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View Full Version : Probability of winning 9 straight coinflips

11-09-2005, 09:40 PM
I'm currently on a heater where I have won the last 9 overcard vs pocket pair gambles (me being a 45% favorite going allin preflop). Is there any way to calculate the probability of this happening?
11-09-2005, 09:42 PM
I'm not 100% certain (in fact it is a coin flip lol) but isn't it (1/2)^9?
JinX11
11-09-2005, 09:43 PM
[ QUOTE ]
I'm not 100% certain (in fact it is a coin flip lol) but isn't it (1/2)^9?

[/ QUOTE ]

This is correct if flipping coins.

If a 45% favorite, it is similarly .45^9.
11-10-2005, 06:47 PM
[ QUOTE ]
[ QUOTE ]
I'm not 100% certain (in fact it is a coin flip lol) but isn't it (1/2)^9?

[/ QUOTE ]

This is correct if flipping coins.

If a 45% favorite, it is similarly .45^9.

[/ QUOTE ]

yes, that's correct!. i don't understand the wording. is 45% the favorite? unless there's a big push potential, you'd be an underdog.

but yes, .45^9.... i know 50% to the 9th is 1 in 512. yours would be even longer-shot
11-10-2005, 06:58 PM
I assume you meant your odds of winning were 45% on each hand. The odds of it happening are 1322-1.

The odds of it happening to you are 100%
11-11-2005, 09:03 AM
[ QUOTE ]
I assume you meant your odds of winning were 45% on each hand. The odds of it happening are 1322-1.

The odds of it happening to you are 100%

[/ QUOTE ]

Small correction to the second statement: there are no such odds that say that you are the exception and that you have no chance of being the 1 in 1322.
11-11-2005, 09:51 AM
on the contrary, the odds are 1:1491.33333333, as long as you take the inverse of the reverse coralation....
11-11-2005, 09:58 AM
Drew Da Donk,

Are you sure??? I always assumed that if there was a chance in a 100 to win (or loose) I could be the one! lol /images/graemlins/smile.gif
11-11-2005, 10:03 AM
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Drew Da Donk,

Are you sure??? I always assumed that if there was a chance in a 100 to win (or loose) I could be the one! lol /images/graemlins/smile.gif

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I must re evaluate your statement, as the negative impact of the variance of the negative axis recordulates the inverse, you get your answer.... think about it
11-11-2005, 10:11 AM
Drew Da Donk,

Thanks for your clarification. You must be right /images/graemlins/smile.gif

The point I was trying to make is, that often, people will say there is only 1 chance in a million... therefore it must be rigged /images/graemlins/smile.gif And I know that is wrong.

Anyway, enough wine and calvados for the night... time to call it a day...

Sorry for the uncalled for comment, and keep on enjoying /images/graemlins/smile.gif
OrangeKing
11-11-2005, 02:19 PM
It's also possible the OP meant he was a 55-45 favorite on each hand. In that case, it was just 1/217 or so that this would happen. /images/graemlins/smile.gif
11-11-2005, 02:41 PM
[ QUOTE ]
[ QUOTE ]
I assume you meant your odds of winning were 45% on each hand. The odds of it happening are 1322-1.

The odds of it happening to you are 100%

[/ QUOTE ]

Small correction to the second statement: there are no such odds that say that you are the exception and that you have no chance of being the 1 in 1322.

[/ QUOTE ]

You missed the point---
It already happened to him---hence, 100%