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The Magic Cafe Forum Index » » Magical equations » » Odds and 'Up The Ante' Printer Friendly Version
captainsmiffy

Elite user
UK, resident UAE
465 Posts
Posted: Oct 7, 2011 2:11am    Reply with quote   View Profile of captainsmiffy  

I have a simple question that has just started to intrigue me - I have never been very good with probability. When I perform UTA and the first pair is dealt to the table, face down after the spectator shuffle I make a wager that this is a red and a black card. I tell the audience that there can only be 4 possible outcomes - RR, BB, RB or BR. Since I would win on 2 out of these 4 outcomes then the resultant odds should be 50/50? That has stood me in good stead for a few years and nobody has really challenged my thinking here. However, recently, two facts have been disturbing my thoughts. The first one is this: when the first card is dealt to the table there is a 26/51 chance that the next card is of the opposite colour (ie fractionally more than a 1 in 2 chance), thus skewing my earlier premise. The second thought, as I write is this: RB and BR are, effectively, the same thing so Should I be pattering that the odds are stacked in the audiences favour of the pair not being different colours? IE they have a 2/3 chance of winning with the cards the same colour and I only have a 1/3 chance with them being opposite colours?

In the grand scheme of things it won't really matter as nobody has challenged my initial assertions anyway and the odds are truly dwarfed by the odds of the outcomes as the routine develops (dealing 1 of each suit and, then, later a full run of A to K). It is just a point of interest now I got to thinking on probabilities. I appreciate that I have not got much of an idea probability-wise but would very much appreciate mathematical input here to put my frazzled brain straight.

Have you tried 'Up The Ante' yet?? The ultimate gambling demo....a self-working wonder! See the reviews here.
S2000magician

Inner circle
Yorba Linda, CA
3597 Posts
Posted: Oct 7, 2011 9:43pm    Reply with quote   View Profile of S2000magician  

Quote:
On 2011-10-07 02:11, captainsmiffy wrote:
The first one is this: when the first card is dealt to the table there is a 26/51 chance that the next card is of the opposite colour (ie fractionally more than a 1 in 2 chance) . . . .


True.

Quote:
On 2011-10-07 02:11, captainsmiffy wrote:
The second thought, as I write is this: RB and BR are, effectively, the same thing so Should I be pattering that the odds are stacked in the audiences favour of the pair not being different colours? IE they have a 2/3 chance of winning with the cards the same colour and I only have a 1/3 chance with them being opposite colours?


Nope.

Quote:
On 2011-10-07 02:11, captainsmiffy wrote:
In the grand scheme of things it won't really matter as nobody has challenged my initial assertions anyway and the odds are truly dwarfed by the odds of the outcomes as the routine develops (dealing 1 of each suit and, then, later a full run of A to K). It is just a point of interest now I got to thinking on probabilities. I appreciate that I have not got much of an idea probability-wise but would very much appreciate mathematical input here to put my frazzled brain straight.


The number of ways you can choose 2 cards from a deck of 52 is 52! / (2! * 50!) = 52 * 51 / 2 = 1,326.

The number of ways you can get a pair of the same colour is 2 * 26! / (2! * 24!) = 2 * 26 * 25 / 2 = 650.

Thus, the probability of getting a pair of the same colour is 650 / 1,326 = 25 / 51 = 0.4902. The probability of getting one of each colour is 26 / 51 = 0.5098.

I'd go with 50 / 50: it's easier on the brain.

BCIII
The Polite Side of Magic
Risk Mitigation Associates
Bottom Line Gurus
I took the Pledge
LobowolfXXX

Inner circle
La Famiglia
1598 Posts
Posted: Oct 7, 2011 9:49pm    Reply with quote   View Profile of LobowolfXXX  

Quote:

On 2011-10-07 21:43, S2000magician wrote:
Quote:
On 2011-10-07 02:11, captainsmiffy wrote:
The first one is this: when the first card is dealt to the table there is a 26/51 chance that the next card is of the opposite colour (ie fractionally more than a 1 in 2 chance) . . . .


True.



Which is why, when you have 11 trump missing the king, if you lead up to the AQ and your left-hand opponent follows with a low card, other things being equal, play the ace...

-DFO

"All I wanted to do was work with John for the rest of my life."

LSAT tutor & author of "LSAT 60 Dissected," available online.
S2000magician

Inner circle
Yorba Linda, CA
3597 Posts
Posted: Oct 7, 2011 10:57pm    Reply with quote   View Profile of S2000magician  

Quote:
On 2011-10-07 21:49, LobowolfXXX wrote:
Quote:
On 2011-10-07 21:43, S2000magician wrote:
Quote:
On 2011-10-07 02:11, captainsmiffy wrote:
The first one is this: when the first card is dealt to the table there is a 26/51 chance that the next card is of the opposite colour (ie fractionally more than a 1 in 2 chance) . . . .


True.


Which is why, when you have 11 trump missing the king, if you lead up to the AQ and your left-hand opponent follows with a low card, other things being equal, play the ace . . . .


RKCB

BCIII
The Polite Side of Magic
Risk Mitigation Associates
Bottom Line Gurus
I took the Pledge
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