71 of 100: A Pair of Pairs?

We want to count how many pairs are remaining in the deck after Amy's cards are given in Scenario 1 and Scenario 2. Since the number of cards remaining is identical in each scenario, the deck with the most pairs will be the most likely to deal Blake a pair.

We can also ignore all the cards other than the queens and 5s; since there are four of every other type, they will provide an identical number of pairs in both scenarios.

Scenario 1 consists of two queens and four 5s. We can make 1 pair from the former and 6 pairs from the latter. (In combinatorics terms, because 4 choose 2 = 6.)

Scenario 2 consists of three queens and three 5s. We can make 3 pairs from the former and 3 pairs from the latter. (In combinatorics terms, 3 choose 2 = 3.)

Since Scenario 1 allows $1 + 6 = 7$ pairs and Scenario 2 allows $3 + 3 = 6$ pairs, Scenario 1 is more likely to deal out a pair!

There are total $13\times \binom{4}{2}=78$ pairs in a standard, 52-card deck.

In scenario 1, there are $73$ remaing pairs except 5 pairs which contain red Queens.

In scenario 2, there are $72$ remaing pairs except 3 pairs which contain Queen of hearts and 3 pairs which contain 5 of spades.

So, scenario 1 is more likely that Blake has a pair.

The probability of getting pairs of As, Ks, 2s, 3s and other cards is the same in both scenarios, so we can isolate them, and imagine we have a deck of only Qs and 5s and compare the scenarios. In the first scenario, Blake can have a pair if he draws the two remaining Qs or a combination of the four remaining 5s (6 combinations of pairs). So, we have 7 possible combinations of cards. In the second scenario, Blake can have a pair if he draws a pair out of the three remaining Qs (3 possible pairs) or out of the three remaining 5s (also 3 possible pairs). It's a total of 6 combinations of cards. So, in the first scenario it is more probable that Blake will have a pair.

Just thinking based off of logic, since scenario 1 shows a pair established. That means that Blake has n-1 possible pairs from n possible pairs.

However, scenario 2 shows not a pair of cards, and we know that since a queen and five are not pairs, their respective ranks (I.e. other queens and fives) can each only have one pair, decreasing the number of possible pairs to n-2 pairs.

Therefore, scenario 1 is more likely to get a pair from Blake.

Note: It does not matter the rank of cards, as long as one scenario shows a pair and the other doesn't.

The difference between the 50 unknown cards of the two scenarios are:

• Scenario 1: $\boxed{Q^\clubsuit} \boxed{Q^\spadesuit} \boxed{5^\diamondsuit} \boxed{5^\heartsuit} \boxed{5^\clubsuit} \boxed{5^\spadesuit}$ and the number of pairs possible from them $\displaystyle p_1 = {2 \choose 2} + {4 \choose 2} = 1 + 6 = 7$ .

• Scenario 2: $\boxed{Q^\diamondsuit} \boxed{Q^\clubsuit} \boxed{Q^\spadesuit} \boxed{5^\diamondsuit} \boxed{5^\heartsuit} \boxed{5^\clubsuit}$ and the number of pairs possible from them $\displaystyle p_2 = {3 \choose 2} + {3 \choose 2} = 3 + 3 = 6$ .

Therefore, in scenario 1 Blake is more likely to have a pair.

In the first case, one pair is occupied and hence Blake can't make one pair.

In the second case, one pair of a queen is disturbed and one pair of 5 is disturbed. So, 2 pairs are not possible

Hence, in scenario $\boxed {1}$ , it is more likely that Blake has a pair.

In the first case, there are 4C2 ways of blake selecting a pair from all the suits except the queens.There is only 1 way of choosing 2 queens from the remaining 2 queens. So the likelihood of blake having a pair in the first case is : (12*(4C2) + 1)/(50C2) = 73/(50C2)

In the second case, there are 11 suits with 4 cards remaining and 2 suits (Queen & 5) with 3 cards remaining. So, the the likelihood of blake having a pair in the second case is : (11 (4C2) + 2 (3C2))/(50C2) = (66+6)(50C2) =72/(50C2).

