A

and

B

P (A \cap B) = P (A) P (B) .

For a possible sum

2 \leq t \leq 12

, let

N_{t}

be the number of ordered die pairs with sum

t

. Then

P (A_{t}) = \frac{N _{t}}{36}, P (B_{i}) = \frac{1}{6} .

t - i \in {1, \dots, 6}

, then

P (A_{t} \cap B_{i}) = 1/36

; otherwise it is zero. Independence for a possible sum therefore requires

\frac{1}{36} = \frac{N _{t}}{216},

N_{t} = 6

. This occurs only for

t = 7

, and then

7 - i

is valid for every

i = 1, \dots, 6

. Thus

A_{t} and B_{i} are independent exactly when t = 7, 1 \leq i \leq 6 .

If impossible sums are admitted as events of probability zero, they are trivially independent as well.

ii

Solution

Call the coins

1, 2, 3

in the order given. Coin 1 beats the constant score of coin 2 exactly when it shows a head, so coin 1 wins with probability

3/5

. Coin 2 beats coin 3 exactly when coin 3 shows a head and scores three, again with probability

3/5

Coin 3 beats coin 1 whenever coin 3 shows a tail, or when coin 3 shows a head and coin 1 shows a tail. Its winning probability is therefore

\frac{2}{5} + \frac{3}{5} \frac{2}{5} = \frac{16}{25} > \frac{1}{2} .

Thus the preferences form a nontransitive cycle:

1 beats 2, 2 beats 3, 3 beats 1.

The second chooser can always select a coin that has winning probability greater than one half against the first choice. Therefore

it is preferable to choose second .

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