Ask the Wizard #124
 Player picks door 1 > shown 2 > switch to 3, lose
 Player picks door 1 > shown 3 > switch to 2, lose
 Player picks door 2 > shown 3 > switch to 1, win
 Player picks door 3 > shown 2 > switch to 1, win
As you can see the probability of winning is 50% whether you switch or not. Furthermore it just goes against common sense that switching would be better.
Anonymous
Your mistake is to assume each of these events has a 25% possibility. Following is the correct probability of each event.
 Player picks door 1 (1/3) * shown 2 (1/2) = player loses (1/6)
 Player picks door 1 (1/3) * shown 3 (1/2) = player loses (1/6)
 Player picks door 2 (1/3) * shown 3 (1/1) = player wins (1/3)
 Player picks door 3 (1/3) * shown 2 (1/1) = player wins (1/3)
So losing events have a total probability of 2*(1/6) = 1/3 and winning events have a total probability of 2*(1/3)=2/3.
Anonymous
We can see from my deuces wild section that the probability of four deuces in any one hand is 0.000204. So the probability of not getting four deuces in any one hand is 10.000204 = 0.999796. The probability of going 14000 hands without four deuces is 0.999796^{14000} = 5.75%.
Anonymous
It is not that unusual. Sometimes Vegas casinos have a promotion in which the second royal hit in a 24hour period pays double. Let’s assume you play for 8 hour at a speed of 400 hands per hour, or 3200 hands total. The probability that one hand is a royal flush is 0.00002476. The probability of getting zero royals in 3200 hands is (10.00002476)^{3200} = 0.923825. The probability of getting one royal is 3200*0.923825*(10.923825)^{3199} = 0.073198. So the probability of getting two or more is 1 0.923825  0.073198 = 0.002977, or about 1 in 336.
Anonymous
Most machines I’ve seen that tell you what to hold do use the proper strategy, but the better the paytable, the less likely the machine will be to offer advice in the first place. And I’ve never seen a machine with a positive expectation that told you what cards to hold.
As for the accuracy of the advice  Microgaming Internet casinos do follow optimal video poker strategy. However I’ve played some machines at a racetrack in Delaware that advised the player on which cards to hold, and the advice was clearly incorrect.
Anonymous
There are still video poker games that with proper strategy pay over 100%. I’ve also seen a blackjack game at the Fiesta Rancho and SlotsaFun in Las Vegas that had a basic strategy advantage. As I argue in my sports betting section betting NFL underdogs at home against the point spread also has resulted in a historical advantage. So 100x odds in craps is still one of the best bets out there, but not the very best. Yes, 0.014% means that per $100 bet you lose 1.4 cents on average.
Anonymous
Between two players there are 9 total cards. These must consist of two four of a kinds and one singleton. The number of combinations for this is combin(13,2)*44 = 3432. The total number of ways to pick 9 cards out of 52 is combin(52,9) = 3,679,075,400. So the probability you have the right cards, but not necessarily in the right order, is 3432/3,679,075,400 = 1 in 1,071,992.
However just because the cards are AAAABBBBC doesn’t mean both players will have different four of a kinds. The number of ways to arrange them into a 5card hand and two 2card hands is 9!/(5!*2!*2!) = 756. Following are the ways those 9 cards can fall.
Four of a Kind Bad Beat Combinations
36
Player 1 
Player 2 
Flop 
Mirror Patterns 
Combinations per Pattern 
Total Combinations 
AA 
BB 
AABBC 
2 
72 

AA 
AB 
ABBBC 
4 
48 
192 
AA 
AA 
BBBBC 
2 
6 
12 
AA 
AC 
ABBBB 
4 
12 
48 
AA 
BC 
AABBB 
4 
24 
96 
AB 
AB 
AABBC 
1 
144 
144 
AB 
AC 
AABBB 
4 
48 
192 
Of these only the first and the fifth group result in both players having a different four of a kind. So the probability that an AAAABBBBC set of cards results in two different four of a kinds is 168/756 = 22.22%.
So the answer to your question is (3432/3,679,075,400)*(168/756) = 1 in 4,823,963. On a more practical note Party Poker has a bad beat jackpot for a losing hand of four eights. Given that there are two four of a kinds the probability that both are eights or greater is combin(7,2)/combin(13,2) = 21/78 = 26.92%. So the probability that any one hand of two players will result in this bad beat jackpot is 1 in 17,917,577.
Anonymous
Actually, if the dealer has been winning it is slightly likely that it is because lots of small cards have come out, which would mean the deck is rich in large cards, in which case the odds would actually bend in your favor the next hand. But this is a very slight effect and nothing you should be trusting in. I think in these situations you have just been having bad luck and switching tables will not help. Lest some perfectionist correct me I will say that between shuffles blackjack hands do have a slightly negative correlation. If you had asked about roulette or craps I would say the past makes no difference at all. It would also say that about blackjack if a continuous shuffler were used. However I can’t absolutely say blackjack hands are independent for the reason I just explained.