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Ask the Wizard #124

I disagree with your answer to the Monty Hall question in the November 19, 2004 column. Assuming the car is behind door one there are actually four possibilities as follows, where the prize is behind door 1.

  • 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.

What is the probability of playing 14,000 hands of deuces wild without getting four deuces?

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 1-0.000204 = 0.999796. The probability of going 14000 hands without four deuces is 0.99979614000 = 5.75%.

My friend and I went gambling and she got a Royal Flush on Bonus video poker in the morning. Later on in the same day she got another Royal Flush on a different machine but in the same row of machines. I was wondering what the odds were to get two Royal Flushes in the same day?

Anonymous

It is not that unusual. Sometimes Vegas casinos have a promotion in which the second royal hit in a 24-hour 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 (1-0.00002476)3200 = 0.923825. The probability of getting one royal is 3200*0.923825*(1-0.923825)3199 = 0.073198. So the probability of getting two or more is 1- 0.923825 - 0.073198 = 0.002977, or about 1 in 336.

Do video poker machines that tell you what to hold, use the optimal strategy? If so, isn’t it inevitable that the machine will eventually lose money?

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.

Is the combined house edge in craps of 0.014% (taken from your chart) on don’t pass and laying 100x odds the lowest house edge of any casino game? And, does 0.014% casino edge mean that for every $100 you wager you will lose 1.4 cents?

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 Slots-a-Fun 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.

What is the probability that two players will have different four-of-a-kinds in Texas Hold’em?

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 5-card hand and two 2-card 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.

We have all been at blackjack tables where it appears the dealer cannot seem to lose. Assuming you cannot count cards and the dealer is winning 3, 4 or 5 hands in a row, is there any assumptions one can make about the count or is all just random? Do you get up and leave (and/or reduce your bet) and go to another table on the theory that the count is against you and that is why you are losing. Or, do you just assume that the past has no influence on the next hand and continue on. What would the Wizard do? I know hunches have nothing to do with it but, particularly in Blackjack, are there any mathematical conclusions one can draw about the future from the fact that the dealer has been winning (or losing for that matter) for what seems like an inordinate amount of time.

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.