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Probability of 11 cards

 
 
Reply Sat 16 Nov, 2013 07:07 am
Hello, this problem is a little tricky for me. Can you help me?

I have a 11 cards deck composed of ten cards from 0 to 9 and one Joker. (that can represent any number from 0 to 9)

I pick two cards to make a two digit number, the first one will represent the tenths and the second one is the units. The Joker will be the same number as the other card that is picked, so if i draw a 6 and a Joker i get 66.

I am not sure of how many combinations do i have here. Do i have 100 or 110 combinations?. All the combinations of the Joker beign Card Nº1 appear in the other combinations with cards from 0 to 9.

The probability of picking the Joker is 1 /11 but the joker will not have a value until we pick the second card, so after picking the Joker it has 1/10 probability of being any number from 0 to 9.

Can i say that picking the Joker as first card doenst make any difference in the probabilities ? Picking the Joker first or second does change the probabilities?

The problem is that i just want to have 100 different results (0 to 99) and not 110.

I made a table with all the possibilities, the first column is Card Nº1, second column is Card Nº 2, third column is Result.

1 2 12
1 3 13
1 4 14
1 5 15
1 6 16
1 7 17
1 8 18
1 9 19
1 0 10
1 J 11

2 1 21
2 3 23
2 4 24
2 5 25
2 6 26
2 7 27
2 8 28
2 9 29
2 0 20
2 J 22

3 1 31
3 2 32
3 4 34
3 5 35
3 6 36
3 7 37
3 8 38
3 9 39
3 0 30
3 J 33

4 1 41
4 2 42
4 3 43
4 5 45
4 6 46
4 7 47
4 8 48
4 9 49
4 0 40
4 J 44

5 1 51
5 2 52
5 3 53
5 4 54
5 6 56
5 7 57
5 8 58
5 9 59
5 0 50
5 J 55

6 1 61
6 2 62
6 3 63
6 4 64
6 5 65
6 7 67
6 8 68
6 9 69
6 0 60
6 J 66

7 1 71
7 2 72
7 3 73
7 4 74
7 5 75
7 6 76
7 8 78
7 9 79
7 0 70
7 J 77

8 1 81
8 2 82
8 3 83
8 4 84
8 5 85
8 6 86
8 7 87
8 9 89
8 0 80
8 J 88

9 1 91
9 2 92
9 3 93
9 4 94
9 5 95
9 6 96
9 7 97
9 8 98
9 0 90
9 J 99

0 1 1
0 2 2
0 3 3
0 4 4
0 5 5
0 6 6
0 7 7
0 8 8
0 9 9
0 J 0

J 1 11
J 2 22
J 3 33
J 4 44
J 5 55
J 6 66
J 7 77
J 8 88
J 9 99
J 0 0
 
View best answer, chosen by ptkaisen
engineer
  Selected Answer
 
  3  
Reply Sat 16 Nov, 2013 08:11 am
@ptkaisen,
You have 100 combinations, 110 permutations. The possible problem is that the probability of a double digit number (11,22,etc) is 2x that of the other numbers. If you want exactly 100 equally probable numbers, you need two decks of 0-9 and you draw one from each.
ptkaisen
 
  1  
Reply Tue 19 Nov, 2013 03:40 am
@engineer,
Thanks for your help
0 Replies
 
Miss L Toad
 
  1  
Reply Wed 27 Nov, 2013 04:06 am
@engineer,
It would be simpler to only count the joker if it appears second.
Miss L Toad
 
  1  
Reply Wed 27 Nov, 2013 05:36 am
@Miss L Toad,
Or to avoid replacement only count the joker if it appears first.

An alternative is to discard the joker from the pack of 11 and select cards with replacement.
0 Replies
 
 

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