

captainsmiffy Special user UK, resident UAE 588 Posts 
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 probabilitywise 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 selfworking wonder! See the reviews here on the cafe.

S2000magician Inner circle Yorba Linda, CA 3465 Posts 
Quote:
On 20111007 02:11, captainsmiffy wrote: True. Quote:
On 20111007 02:11, captainsmiffy wrote: Nope. Quote:
On 20111007 02:11, captainsmiffy wrote: 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. 
LobowolfXXX Inner circle La Famiglia 1192 Posts 
Quote:
On 20111007 21:43, S2000magician wrote: Which is why, when you have 11 trump missing the king, if you lead up to the AQ and your lefthand opponent follows with a low card, other things being equal, play the ace...
"Torture doesn't work" lol
Guess they forgot to tell Bill Buckley. "...as we reason and love, we are able to hope. And hope enables us to resist those things that would enslave us." 
S2000magician Inner circle Yorba Linda, CA 3465 Posts 
Quote:
On 20111007 21:49, LobowolfXXX wrote: RKCB 
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