uk lotto odds ?

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exxos
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uk lotto odds ?

Post by exxos »

Was just wondering, if its 14 million to 1 to win the lotto... then what would be the odds in getting 3 numbers in exactly the same order as they are drawn aswell ?

Chris
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Post by bullis1 »

Break out the TI-83 dude.

I'll assure you that your odds aren't very good.
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Post by exxos »

Well was wondering exactly what the odds would be :)
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Post by bullis1 »

I know, but I'm not about to calculate it.

However, it's probably not a hard calculation at all.
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Post by Sarek »

calculate B = number of permutations of 6 balls (=6*5*4*3*2*1)
calculate A = number of permutations of the six in which any three balls appear in sequence.

Required Probability = A/B * Probability of winning the lotto.

Hope that gets you started. :)

suppose the drawn balls are 1,2,3,4,5,6 in that order.

The possible draw orderings with 1,2&3 in sequence:
1,2,3,x,x,x
x,1,2,3,x,x
x,x,1,2,3,x
x,x,x,1,2,3

xxx can be permuted in 6 ways: 456, 465, 645, 654, 546, 564

ie there are 6*4 = 24 ways of doing it with 1,2,3 in order.

Other sequences are: (2,3,4), (3,4,5), (4,5,6)

which gives 4 * 24 = 96 ways.

A/B = 96 / (6*5*4*3*2*1) = 2 / 15

So there is a 2/15 chance of having this outcome, on top of the odds of winning the lottery.

I hope. I now feel my reputation in fragile balance :?
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Post by Maartau »

Sarek wrote:calculate B = number of permutations of 6 balls (=6*5*4*3*2*1)
calculate A = number of permutations of the six in which any three balls appear in sequence.

Required Probability = A/B * Probability of winning the lotto.

Hope that gets you started. :)

suppose the drawn balls are 1,2,3,4,5,6 in that order.

The possible draw orderings with 1,2&3 in sequence:
1,2,3,x,x,x
x,1,2,3,x,x
x,x,1,2,3,x
x,x,x,1,2,3

xxx can be permuted in 6 ways: 456, 465, 645, 654, 546, 564

ie there are 6*4 = 24 ways of doing it with 1,2,3 in order.

Other sequences are: (2,3,4), (3,4,5), (4,5,6)

which gives 4 * 24 = 96 ways.

A/B = 96 / (6*5*4*3*2*1) = 2 / 15

So there is a 2/15 chance of having this outcome, on top of the odds of winning the lottery.
Yes... :) 8)

From a 1st glance you're correct :P
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Post by Sarek »

Maartau wrote: Yes... :) 8)

From a 1st glance you're correct :P
No I'm not Maartau! :lol:

I should know better. God I hate probability. There are overlaps in this method.


123xxx A, 234xxx B, 345xxx C, 456xxx D,
x123xx E, x234xx F, x345xx G, x456xx H
xx123x I, xx234x J, xx345x K, xx456x L
xxx123 M, xxx234 N, xxx345 O, xxx456 P

Overlaps are:
A+F 123456, 123465
A+K 123456
A+P 123456
B+G 234561, 234516
B+L 234561
C+H 345612, 345621
D+M 456123
E+J 512346, 612345
E+O 612345
F+K 123456, 623451
F+P 123456
G+L 134562, 234561
I+N 561234, 651234
J+O 162345, 612345
K+P 123456, 213456

So 15 permutations are duplicated:

a 123456 - A F K P
b 123465 - A F
c 234561 - B G L
d 234516 - B G
e 345612 - C H
f 345621 - C H
g 456123 - D M
h 512346 - E J
i 612345 - E J O
j 623451 - F K
k 134562 - G L
l 561234 - I N
m 651234 - I N
n 162345 - J O
o 213456 - K P

These 15 permutations (a-o) are seen to appear 34 times in the templates A-P, though we only need them once. Therefore 96 - 34 + 15 = 77 distinct permutations

Required Probability = 77/720

which is about 1 in 9 :)

This is beginning to remind me of those unpleasant competitions they put in PC Plus magazine.


I hope you've not posted your competition slip in yet Chris. ;)
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Post by Maartau »

@Sarek & Exxos : my bad if I took things to speed :oops: .

I haven't used proba. for many years from now : the best I can do is rewriting the whole by my own [better than a quick glance :wink: ].

[Promise next time "I don't open my mouth" :lol: :wink: ]
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