[daneault23]

2 cards are drawn without replacement from a standard 52 card deck. What is the probability that the 1st card is larger than the 2nd card?

[stuntin909]

50%

[rm18]

24/51?

[Inkwell77]

Is the Ace worth 1?

[stuntin909]

without knowing the value of the first card you can't make any sort of accurate judgement. Sometimes the first card will be a 2 and sometimes it will be an Ace. Obviously sometimes you realistically have 0% (when it's an A) but 100% when it's a 2. After you do the math for each scenario you should end up at 50%.

Thats my theory. Too lazy to do math right now.

Forgot to add, you'll have to subtract the 3 other cards

Actually my answer is 46%

Pick a random card and subtract the 3 other similar cards of the same value. Leaves you with 48 left. From here on out you have a 50% chance in the long run. 24/52 = 46%

Am i right?

[daneault23]

Yes the ace is worth one.

[SBR_John]

50% assuming the first card selected is selected at "random"

[DiggityDaggityDo]

Is this algebra??? Fukk, my brain hurts!!!

[daneault23]

and Stuntin, you have to remember when you pick ur first card that you have 3 other cards that are equal value, thus it can't be 50%

[rm18]

it is lower than 50 because both cards can be the same

[stuntin909]

read my last post dane

[MeatWad]

Also when you pick the lowest card 1st your chance is 0% when you pick the highest card 1st your chance is only 48/51. The average card 7, is 24/51, but you have to factor in the top and bottom points.

[hokim]

312/(51*13) = 0.470588235...

[daneault23]

Stuntin, your close I think but I believe hokim is correct. Not gonna lie men, this was one of my HW problems for my Probability Theory class and I couldn't figure it out after doing hours of other HW so I figured I'd ask here.

Hey Hokim, where did the 312 come from?

[NYSportsGuy210]

I agree with Stuntin's answer. It all depends on what the first card is.

[MeatWad]

yeah 47.05882% is correct, the top and bottom points average cancel each other out to the mean, not sure why I thought they were unique data points.

[stuntin909]

I also want to know where the 312 is coming from...

[MeatWad]

All outcomes are evenly distrubuted around the mean. The mean is 24/51=.4705882

[stuntin909]

Aww damn I was on the right track. I should've divided by 51 not 52. Thats why i got 46%

I got it figured out now.

[EmpireMaker]

312/(51*13) = 0.470588235...

you didn't account for ties

[Inkwell77]

This is probably wrong, but the 312 could be from the average of the cards, 6.538
And then multiply that 6.538*48 cards, but that is 313, so I'm not sure and I don't think it is right.... (oh wait 6.5 times 48 is 312)

[NYSportsGuy210]

Another way of looking at it is initially suppose you choose an Ace one the first card (the lowest card in the deck).

The odds of the second card being picked as smaller is 0/51.

If you picked a king on the first card (the highest card in the deck) the odds of you picking a smaller card in the deck a second time around is 48/51.

What meatwad did was take two probabilities and got their average which is (0/51) + (48/51) = (48/51)/2

= 24/51.

[Blax0r]

This is probably wrong, but the 312 could be from the average of the cards, 6.538
And then multiply that 6.538*48 cards, but that is 313, so I'm not sure and I don't think it is right.... (oh wait 6.5 times 48 is 312)

This is where the "312" comes from:

You are drawing two cards from a 52-card deck w/o replacement. This means you will have 52 * 51 possible combos.

We want to sum up all possible combos that meet the "1st card larger than 2nd card" criteria. Let's start listing some:

A 2, A 3, A 4, ... , A K --> 4 (aces) * 48 (other cards greater than A)
2 3, 2 4, 2 5, ... , 2 K --> 4 (twos) * 44 (other cards greater than 2)
3 4, 3 5, 3 6, ... , 3 K --> 4 (threes) * 40 (other cards greater than 3)

There's a pattern here; we're summing the products for 4 and multiples of 4 (up to 48). So the total number of combos that fit our criteria is:

4*48 + 4*44 + 4*40 + ... + 4 * 4 --> 4 * (48 + 44 + 40 + ... + 4).

So now we simply take this number and divide it by the total number of combos (52 * 51):

4 * (48 + 44 + 40 + ... + 4) / (52 * 51)

(48 + 44 + 40 + ... + 4) / (13 * 51)

312/(13*51)

You can simplify this further, but hokim left it at this point. This is how I would solve the problem; it's straightforward and doesn't require any complicated logic. Hope this helps.

[MeatWad]

I prefer my method, much easier to calculate and apply in practical situations. I never liked memorizing formulas, preferred to understand the principles behind the formulas. But combination and permutation formulas are useful tools for probability problems of this type.

[stuntin909]

My way.

52 cards in deck. Pick 1 which eliminate the 3 others of similar value. This gives you 48 cards left. At this point the original premise we believe of 50% probability kicks in. This gives you 24. Then divide that 24 by 51 (since you pulled a card already) and get 47%