Ask the Wizard! (No. 122)
November 19, 2004 Column
On the game show Let's Make a Deal, there are three doors. For the sake of example, let's say that two doors reveal a goat, and one reveals a new car. The host, Monty Hall, picks two contestants to pick a door. Every time Monty opens a door first that reveals a goat. Let's say this time it belonged to the first contestant. Although Monty never actually did this, what if Monty offered the other contestant a chance to switch doors at this point, to the other unopened door. Should he switch?
Yes! The key to this problem is that the host is predestined to open a door with a goat. He knows which
door has the car, so regardless of which doors the players pick, he always can reveal a goat first. The question is known as the "Monty Hall Paradox." Much of the confussion about it is because often when the question is framed, it is incorrectly not made clear the host knows where the car is, and always reveals a goat first. I think put some of the blame on Marilyn Vos Savant, who framed the question badly in her column.
Let's assume that the prize is behind door 1. Following are what would happen if the player (the second contestant) had a strategy of not switching.
- Player picks door 1 --> player wins
- Player picks door 2 --> player loses
- Player picks door 3 --> player loses
Following are what would happen if the player had a
strategy of switching.
- Player picks door 1 --> Host reveals goat
behind door 2 or 3 --> player switches to other
door --> player loses
- Player picks door 2 --> Host reveals goat
behind door 3 --> player switches to door 1 -->
player wins
- Player picks door 3 --> Host reveals goat
behind door 2 --> player switches to door 1 -->
player wins
So by not switching the player has 1/3 chance of
winning. By switching the player has a 2/3 chance of
winning. So the player should definitely switch.
For further reading on the Monty Hall paradox, I recommend the article at Wikipedia.
In Boss Media single-deck blackjack the player has the
edge? What's the catch? So I could go to one of the online
casinos and play it using optimal strategy AND win over the
long run? What am I missing?
I'm sure they still make money on the game due
to player mistakes. There are also forms of video poker
here in Vegas that return over 100% with optimal
strategy. Again, the casino counts on player mistakes to
bring them under 100%. With most games that do pay over
100% the edge is so small it isn't worth ones time to
play the game as a living. However if you are going to
play anyway you may as well get the best odds possible.
First of all, if there's a better site on gambling on
the web, I sure haven't seen it! It was also cool to put a
name to a face when watching the Travel Channel. We always
bring this question up at my monthly game, and decided it
was time for an answer. In a 5-card draw game of "trips to
win", where you must have 3 of a kind or better to win the
pot, if I get dealt 2 pair, is it better to keep only one of
the pair, and get 3 new cards to try and match the first
pair, or should I keep the 2 pair and get one card to try to
match either pair? Assume 6 players at the table, no wild
cards, that players can draw 3 cards, four with an Ace, and
experience shows that any 3 of a kind will probably win the
hand, making pulling the full house not that much more
advantageous than just the 3 of a kind. Thanks! - Dave A.,
Cincinnati, Ohio
Thanks for the kind words. I'm familiar with
this game. Let's assume your intial hand was JJQQK and
you keep the two jacks. The number of ways to get one
jack and two other cards on the draw is 2*combin(45,2) =
1980. The number of ways to get two jacks on the draw is
45. The number of ways to get a three of a kind on the
draw is 10*4+1 = 41. So the number of ways to improve the
hand to a three of a kind or better is 1980+45+41 = 2066.
The total number of ways to choose 3 card out of the 47
left is combin(47,3) = 16215. So the probability of
improving the hand to three of a kind or better is
2066/16215 = 12.74%. If you kept the two pair the
probability of improving to a full house is 4/47 = 8.51%.
So assuming a three of a kind would probably win I agree
that keeping just one pair (the higher one) is the better
play.
Mr. Wizard, First, let me say you have a terrific
site! I've been reading it for a while now. I checked the
poker questions, and didn't see this one. Another site
claims this, "In Texas Hold'em, the probability of AK dealt
pre-flop and hitting an A or K by the river is 1 in 2
(even)." This seems intuitively way too high. What are your
thoughts? Thanks again! - John
Thanks to you too for the kind words. For those
not familiar with hold'em this question is akin to asking
if a player were dealt an ace and a king plus five random
cards from the remaining 50 cards, what is the
probability the player would pair up the king and/or ace.
Of the other 50 cards 44 of them are not kings or aces.
The number of ways to draw any five cards out of 44 is
combin(44,5) = 1,086,088. The number of ways to draw any
five cards out of all 50 is combin(50,5) = 2,118,760. So
the probability of not pairing up the ace and/or king is
1086088/2118760 = 51.26%. Thus the probability you will
pair up is 1-51.26% = 48.74%. This is pretty close to 1
in 2.
Some of the online progressive video poker games, like
Playtech's MegaJacks, reset to a base after a win (I seem to
recall they reset to $325). But others drop down but not to
a set value. For example, the Viper game Jackpot Deuces
seems to drop back a different amount each time, often to a
still sizeable new level. I don't see the "algorithm" behind
this. Any insights into what they (and others) might be
thinking/doing? - Gary K.
Often with progressives part of each dollar bet
goes to seeding the next meter. This way when somebody
pops the jackpot the next meter does not start at a small
amount but the secondary meter has already grown to a
respectable amount. The percentage devoted to the second
meter is not necessarily constant but sometimes increases
as the primary meter grows. Not that you asked, but in
some games like those at Be the Dealer there is a
different jackpot for each coinage, and each jackpot is
proportional to the coinage. The way I think they do that
is what I call a "super meter" that all coinages
contribute to. Then each specific coinage gets a share of
the super meter in proportion to that coinage divided by
the sum of all coinages. For example if they had a
progressive video poker game in coinages of 5 cents, 25
cents, $1, and $5 and the super meter had $100,000 then
the $1 game meter would have (1/6.75)*100,000 =
$14,814.81.
I wanted to know if casinos put caffeine in their
drinks they serve, to keep people awake, or is this an urban
legend like the oxygen? - Jennifer
I've never heard this one but I'm sure they do
not spike drinks with caffeine.
Do you have any comment on the Blackjack
Pro device for card counting?
Interesting. Basically this looks like a two-way
clicker to help the player keep track of the running
count in blackjack. From what I read there is no true
count conversion or index number help. Still knowing the
running count and betting accordingly is much better than
not counting at all. It is also a clever disguise.
However be aware that using any device to help calculate
the probabilities on any game in a Nevada casinos is a
felony and carries a punishment comparable to bank
robbery.
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