Blackjack

King_J · #1 08-10-2003, 11:14 AM

Me and my friend argues about what the chances are that there will be black or red 8 times in a row in blackjack (to make it easier, say that every time it hits green we will spin again.)

Plz give me the answer in odds. (ex 187-1)

Wake up CALL · #2 08-10-2003, 12:03 PM

Do you mean roulette?

King_J · #3 08-10-2003, 12:16 PM

Terry · #4 08-10-2003, 01:19 PM

If you’re going to ignore green it becomes a 50/50 proposition, same as flipping a coin.

(1/2)^8 = 1/2*1/2*1/2*1/2*1/2*1/2*1/2*1/2 = 1/256

Chances are 1/256 or 255 to 1.

Bozeman · #5 08-10-2003, 03:51 PM

Since he said red or black, it is 1/128 127:1.

Wake up CALL · #6 08-10-2003, 06:54 PM

[ QUOTE ]
Since he said red or black, it is 1/128 127:1.

[/ QUOTE ]

I don't think so!!!!!

Care to reconsider?

Terry · #7 08-10-2003, 07:07 PM

[ QUOTE ]
127:1

[/ QUOTE ]

If you want to bet at those odds I'd be happy to book all your action. [img]/images/graemlins/wink.gif[/img]

Homer · #8 08-10-2003, 10:50 PM

[ QUOTE ]
I don't think so!!!!!

Care to reconsider?

[/ QUOTE ]

Why don't you think so? The first spin can be either red or black, which has a probability of 1. The next 7 spins must be all red or all black depending on the outcome of the first spin. Either way the probability is (1/2)^7 = 1/128 or 127:1.

-- Homer

Terry · #9 08-11-2003, 02:03 AM

Ah, yes. If you intrepret the question in that way, then 1/128 is correct since you are determining the chance of the next seven events being the same as the previous random event.

If you must "first" choose either red or black and are betting to get eight in a row then it is 1/256.