Does It Matter?

Just for the sake of keeping track let's assume Amy has already drawn a pair of 4s. So we have 50 cards left: having 2 4s and rest 12x4 cards.

Now its Blake's turn & we have 2 cases,

1. Blake draws a 4 first, then again a 4

The probability of this is $\frac{2}{50} × \frac{1}{49}$

2. Blake draws anything except a 4 & then again the same number to form a pair

The probability of this is $\frac{48}{50} × \frac{3}{49}$

So the total probability is $\frac{144+2}{49×50} = \frac{1}{16.78}$

As there are $\binom{4}{2} = 6$ pairs within each of the $13$ denominations, there are a total of $13*6 = 78$ different pairs that can be drawn. As Amy has a pair of some denomination, there are just $2$ cards, and hence just one pair, left within this denomination, and so there are still $12*6 + 1 = 73$ pairs left to be drawn out of the $50$ remaining cards. Thus the probability that Blake draws a pair given that Amy has drawn a pair is

$\dfrac{73}{\binom{50}{2}} = \dfrac{73}{25*49} \approx \dfrac{1}{16.78} \gt \dfrac{1}{17}$ .

So, somewhat counterintuitively, Blake's chance of having drawn a pair is greater than Amy's, given that Amy did in fact draw a pair.

Amy has already been dealt a pair. The probability that the first card dealt to Blake shall be of the same type is $\frac{1}{25}$ . Only 1 out of 49 left cards could form a pair for Blake : $\dfrac{1}{25}*\dfrac{1}{49}$

The probability that the first card dealt to Blake shall NOT be of the same type is $\frac{24}{25}$ . Only 3 out of 49 cards could form a pair for Blake: $\dfrac{24}{25}*\dfrac{3}{49}$

And all in all: 24/25 3/49+1/25 1/49=73/(25*49) So all in all we have: $\dfrac{1}{25}*\dfrac{1}{49}+\dfrac{24}{25}*\dfrac{3}{49}=\dfrac{73}{25*49}= \approx0.05959\gt \approx0.0588=\dfrac{1}{17}$

So the answer is: $\fbox{Greater than 1/17}$ .

What is the probability to have a pair?

In total there are ${52\choose 2}$ ways to get 2 cards.

There are 4 cards of each rank and 13 different suits, so there are ${4\choose 2}*13$ ways to get a pair.

$P(Pair_{Blake})=P(Pair_{Amy})=\frac{{4\choose 2}*13}{{52\choose 2}}=\frac{1}{17}\approx 0.0588$

Given that Amy is dealt a pair, what is the probability that Blake has a pair?

If we know that Amy has a pair there are ${50\choose 2}$ ways to get 2 cards with ${4\choose 2}*12$ ways to get a pair of a different rank and ${2\choose 2}$ ways to get a pair of the same rank.

$P(Pair_{Blake}|Pair_{Amy})=\frac{{4\choose 2}*12+{2\choose 2}}{{50\choose 2}}=\frac{71}{1225} \approx 0.0596$

Solution:

$\frac{1}{17}\approx 0.0588 < \frac{71}{1225} \approx 0.0596$

Consider the events: A:<<Amy was dealt a pair>> and B:<<Blake was dealt a pair>>

Given that Amy is dealt a pair, the probability that Blake has a pair=p(B|A)= $\frac{p(AandB)}{p(A)}$ = $\frac{p(A|B).p(B)}{p(A)}$ = $\frac{p(A|B)}{p(A)}$ x $\frac{1}{17}$ , with p(A|B)>p(A), so probability required > $\frac{1}{17}$ .

Remove 2 more aces, and you've removed all aces. It's as if you started with only 12 different values, instead of 13. The fewer values, the more likely to get a pair.