Three of a kind!

There are 2 possibilities. Either you get a match on all three turns, or on only one of them.

If its all three, then there are six ways to do it. 3 permutations on the order of Ace, two and three.

If its just one then there are 54 ways to do it:

• 3 ways to choose which card matches
• 3 ways to choose on which turn they match
• 6 ways to choose the remaining cards (not to match since that would be double counting)

$3 \cdot 3 \cdot 6 + 6 = 60$ ways

And there are $6^3 = 216$ ways the cards could be played. (Each player has 6 permutations for their 3 cards)

$\dfrac{60}{216} = \dfrac{5}{18}$

$5+18 = \boxed{23}$

Regardless of the order of Slim's cards, the probability of the other cards matching Slim's is the same. (Consider, even if Slim's cards are not in order $A\ 2\ 3$ , we can change the flip order so this is the case, and the number of matches will be the same).

There are $6$ ways of ordering $3$ cards. This means there are $6\times6=36$ ways of ordering Jim & Tim's cards in relation to Slim's cards.

If there is exactly $1$ matching flip, there are $3$ possible ways to order the remaining $4$ cards without resulting in more matches. as there are $3$ flips, the total number of permutations of exactly $1$ matching flip is $3\times3=9$ .

There cannot be exactly $2$ matching flips. If $2$ flips are matching, the remaining cards are all the same.

There is only $1$ way of having $3$ matching flips.

This gives us a total of $10$ permutations with at least one matching flip, out of a total $36$ permutations. As such, the probability of this happening is $\frac{10}{36}$ which cancels down to $\frac{5}{18}$ , giving us our answer of $5+18=\boxed{23}$ .

WLOG*, Slim turns over $(1,2,3)$ in order. There are $3$ cases based on what Tim turns over

• With probability $1/6$ , Tim also turns over $(1,2,3)$ in that order. In this case Jim matches at least one of the three with probability $2/3$ .

• With probability $1/2$ , Tim matches exactly one of the three cards: $(1,3,2)$ or $(3,2,1)$ or $(2,1,3)$ . In this case Jim matches the same card with probability $1/3$ .

• With probability $1/3$ , Tim doesn't match any of the three: $(2,3,1)$ or $(3,1,2)$ . In this case no match is possible.

Thus the probability of at least one match is $(1/6)*(2/3) + (1/2)*(1/3) + (1/3)*0 = 5/18$ . Thus $a = 5$ and $b = 18$ and $a+b = 23$ .

*WLOG = Without loss of generality

The permutation group S3 contains six elements: in cycle notation, (), (A2),(23),(A3),(A23) & (A32). So there is one cycle of length 0 (the identity), three cycles of length 2 & two cycles of length 3.

There is a 13/18 probability that there is no three-way match: we can show this using a probability tree with three branches, and only a depth of two.

We can let Slim’s cards come up in any order.

I) Jim’s cards have a 1/6 probability of turning up the same as Slim’s, with the identity permutation (). Then the permutation to Tim’s cards, for all of his to be different, would have to be a cycle of length three, (A23) or (A32), adding up to a 1/3 probability. So the joint probability of no match with this starting case would be $\frac {1}{6}$ * $\frac {1}{3}$ = $\frac {1}{18}$ .

II) Jim’s cards have a 1/2 probability of turning up one the same and two different from Slim’s, with the length-two cycles (A2),(23) or (A3). At the turn where the two previous players’ cards were the same, Tim’s card would have to be different (the other turns don’t matter), and that would happen with a $\frac {2}{3}$ probability. So the joint probability of no match with this starting case would be $\frac {1}{2}$ * $\frac {2}{3}$ = $\frac {1}{3}$ .

III) Jim’s cards have a 1/3 probability of turning up all different from Slim’s—with the length-three cycles (A23) or (A32)—in which case Tim’s cards could come up in any order and there would be no three-way match. So the probability of a match not happening this way is $\frac {1}{3}$ * 1, so also $\frac {1}{3}$ .

The total probability of a three-way match not happening, since the three cases above are independent, is $\frac {1}{3}$ + $\frac {1}{3}$ + $\frac {1}{18}$ = $\frac {13}{18}$ . So the probability of the match happening is $\frac {5}{18}$ , and the answer, 5+18, is $\boxed{23}$ .

There are three possible cases for the first flip:

• Jim's and Tim's card will both match Slim's, with a $\frac{1}{9}$ probability. This is a 3-way match already. It doesn't matter what happens on the subsequent flips.

• Jim's and Tim's card will both be different from Slim's, with a $\frac{2}{9}$ probability. This leaves zero probability of a 3-way match on either of the subsequent flips.

• Exactly two of the three cards match. The possibility of this is the residual 1 – $\frac{1}{9}$ – $\frac{2}{9}$ = $\frac{2}{3}$ .

In the third case, the players hold two cards each, and within those three pairs there is only one matching set. There are 2 x 2 x 2 = 8 possible plays, and 2 of those include a 3-way match (on the second flip, or on the third flip). So, given the third case for the first flip, the chance of a subsequent 3-way flip is $\frac{2}{8}$ = $\frac{1}{4}$ .

Therefore the total probability of getting a 3-way match is the sum of this happening by the first case or the third case:

$\frac{1}{9}$ + $\frac{2}{3}$ • $\frac{1}{4}$ = $\frac{1}{9}$ + $\frac{1}{6}$ = $\frac{2}{18}$ + $\frac{3}{18}$ = $\frac{5}{18}$ .

The answer is therefore 5 + 18 = 23.