Math Question

[Mr J]

I hope I ask this correctly. (38 numbers) what are the chances that I choose ANY two numbers and BOTH of those numbers end up being the furthest back unhit. Meaning, I start crossing off every hit number until I have two left and it just so happens its the two I pre-chose. Ken

[RoulettePlayer]

Mr. J,
Firstly, I hope this is a hypothetical question and that you didn't loose one of your long progressions.
The way you've worded it, I'm not sure that I'm giving you what you want.  It might require KFS or =, if indeed he has an equal. 

chances of 2 numbers hitting in 1 spin    5.26%
I think this is the answer to your question, but ?


other info

chances of 2 numbers hitting at least once in 2 spins 10.25%
                                                             10        41.76%
                                                             38        87.19%
                                                             50        93.30%
                                                             75        98.27%
                                                           100        99.55%       


RP
               

[Mr J]

Lets say I choose any two numbers, the 4 and the 31. I then start crossing off numbers as they hit until only TWO are left unhit. The 4 and the 31 just so happen to be the last two. What are the chances of this happening?  Ken

[madupz4]

Lets say I choose any two numbers, the 4 and the 31. I then start crossing off numbers as they hit until only TWO are left unhit. The 4 and the 31 just so happen to be the last two. What are the chances of this happening?  Ken

To state the obvious, I would say, "SLIM to NONE!"

[hoper35]

I think it's the same % as them being the first two numbers spun?   2 x 1 / 38 x 37

I think you can ignore duplicates.

Ron.

[hoper35]

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