Maths Geeks - assistance please :)

[Nifty29]

Hey gang

Question for the maths heads out there....

In video poker, what are the odds of the following:

1. You hold 2 to a RF, and hit only 2 of the required cards (e.g. no RF)

1. You hold 3 to a RF, and hit only 1 of the required cards

1. You hold 1 to a RF and hit only 3 of the required cards

I've been curious for a while, but I also have a sneaking suspicion about something and would like to possess some facts

Thankyou in advance!!

 

[3Dice]

Hi Nifty,

holding 3 cards to royal flush, drawing 2 out of 47 remaining cards :

odds RF : 1 / ( 47*46/ 2 ) = 1 / 1081 = 0.09 %

odds one of : 2 * 45 / ( 47*46 / 2 ) = 90 / 1081 = 8.3 %

magnitude .. 1 in 10

holding 2 cards to royal flush, drawing 3 out of 47 remaining cards :

odds RF : 1 / ( 47*46*45/ (3*2) ) = 1 / 16215 = 0.006 %

odds one of : 3 * 44 / ( 47*46*45 / (3*2 ) ) = 132 / 16215 = 0.81 %

magnitude .. 1 in 100

holding 1 card to RF, drawing 4 out of 47

odds RF : 1 / ( 47*46*45*44/(4*3*2)) = 1 / 178365 = 0.00056 %

odds one of : 4 * 43 / ( 47*46*45*44/(4*3*2)) = 172 / 178365 = 0.09 %

magnitude .. 1 in 1000

Cheers,

Enzo

 

[3Dice]

Was just rereading this .. and realized I'm a geek

let me try and put it in a more understandable way.

If you hold 3 cards of a RF in first deal, then you should expect to see 90 'near RF's' for every RF (which you should expect 1 in 1081 after holding 3 RF cards).

If you hold 2 cards of a RF in first deal then you should expect to see 132 'near RF's' for every RF (which you should expect 1 in 16215 after holding 2 RF cards)

if you hold 1 card of a RF in first deal then you should expect to see 172 'near RF's' for every RF (which you should expect 1 in 178365 after holding 1 RF card)

So in general expect to be seeing the 'near RF' 100 or even 200 times before you get the full RF.

Cheers,

Enzo

 

[rainmaker]

I bet that someone here did well in school

Good replies Mister Enzo.