Since you guys love doing these problems here is another one for you to sink your teeth in. My only request is that our resident math guru's take some time before they answer it. I know you will be able to figure it out. OK here is the problem This has nothing to do with counting This is the start of a new shoe/deck Which are you most likely to receive a hand total of 11 or a Blackjack ? The hand total of 11 must consist of 5-6 or 8-3 or 9-2 or 7-4 ONLY So which one will you get the most often 11 or a Blackjack ? Joep2
And the wrong answer is... Still waiting for my copy of "Probabilty for Dummies" to arrive form Amazon but until then... I think you've go a 50:50 chance of BJ or 11. Odds of BJ: Odds of getting a Ten followed by an Ace = 4/13 * 1/13 = 2.3669% Odds of getting an Ace followed by a Ten = 1/13 * 4/13 = 2.3669% Odds of BJ = 2.3669 + 2.3669 = 4.7337% Odds of 11: Odds of any of the 2 card combinations totaling 11 are 1/13*1/13= 0.5917% There are 8 different 2 card combinations totalling 11 therefore the odds of hitting 11 are 8*0.5917%=4.7337% I expect I've fallen into some kind of trap here but hey, I like to put my neck on the line. Cheers Reachy
They're the same... ...for being dealt 11 and BJ for your first two cards For N number of decks the probability of getting each is 32N/13(52N-1) P(BJ) = P(A,T) + P(T,A) = 1/13 * 16N/(52N-1) + 4/13 * 4N/(52N-1) = 32N/13(52N-1) P(11) = P(9,2) + P(8,3) + P(7,4) + P(6,5) + P(5,6) + P(4,7) + P(3,8) + P(2,9) = 8 * [1/13 * 4N/(52N-1)] = 32N/13(52N-1) For infinite decks the probability of getting each is 8/169 P(BJ) = P(A,T) + P(T,A) = (1/13 * 4/13) + (4/13 * 1/13) = 8/169 P(11) = P(9,2) + P(8,3) + P(7,4) + P(6,5) + P(5,6) + P(4,7) + P(3,8) + P(2,9) = 8 * (1/13 * 1/13) = 8/169 Then you hit or DD your 11.