So, it is more likely that blake forms a pair in the 1st case

In comparing the two scenarios, the cards that Amy already has only affect the probabilities of Blake getting a pair of queens and a pair of 5s, therefore one need not be concerned about the probabilities of Blake getting pairs of other types of cards.

Hence, we only need to look at the probabilities of Blake being dealt a pair of queens and a pair of 5s for each scenario.

Scenario 1: P(a pair of Qs or a pair of 5s) = (2/50)(1/49) + (4/50)(3/49) = 14/2450

Scenario 2: P(a pair of Qs or a pair of 5s) = (3/50)(2/49) + (3/50)(2/49) = 12/2450

It is now obvious Scenario 1 has the higher probability.

Here is a way to calculate the numerical values:

finding a pair in case QQ--> (48/50) (3/49)+(2/50) (1/49)=0,059376-->5,9376%

finding a pair in case Q5--> (44/50) (3/49)+(6/50) (2/49)=0,05877-->5,877%

explanation:

48/50-your first card is nonQ

3/49- your second card is it's pair

2/50- your first card is Q

1/49 your second card is the last Q

44/50- your first card is nonQ and non5

3/49- your second card is it's pair

6/50- your first card is Q or 5

2/49- your second card is it's pair

In scenario 1, there are 2 queens left for Blake to choose, which is a possibility of 1 pair, which earlier was 6 possible pairs, so a difference of 5 pairs.

In scenario 2, there are three more queens, so there are 3 possible pairs for what you get. The same goes for the 5. There were originally 6 pairs for queens, but now there are 3. The same goes for the 5s. In scenario 2, you have eliminated 6 possible pairs, as for scenario 1 you have eliminated 5 possible pairs.

Thus scenario 1 has a higher probability of a pair.

Number of ways of getting a pair in $Scenario 1$ :

$\begin{aligned} & =\binom{12}{1}\binom{4}{2}+\binom{2}{2} \\ &=12\times 6 + 1 \\ &=73\end{aligned}$

Number of ways of getting a pair in $Scenario 2$

$\begin{aligned} & =\binom{11}{1}\binom{4}{2}+\binom{3}{2}+\binom{3}{2} \\ &=11\times 6 + 3+3 \\ &=72\end{aligned}$

Thus, it is more likely that Blake has a pair in the first scenario.

First of all, ignore whether or not Amy has a pair. It's completely irrelevant. Second, you must figure out how many possible pairs are there: For every number in the suit, there are 6 pairs (DH, DC, DS, HC, HS, CS), so that's a total of 78 pairs in a standard deck. The total number of possible combination of cards is also irrelevant, as that remains the same across scenarios (for the record, it's 1326) Third, figure out how many pairs are no longer possible given Amy's hand.

Scenario 1 removes QDH, QDC, QDS, QHC, and QHS. 5 pairs are no longer possible. 73 remain. Scenario 2 removes QHD, QHC, QHS, 5SD, 5SH and 5S. 6 pairs are no longer possible. 72 remain.

Therefore, Scenario 1 is more likely to have Blake play a pair. I hope that makes sense.

Since it’s two different cards in scenario 2 that means that’s another pair you can’t get

Imagine the deck is 4 cards 2 5's and 2 Q's. If Amy has 2 Q's than Blake has 2 5's, and if Amy has a Q and 5 than Blake has a Q and a 5. There are more cards in a 52 card deck and the odds are less drastic, but the principal is the same. Because Amy has 2 cards of the same value Blake is more likely to also get 2 cards of the same value, but the more valuable question is who has the better odds of winning the hand. That would be Amy as she has a pair of Q's and even if Blake gets a pair odds are it is less valuable than Amy's.

that's why in TX hold em', you should alway bet low when you get a pair cuz there's a higher chance someone else also have pair. mike drop

If amy had two different cards, there would be two pairs of cards that amy has one half to. Due to this, the first scenario is correct.

