hophoofer wrote: ↑Mon Nov 11, 2019 9:00 am
Thank you all guys!
For that 5 aces thing...the “easiest” way to win is get lucky to be dealt with 5 aces. No strategy is needed. Chance is 1 in 2,869,685. Hitting the jackpot by drawing the last ace on all 3 hands when 4 aces are dealt is much harder. The chance is 1 in 1,295,364,096 (2,869,685 / 5 /
49 * (48^3)), if I am not mistaken...
Hey hh, Not that anyone is going to make any changes based on this (nor will most even try and understand), but the chances are slightly worse at 1 in 1,322,350,848. If you change the number 49 in your calculation to 48, I think you will get the same answer. There are 48 cards from 2 thru king (non ace/joker), not 49.
For those who don't follow what you did, if you let the joker stand as the 5th ace, there are 240 ways to get 4 aces on the deal; i.e., 5 different ways to get four aces by leaving out one of the five ace/joker cards multiplied by the 48 ways to get the single card that is from two through king. 5 x 48 = 240. So the probability for this type of
deal is 240 / 2,869,685. Then, as you correctly note, you need to draw the last ace/joker from the 48 remaining cards and you need to do that on all three hands, so that probability of the necessary
draw is as you indicated (1/48)*(1/48)*(1/48) or 1/(48^3) = 1 / 110,592.
Multiplying those together gets you 240 / 317,364,203,520. Dividing both the numerator and denominator by 240 leaves you with 1 / 1,322,350,848.
So, Carcounter's draw was one of the 110,591 "non-jackpot" draws! 103,823 of the 110,592 draws would have yielded zero (of 3) hits of five aces. He still did pretty good by hitting even one of the hands!!
And we will not try and deal with how you can get 3 hands of 5 aces by drawing 2, 3, 4, or even 5 cards after the deal using max EV (or any other) strategy !!
