right.. but how do the the odds of getting the first 50/50 choice AND the second 1/3 choice correct add up?

Thanks!

Answer 1

You need to multiply the probabilities with "AND" (and you would add them instead of multiplying them if it was "or"). Here, as you point out, at the first pick he has 1 chance out of 2 of picking one of his sandwiches; at the second pick, only 1 out of 3 chances. So the result is 1/2*1/3 = 1/6, or 0.16666667.

Answer 2

Line up the four sandwiches in a row where Joe will get the first two and Frank the last two. The possibilities are $$SSCC, SCSC, SCCS, CCSS, CSCS, CSSC$$ and only in one out of these six cases does Joe get the two Spicy Italian sandwiches.

Answer 3

This grew too long for a comment, but I think something needs to be cleared up about the question and the answers (+1 to both by the way, both fine answers):

While the answers you have here give probabilities, your question asks about odds, which are not the same thing (See here).

You're no doubt using the terms interchangeably (as is commonly done, and usually understood as many people convert between the two so happily the process isn't even conscious), so both answers would respond to your question. However, I'd like to inject some additional clarity (especially for readers less familiar with moving between the two ways of describing chance events).

Your question actually gives values in terms of one and then the other: "50/50" is effectively odds while "1/3" is probability.

In terms of probability, the events you mention would under random choice have (by symmetry) probabilities $\frac{1}{2}$ and $\frac{1}{3}$. Both answers correctly describe the calculation of the probability of the combined event ("first choice is correct -and- second choice is correct given first choice was correct").

In terms of odds, the two events have odds of 1-1 and 2-1 against (1-2 for), and the combined event has odds of 5-1 against (1-5 for).

So for the sake of completeness (and some clarity for those who might get muddled), you obtain the odds of the combined event by converting any odds of the original events to probabilities, computing the resulting probability (either from first principles by simply counting the events, or using the probability rules for such events, like $P(AB) = P(A)P(B|A)$) -- since we have extensive rules for dealing with probability -- and then you convert the answer back to a statement about odds.

Answer 4

It's a classic application of the combination formula. Generally speaking, if we have n choices, of which, we can only choose r, then the possibilities are counted thus:

$${n \choose r} = \frac{n!}{r!(n-r)!}$$

In your specific question, you have 4 sandwiches, of which you choose 2, resulting in 6 possibilities:

$${4 \choose 2} = \frac{4!}{2! (4-2)!} = 6$$

Of these 6 possibilities, there is exactly 1 correct combination, then, the probability is $\frac{1}{6}$.