Probabilities Question

Can anyone help me figure the following probabilities questions?

Using a standard 52 card deck of playing cards, a player turns one of the cards face-up, the card is a 2(suit is irrelevant).
What are the probabilities that the next card turned face-up will not be a 2 card?
What are the probabilities that the third card turned face-up will not be a 2 card or the last card turned face-up?
What are the probabilities for the fourth, fifth and so-on cards turned face-up, without repeating one of the cards previously turned face-up, until all 13 different cards are turned face-up, out of 13 cards turned face-up?

I'm thinking the way I'm figuring it out is too easy, so it has to be wrong...

Thanks

Some of them don't require difficult math.

I get 1 in 123,011.71 or 0.0000081293 for the last question. I solved it two different ways, so I'd hope it would be right. :) This is the probability that a 2 comes first, and then the other 12 ranks are dealt in the next 12 cards. If you got that right, you likely got them all.

well the 2 doesn't have to come first, I guess what I'm trying to figure out what are the probabilities that you can draw up to 13 cards without repeating a card by value. And all the probabilities along the way, drawing two cards without repeating, then three and so on all the way to 13 cards being drawn. My number is no where close to yours. I thought I might be figuring it wrong.. Thanks..

Quote: allinriverking

well the 2 doesn't have to come first, I guess what I'm trying to figure out what are the probabilities that you can draw up to 13 cards without repeating a card by value. And all the probabilities along the way, drawing two cards without repeating, then three and so on all the way to 13 cards being drawn. My number is no where close to yours. I thought I might be figuring it wrong.. Thanks..

Well, the way the question was written, I assumed it was building on the previous parts. If any card can be first, then it would be: 1 in 9462.44 or 0.000105681. This is to get all 13 cards of different ranks.

I feel like the question is a bit unclear on what it wants, unfortunately.

I'm sorry... I re=read the way I worded it...

this is what I figured, but I don't know if the other numbers are right, I calculated drawing the next card and when a repeating number is drawn it stops the draw. So I calculated it that way... Do you have same numbers?



3954242643911240000000
417888291992371000
0.0001056810

3954242643911240000000
3760994627931340000
0.0009511

98856066097781000000
430947301117133000
0.0043593

2411123563360510000
32647522811904000
0.0135404

57407703889536000
1836423158169600
0.0319891

1335062881152000
81618807029760
0.0611348

30342338208000
2975685672960
0.0980704

674274182400
91092418560
0.1350970

14658134400
2372198400
0.1618349

311875200
52715520
0.1690276

6497400
988416
0.1521248

132600
14976
0.1129412

2652
156
0.058824

Well, the top line, I have the same answer, so I guess so.

For 13 cards of different rank it's just:

(52/52)*(48/51)*(44/50)*(40/49)*(36/48)*(32/47)*(28/46)*(24/45)*(20/44)*(16/43)*(12/42)*(8/41)*(4/40) = 1 in 9462.44

or

4^13/C(52,13) = 4^13 / [52!/13!/39!] = 1 in 9462.44 = 0.000105681