# Ask the Wizard #122

*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?

Anonymous

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.

Anonymous

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.

Dave A. from 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.

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.

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.

Jennifer

I’ve never heard this one but I’m sure they do not spike drinks with caffeine.

**Blackjack Pro**

**device for card counting?**

Anonymous

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.