I think it is because... If Amy already had a pair, then that leaves all other pairs untouched.

If Amy had two different cards, then it takes out two different pairs.

Say we only had 10 cards. If Amy took a pair, then all 8 could end up with something they match. If Amy had different cards, 2 cards (The twins of the ones she took out) in your 8 cards would be useless, having no partner and lowering your chances of getting a pair.

Hope you find this useful.

In a standard deck of cards there are 4 suits, each with 13 cards in it, so the number of possible pairs is 26. If Amy has a pair it takes one of those possibilities away, but if she has two different cards, it takes the possibility for two pairs away. Because either of the cards she has can't be part of a pair that Blake could get.

so lets say you have 10 fruit, 2 of each kind if you take the 2 apples, you'll have 4 pairs instead of 5 but if u take 1 apple and 1 orange, the other apple and orenge are useless because they don't have a pair so your left with 3 pairs

The number of pairs in scenario 1 = 48 3 + 2 1 = 146. Blake can draw 48 non-Q cards that have 3 possible matches. Blake can also draw one of the 2 remaining queens remaining but then must draw the last queen.

The number of pairs in scenario 2 = 44 3 + 6 2 = 144. Blake can draw 44 non-Q & non-5 cards that have 3 possible matches. Alternatively, there are 6 queens/fives remaining (3 each) and Blake must draw one of the remaining 2 matches after drawing a queen/five.

Blake has more ways to draw a pair in scenario 1 and the number of possible hands in both scenarios is the same.

I answer scenario 1.

It is probably not the fastest method, but I calculcated the probability. Here is some quick maths.

Scenario 1, After Amy's draw there are 50 cards left in the deck. There are two Queens remaining in the deck. If you draw a Queen, there are 49 cards left and there is a $\frac {1}{49}$ eq. 2.04% chance you get that last Queen to make your pair. If it is not a Queen, there is a $\frac {3}{49}$ eq. 6.12% chance you get a pair. Considering there was only a $\frac {2}{50}$ chance you get that Queen and a $\frac {48}{50}$ chance you get a start on a fresh pair you can multiply the fractions resulting into $\frac {48}{50}\ \times \frac {3}{49}\ + \frac {2}{50}\ \times \frac {1}{49}$ which is roughly 5.96%

Scenario 2, Quite, similar to scenario one but this time there are three Queens and three Card-5 in the deck, which means you have a $\frac {6}{50}$ chance to get either a Queen or Card-5 and $\frac {44}{50}$ chance for that start on a fresh pair. Multiplying the fractions once more: $\frac {44}{50}\ \times \frac {3}{49}\ + \frac {6}{50}\ \times \frac {2}{49}$ which is roughly 5.88%

Conclusion There is a 0.08% higher chance to get a pair following Scenario 1.

If Amy's cards form a pair, it will eliminate the chance that Blake got that pair, giving him a higher chance of getting a pair. But if Amy had two cards that do not form a pair, Blake has a chance to get the cards that Amy has, reducing the chance of him getting a pair

This won't be good for some sort of proof, but hopefully it communicates things intuitively. In scenario 1, Amy has an entire pair. This leaves Blake with no unmatchable pairs. However, in scenario 2, Amy has two unmathched cards, leaving Blake with two unmatchable cards to potentially be dealt. Thus, it is more likely for Blake to get a pair in scenario 1.

To be very honest, i'm a high school student and I study probabilities in maths but for this one I went with logic more than probability. So if that person had a fixed card, the queen in this case, means that for me to get a pair will imply that I shouldn't get a queen isn't it? So the case where she gets her queen pair will give me even more chances of completing my pair and in the same time not getting the queen she missed.

In scenario 1, Blake has 2/50 probability of drawing a Queen and 48/50 probability of drawing a non-queen. If Blake draws a Queen, then the probability of drawing the other queen is 1/49. If Blake draws a non-Queen, then the probability of drawing a matching non-Queen is 3/49. So, the overall probability of drawing a pair in scenario 1 is 2/50*1/49 + 48/50*3/49.

