Five Cards, Two Draws

Every possibility for the first and second draw is shown above. "x" marks are not possible (since it would be the same card twice). Checkmarks are where the second card is larger than the first. Of the 20 possible spaces, 9 of them have the second card be larger, so the probability the second card is larger than the first is $\frac{9}{20} = 45\% .$

Well, the first card being greater than the second is symmetric to the first card being less than the second, and since there exists a possibility where the two numbers are the same, we must have the probability the second card is greater than the first is less than 50%.

Since the probability of drawing a greater number second is the same as drawing a smaller number, they have equal probabilities. Also, we know that you can draw an equal amount with two fours, so the chance of drawing a greater card second would be half of 1-P(same card), so it would be 1/2-something, so it would be less than 50%.

There are 5 independent choices of the 1st card.

Once the first card is chosen we have 4 cards to choose from.

Therefore we have 4 * 5 = 20 choices for the second card (while we always choose from the set of 4 cards, these sets are always different by one member and can be seen as different "universes", e.g. universe in which 1st card is "8", the universe of the 1st "6" and so on).

Now, the chance to take a greater card then the first one is:

There are two "4" cards, so the sum of successful choices is 3*2 + 2 + 1 = 9 .

Which means that the chance to get a greater 2nd card is 9 out of 20 or 0.45 .

Number of ways to select two cards out of five is $C_{2}^{5}$ . Given a pair of cards, there's only one way to arrange them so that second is greater than the first with the exception of a pair 44. Hence, the probability is $\dfrac{C_{2}^{5}-1}{2\cdot C_{2}^{5}} = 45\% < 50\%.$

We add up the percentages for each card being drawn out and then divide by 5 to find the average: if 4 then 75% x 2 = 150%, 6 then 50% if 7 then 25% and if 8 then 0% then we get 225/5 = 45% hence the correct answer less than 50%.

There are 2 number 4 cards. Probabilty of choosing number 4 cards: 2/5

For second card, probability of choosing card that bigger than number 4: 3/4

Probability is 2/5*3/4=6/20

By this way, total probability is

2/5 * 3/4 + 1/5 * 2/4 + 1/5 * 1/4 = 9/20

if you pick a 4 you have 75% chanches of takeing a greater number if you pick a 5 then 50% a 6 25% and an 8 you have 0% so

(75×2+50+25+0)÷5=45% so less than 50%

There are 5 cards and only 2 of them are the smallest number

this means= 5/2 = %40

You have a 20% chance of getting each one of those cards on your first pick which totals a 40% chance to get a 4 then a 20% chance for 6, 7 and 8. For a 4 you have a 75% chance of getting a card higher than 4, for a 6 you have a 50%, for a 7 you have a 25% and an 8 you have 0%. By multiplying the % chance of you getting x card number by the percent chance of getting a second card bigger than it, for example for a 6 there is a 20% chance of getting it times a 50% chance of getting a card bigger than it so a 10% total chance, than add all of the percentages together you get 45%, which is less than 50% =>QED.

Each card has a 25% percent of being drawn, therefore its less than 50%

The first drawn card has a probability of occurrence of 20% (1/5=0.2). If we consider each case separately we get:

• 1) The probability that the first card which will be drawn is 8 is 0.2 (as noticed above). The probability that the next card is higher than 8 is simply 0. So P1=0.2*0=0

• 2) The probability that the first card which will be drawn is 7 is 0.2 (as noticed above). The probability that the next card is higher than 7 is simply 0.25 as only card 8 satisfies the requirement (1 card out of remaining 4 so 25%) . So P2=0.2*0.25=0.05

• 3)The probability that the first card which will be drawn is 6 is 0.2 (as noticed above). The probability that the next card is higher than 6 is simply 0.5 as only cards 7 & 8 satisfy the requirement (2 cards out of remaining 4 so 50%) . So P3=0.2*0.5=0.1

• 4)The probability that the first card which will be drawn is 4 is 0.4 (2 cards with number 4). The probability that the next card is higher than 4 is simply 0.75 as only cards 6,7 & 8 satisfy the requirement (3 cards out of remaining 4 so 75%) . So P4=0.4*0.75=0.3.

So in total: P1+P2+P3+P4=0.05+0.1+0.3=0.45 (45%)

I hope it makes sense ! :)

The cards that can have a greater amount on them for the second draw is 4,6,7 so the first probability is 3/5 and there are three cards that can greater but without replacement 3/4 giving: (3/5)*(3/4) = 9/20

For having a first reference, we could calculate how many sets of cards can be shown, so there are 5 different cards that can be taken in pairs, this is 5*4=20, 5 represents the first card (any of them), and 4 represents the second card (because it's not possible to repeat one of them), now we can see how many sets are not favorable (the first is greater). There are 5 non-favorable sets which begin with an 8, 4 which begin with a 7, 3 which begin with a 5... But... Wait we've found out 12 non-favorable cases, and that's greater than 50%

Probability of getting 4 itself as the second card is 50 %, if the first card is from 6, 7 or 8 which also is more than 50%.

There is a small possibility that the first two cards are the same (i.e. both $4$ ), then either one is bigger will share equal possibilities, so the answer is Less Than 50%.