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December 5, 2004
Column
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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.
Your mistake is 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?
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?
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?
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?
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?
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.
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Four of a Kind Bad Beat
Combinations
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Player 1
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Player 2
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Flop
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Mirror Patterns
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Combinations per Pattern
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Total Combinations
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AA
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BB
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AABBC
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2
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36
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72
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AA
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AB
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ABBBC
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4
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48
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192
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AA
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AA
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BBBBC
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2
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6
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12
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AA
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AC
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ABBBB
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4
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12
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48
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AA
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BC
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AABBB
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4
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24
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96
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AB
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AB
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AABBC
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1
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144
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144
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AB
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AC
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AABBB
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4
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48
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192
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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.
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.
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