Meanwhile, in scenario 1, Blake has 6/50 probability of drawing a Q or a 5, and 44/50 probability of drawing another card. If Blake draws a Q or a 5, then the probability of drawing a matching card is 2/49; otherwise, the probability of drawing a matching card is 3/49. So, the overall probability of drawing a pair in scenario 2 is 6/50*2/49 + 44/50*3/49.

So, scenario 1 = (2 * 1 + 48 * 3) / (50 * 49) = 146 / 2450, and scenario 2 = (6 * 2 + 44 * 3) / (50 * 49) = 144 / 2450, which is smaller.

52 total cards and 26 possible pairs. when we know Amy has 1 pair : there are 50 cards and 25 pairs available. When Amy has queen and 5: there are 50 cards and 24 pairs available. Therefore it is more likely to have a pair if Amy has a pair.

This doesn't need any calculation as the pairs possible in Scenario 1 is more than Scenario 2 because in Scenario 2 we have already wasted two probable pairs whereas in Scenario 1 there is only 1 pair wasted and rest are available

For each number in the deck there are 6 possible pairs. In scenario 1 there are 12x6 = 72 + 1(for queens) = 73 possible pairs. Scenario 2, there are 11x6 = 66 + 3(for queens) + 3 (for 5s) = 72. So more chances in scenario 1.

Both cases have the same probability to get a pair on 11 cards that are not queen or five. So we only need to check on Queen and Five cases for comparison,

Scenario 1: C(2,2) + C(2,4) = 7 Scenario 2: C(2,3) + C(2,3) = 6

Therefore Scenario 1 is the answer

There are 13 different face values with 4 different suits, so we can have 6 different combinations of pairs for each value (4C2 = 6) and 78 possible pairs of cards with same value (6 * 13 = 78). Now if Amy has got a pair then there remains 77 possible pairs for Blake on the other hand if Amy has got two different valued cards then remaining cards have 76 possible pairs

In scenario 1, Blake's first card can be anything, and then his second card must be of the same rank as his first, so if it was not a queen ( $\frac{24}{25}$ probability), there is a $\frac{3}{49}$ probability of him getting a pair. Otherwise ( $\frac{1}{25}$ ), there would be a $\frac{1}{49}$ chance of him getting a pair. So the overall probability would be $\frac{24}{25}\cdot\frac{3}{49}+\frac{1}{25}\cdot\frac{1}{49} = \frac{73}{25\cdot49}$ In scenario 2, If Blake's first card is of rank Queen ( $\frac{3}{50}$ ), his probability of getting a pair is $\frac{2}{49}$ . This is the same for the case of getting a card of rank 5. If he gets something else ( $\frac{22}{25}$ ), his probability of getting a pair is $\frac{3}{49}$ . So this total probability would be $2\cdot\frac{3}{50}\cdot\frac{2}{49} + \frac{22}{25}\cdot\frac{3}{49} = \frac{72}{25\cdot49}$ . Because $\frac{73}{25\cdot49} > \frac{72}{25\cdot49}$ , Scenario 1 has a higher chance of Blake getting a pair.

The probability of getting a pair is the number of possible pairs divided by the number of possible two-card combinations (including both pairs and non-pairs).

The size of the deck is the same in both scenarios, and this is all that determines the number of possible two-card combinations. So the question "In which scenario is the number of possible pairs divided by the number of possible two-card combinations greater?" reduces to "In which scenario is the number of possible pairs greater?"

The only pairs that are not possible in a scenario are the ones containing a card that Amy drew.

In Scenario 1, Amy's cards eliminate five pairs from Blake's possibilities: QH QC, QH QS, QD QC, QD QS, and QH QD.

In Scenario 2, Amy's cards eliminate six pairs: QH QC, QH QS, QH QD, 5S 5H, 5S 5D, and 5S 5C.

Since Scenario 1 eliminates fewer pairs, it leaves more possible pairs for Blake.

