Exact same bet in casino different odds?  Hello, I am trying to figure
this out and I am stumped.  I am going to place three bets on black on
the roulette wheel.

1. The first bet: I bet for 25 dollars on black and will press it to
fifty dollars if it wins.   (So I have a 1 in 4 chance of winning,
forget the green on the roulette wheel)

The roulette wheel spins 24 black comes out.  I win

2. The second bet: I press to 50 dollars as I said I would do. And I
am feeling lucky so I bet a separate bet of 50 dollars on black. So I
have a fifty dollar bet (the doulbed bet with a 1 in 4 chance of
winning) and now a separate fifty dollar bet with a 1 in 2 chance of
winning?  For a total of 100 dollars on black  with two fifty dollar
bets.  But the two exact same bets have different odds?

I do not understand this?  Can someone show where my thinking has erred. 
If this is true, than couldnt I say I would be smarter to take both
bets off of black and bet on red?

Well thanks for any help.  

This situation occured at Trumps Hotel and Casino and I lost.  So I
hope Donald bought his wife something nice with all my money.  LOL

ANDY  

ANDY

Hi.

Ignoring green (as you do), you do have a 1 in 4 chance of winning
twice in a row on black, but your thinking has erred when you say that
the doubled bet has a 1 in 4 chance of winning.  It doesn't.   It,
like the separate fifty dollar, has a 1 in 2 chance of winning.

Think about it this way...

BEFORE any spins happen, there are four possible outcomes of the two spins:

#1 - red, red
#2 - red, black
#3 - black, red
#4 - black, black

Each of these outcomes is equally likely: 1 in 4.  Thus, you have a 1
in 4 chance of winning twice in a row on black. But, again, those are
the odds BEFORE any spins have happened.

AFTER the first spin comes out black, outcomes #1 and #2 are no longer
possible, right?

Only outcomes #3 and #4 are possible because the first spin has
already happened and it was black:

(3) black, red
(4) black, black

At this point, each of these two outcomes is equally likely: 1 in 2.

Each spin is independent and the fact that you have previously won on
black has no effect on your odds of winning (or losing) on the second
spin or any future spins.  It's like flipping a coin.  Heads and tails
are equally likely no matter which you flipped the last time. For more
information on this, see "Exposing the Gambler's Fallacy" at
Vegasreference.com :
http://vegasreference.com/gambling/fallacy.html

---------
search strategy:
red black "second flip" odds heads tails

I hope this helps. If anything is unclear, please let me know via the
"request clarification" feature. Thanks.

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Request for Answer Clarification by ajlondon-ga on 24 Jan 2005 02:00 PST

Hello.. thank but I disagree.  Can I clarify why (and maybe sneak
another question in)  I should have put this in my original.
Lets say I pick a number on the roulette wheel from 1 - 36.
I pick the number 24 black. The wheel operator secretly spins the
wheel;, but does not tell me what the result is.  He then places an X
on all the numbers except my orignal pick, 24 black, and 2 black.
He tells me either 24 black or 2 black is the winner. 
Another bettor walks up to the table unaware what has just occured and
places his bet on 2 black.  We cannot have even odds?  There is no
way.....

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Clarification of Answer by juggler-ga on 24 Jan 2005 02:16 PST

I'm sorry to be obtuse, but what's your point?  

What do you mean when you say, "We cannot have even odds? There is no way....."?

Are you trying to say that the second bettor has 1-in-2 odds on black
on the second spin, but you have something less than that?

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Request for Answer Clarification by ajlondon-ga on 24 Jan 2005 08:48 PST

If I place a bet on 24 black.  
The roulette operator spins the wheel, but does not show me the result.
He then X's off all the numbers that did not occur on that spin.
Except the number I orignially bet on 24 black and another number say
2 black

The chances of me winning this bet are not now 50 - 50 are they? The
chances of 2 black being the number he spun are much greater than my
original pick correct?  Even though there are only two choices left? 
This is sufficient to answer, and should have been my original
question. I am asking these questions because I am writing a computer
alogrithm.  And I need to understand a key point.


Thanks ANDY

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Clarification of Answer by juggler-ga on 24 Jan 2005 11:47 PST

"The chances of me winning this bet are not now 50 - 50 are they?"

No, they're not.  "2 black" is more likely the winner.  

You're basically describing what's known as "The Monty Hall" problem.
See:
http://people.hofstra.edu/staff/steven_r_costenoble/MontyHall/MontyHall.html
http://mathforum.org/library/drmath/view/52143.html
http://www.mathreference.com/pr,lmad.html

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Clarification of Answer by juggler-ga on 24 Jan 2005 11:50 PST

I notice that Crythias posted a similar comment a minute before me.  I
guess we think alike. :-)

Thanks!!  I am mixing all kinds of things up... But I understand better now!

On the questioner's clarification, he is asking about the "Monty hall" puzzle...

The mathmatician answer is that the "24b or 2b" is 1/2...
The practicle answer is that ... if you chose 24b when the choice is
(pick one) 1/18 or 1/36, and given the opportunity to switch when all
but two (yours and one other, and exactly one is guaranteed to be the
correct one) are shown, you should always switch.

Extrapolate that to your first choice being 1/1,000,000 to pick
correct the first time, and it's easy to see that switching is
correct.

practicle: practical?

Hee, and I must be showing that I'm suffering from the cold/flu... I
think I meant to say "all but two are *hidden*" or "only two are
shown."

I would agree except that there is a human side to this problem.

The spinner is highly unlikely to ask if you want to change your bet
unless your current bet is a winner.  The probability ignoring the
human factor suggests that 2 black would more likely be the winning
bet, however the human side of this suggests that the house wants you
to move your bet or they wouldn't offer you the chance.

I agree with jack_of_few_trades-ga. If the assumption can be made that
you *always* get the choice, switch. If the choice is given "randomly"
(an objective computer decides 50/50 whether to allow you to do so),
switch. If the choice is given "humanly" (your host determines when he
wants to give you the option), stay.