Odds for consecutive events

[joeg]

Hi everyone,

If I know that something will happen 25% of the time then it is obvious that the chance of this happening twice in a row are 0.25 * 0.25 = 0.0625, or 6.25% of the time. However if I know that the odds of something happening are 3-1 (25%) then is there a quick way of working out the odds of this happening twice in a row, without converting from odds to percentage and back again, working it out the long way I can tell that it is 15-1, but is there a simple relationship that will let me calculate that 15?

Thanks in advance

Joe.

[Mike Haven]

add 1 to the something to one (if you have 6 to 1, this gives you 7)

divide it into 1 (this gives you 1/7)

multiply it with the next similarly obtained fraction ( 1/7 x 1/7 = 1/49)

take one from the bottom number (49 -1 = 48)

use the number as your found "something to 1" odds

(the answer in our example is 48 to 1)

therefore if you wanted to see what the odds were for a 3 to 1 event happening three times in a row you would multiply 1/4 x 1/4 x 1/4 = 1/64 = 1 occurrence in 64 = 63 to 1 odds

[mmbt0ne]

After plugging out a few of these, it looks like you can take X:1 odds, and find (X:1)^2 odds by doing ((X+1)^2-1):1

i.e. 3:1
(3+1)^2-1 = (4)^2-1 = 16-1 = 15

[mmbt0ne]

Damn, I was feeling all smart and calculating and then I post and see you beat me.

[Mike Haven]

if that's happened once in 2500 posts, then for it to happen again we might need to stay here for 6,252,500 more [img]/images/graemlins/frown.gif[/img]

[jason1990]

[ QUOTE ]
After plugging out a few of these, it looks like you can take X:1 odds, and find (X:1)^2 odds by doing ((X+1)^2-1):1

i.e. 3:1
(3+1)^2-1 = (4)^2-1 = 16-1 = 15

[/ QUOTE ]

Might as well simplify (X+1)^2-1 to X^2+2X. So if it's n:1 to happen once, it's (n^2 + 2n):1 to happen twice.

3:1 becomes 9+6:1 = 15:1
5:1 becomes 25+10:1 = 35:1
7:1 becomes 49+14:1 = 63:1
etc.

This looks like the quickest and easiest way to do it. I remember a craps dealer once told me that the odds of two consecutive snake eyes were 900:1. He just squared the 30:1 payoff for snake eyes. Of course, I guess he didn't realize the true odds were 35:1, but even if they were 30:1, the odds of it happening twice would be 960:1. This seems a lot easier than computing (30+1)^2-1.

[Mike Haven]

the easiest and quickest way is to "add 1, multiply, and take off 1":

3:1 becomes 4 x 4 becomes 15:1
5:1 becomes 6 x 6 becomes 35:1
7:1 becomes 8 x 8 becomes 63:1

[jason1990]

Well, I suppose "easiest" and "quickest" are subjective, but for large values, I disagree.

30:1 becomes 30^2 + 60 becomes 960:1
2500:1 becomes 2500^2 + 5000 becomes 6255000:1

This is quicker and easier for me than computing 31 x 31 or 2501 x 2501. It's probably also less susceptible to errors, considering an earlier post in which it was claimed that 2500:1 becomes 6252500:1.

[Mike Haven]

2500 + 1 = 2501

2501 x 2501 = 6255001

6255001 - 1 = 6255000

6255000 - 2500 = 6252500

i said "more [posts]"

not "to 1"

why the vibes?

ps: you're probably right for larger numbers - i was thinking in relation to poker, where the numbers involved are usually fairly low - my bad - sorry to cause hassle - i'll get my coat