I listed out all the possible face pairs that Blake could get in each scenario and found that scenario 1 had one more possibility of getting a pair than scenario 2, making it more likely he would have a pair.

Scenario 1: 2-2, 2-2, 3-3, 3-3, 4-4, 4-4, 5-5, 5-5, 6-6, 6-6, 7-7, 7-7, 8-8, 8-8, 9-9, 9-9, 10-10, 10-10, J-J, J-J, Q-Q, K-K, K-K, A-A, and A-A or 25 possible pairs because it is missing the other Q-Q pair.

Scenario 2: 2-2, 2-2, 3-3, 3-3, 4-4, 4-4, 5-5, 6-6, 6-6, 7-7, 7-7, 8-8, 8-8, 9-9, 9-9, 10-10, 10-10, J-J, J-J, Q-Q, K-K, K-K, A-A, and A-A or 24 possible pairs since it still misses the other Q-Q pair, and also a 5-5 pair.

Interesting to note, before making any lists, originally my gut thought that scenario 1 would be better since only 1 whole pair is already taken out by Amy. Perhaps meaningless, but also interesting to note that in both scenarios there are (at least) 24 possible pairs Blake could get... I wonder if this means anything? ...or what would happen if we were playing with triples instead of pairs?

In scenario 1 Amy Leaves say n pair of similiar cards. Whereas in scenario 2 Amy leaves n-1, keeping the sample space same for both cases. Hence there are less pair to choose in scenario 2 and so the answer.

This is more like an obervation problem rather than a probabilty although it can be solved in both ways. First thing first let us compare both scenarios. In both the number of total possible cases is the same 2450. But notice this: in the first scenario Amy was dealt a pair which means that Blake has 25 more pairs available. In the 2nd case Amy screwed up and didn't get a pair. This leaves Blake, to 24 more pairs available. Beacuse instead of forming 2 pairs of 5 like in 1st scenario, he can now form only 1 pair. So in the 1st scenario Blake has more chances of succes.

Each rank, X, has 6 ways being part of a pair. X1-X2, X1-X3,X1-X4, X2-X3, X2-X4, X3-X4. (where the numbers distinguish suit)

Removing two of them reduces this to one pair, e.g. X3.X4, five pair possibilities are gone.

Removing just one leaves three pairs. e.g. X2-X3, X2-X4, X3-X4. three pair possibilities are gone.

Removing two cards of two different ranks the removes three pair possibilities twice, six overall.

The scenario with a pair showing has one more possible pair than the one with two unmatched cards showing.

Therefore the scenario when a pair is visible has a better chance that the two down cards are a pair.

Let the number of pairs possible be x (rather than work out the actual total)

Scenario 1 = x-1

But because Scenario 2 shows two different cards, that means two opportunities for pairs have been eliminated

So Scenario 2 = x-2

x-1 > x-2 therefore it is more likely for Blake to have a pair in Scenario 2

I think the question should say something like "Blake has a pair of the same card " - otherwise my first thought was "a pair of what, cards?", to which the answer is 100% in both scenarios.

We don't need to calculate the full probabilities but only need to compare the Q and 5 cards.

In the first scenario there is 5C (5 of clubs), 5H (hearts), 5D (diamonds), 5S (spades).

Blake can have:

5C & 5H

5C & 5D

5C & 5S

5H & 5D

5H & 5S

5D & 5S (Note that the order doesn't matter.)

There is only one way that Blake can get a pair of Queens:

QC & QS.

Altogether, scenario 1 has 7 ways that Blake can get a pair just using Q or 5.

In the second scenario there are only three ways for Blake to get a pair using the 5s:

5C & 5H

5C & 5D

5H & 5D

There are also three ways to get a pair of Queens, making 6 ways in total.

Scenario 1 has 7 ways (Blake's 7) compared to 6 ways in scenario 2, so scenario 1 is more likely.

Case 1: A PAIR made increases the chance of creating another PAIR. Case 2: A PAIR broken means it breaks the chance of creating another PAIR.

So, getting a PAIR at case 1 is more likely.