What are the odds? (nc)

dfscott · #1 06-04-2005, 02:29 PM

I've been taking some stabs at the 55s. I've played exactly 9 of them. Here is my finish distribution:

1. 1
2. 1
3. 1
4. 1
5. 1
6. 1
7. 1
8. 0
9. 1
10. 1

Needless to say, I'm not very excited about playing my 10th tonight. [img]/images/graemlins/smile.gif[/img]

octaveshift · #2 06-04-2005, 03:30 PM

Assuming all players are equal, I *think* it would be:

.10*.10*.10*.10*.10*.10*.10*.10*.10 = 1 e-9

aka

Just about a 20 bajillion to 1.

1C5 · #3 06-04-2005, 03:41 PM

Finish 8th on purpose the next one just so you can say you did it. [img]/images/graemlins/cool.gif[/img]

jcm4ccc · #4 06-04-2005, 03:48 PM

Assuming you are an exactly average player, the chances of a distribution like that would be:

1 * 0.9 * 0.8 * 0.7 * 0.6 * 0.5 * 0.4 * 0.3 * 0.2 = 0.003

octaveshift · #5 06-04-2005, 03:49 PM

[ QUOTE ]
Assuming you are an exactly average player, the chances of a distribution like that would be:

1 * 0.9 * 0.8 * 0.7 * 0.6 * 0.5 * 0.4 * 0.3 * 0.2 = 0.003

[/ QUOTE ]

Are you sure? Because the odds that he would bust in each spot is would be exactly .10 each time, regardless of the finish before.

jcm4ccc · #6 06-04-2005, 03:56 PM

[ QUOTE ]
[ QUOTE ]
Assuming you are an exactly average player, the chances of a distribution like that would be:

1 * 0.9 * 0.8 * 0.7 * 0.6 * 0.5 * 0.4 * 0.3 * 0.2 = 0.003

[/ QUOTE ]

Are you sure? Because the odds that he would bust in each spot is would be exactly .10 each time, regardless of the finish before.

[/ QUOTE ]

Yeah I'm sure.

First game: He can end up in any spot. The chances of doing that are 100%

Second game: He can end up in any of the 9 remaining spots. The chances of doing that are 90%.

Third game: He can end up in any 8 of the remaining spots. The chances of doing that are 80%.

etc etc etc

octaveshift · #7 06-04-2005, 03:58 PM

[ QUOTE ]

Yeah I'm sure.

[/ QUOTE ]

Thanks for the explanation. Makes perfect sense now.

maddog2030 · #8 06-04-2005, 03:58 PM

[ QUOTE ]
Because the odds that he would bust in each spot is would be exactly .10 each time, regardless of the finish before.

[/ QUOTE ]

Correct. Any combination of finishes has the same probability of any other combination of finishes for the average player.

Voltron87 · #9 06-04-2005, 04:00 PM

no, the chances of finishing in 1st, 4th, 10th, are not equal.

maddog2030 · #10 06-04-2005, 04:02 PM

You missed a key line: "Assuming you are an exactly average player."