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HesseJam
12-22-2004, 01:29 PM
The following game is proposed to you:

The dealer holds a set of three cards. Two of them are black, one is red. The goal is that you pick the red card. The cards are shuffled and spread face down in front of you. The dealer knows which one is the red but you don't. You now have to point to a card that you guess might be the red one. The dealer then takes one of the two other cards (those you are not pointing to) off the table. He can only take away a black card. So now there two cards left, one black and one red. You can now guess again which one is red, i.e. you can stick with your first guess or take the other card.

The game is offered to you as that you have to bet \$100+1 to get paid \$100 if you correctly guessed the red card.

a) are you playing that game?
b) is there a strategy?

12-22-2004, 01:39 PM
b) I believe that the optimal strategy is to pick the card the dealer didn't.

a) If I have to pay more than the payout, then no. It's EV is -\$0.01.

HesseJam
12-22-2004, 01:48 PM
Just to make sure that I was clear enough:

You pick one card. The dealer takes one away of the other two but it has to be a black. The card taken away by the dealer does not play anymore. Do you take the card you originally pointed to, or do you take the other card and does it really matter?

12-22-2004, 02:45 PM
Tricky for me to explain, but I'll give it a go.

The dealer has two different decisions he can make:

-Pick between two black cards, leaving another black card on the table - which will happen 33% of the time.
-Pick between a red and a black card, leaving a red card on the table - which will happen 66% of the time.

The dealer is much more likely to leave a red card on the table, since it's more likely that you didn't pick it initially. You'll lose the 33% of the time that you pick the red card initially, and win the 66% of the time that you don't.

That being said, I think my EV calculation was off. Your EV is +\$33, so it's a good bet.

Cobra
12-22-2004, 03:40 PM
This is known as the Monty Hall trick. You should switch cards all the time. During your first pick you lost 2/3 the time and won 1/3. After the dealer removes the other card you are left with the following. If you don't switch you will win 1/3 time and lose 2/3.(The same as before). If you switch you will win 2/3 the time and lose 1/3. Obviously this is over many hands.

HesseJam
12-23-2004, 05:18 AM

The EV is \$4 of pay outs in 3 games - \$3.03 of bets for three games = \$0.97 for three games = \$.32 per game.

Thanks for quoting the name of the trick, I did not know this.

SpaceAce
12-23-2004, 06:54 AM
This is, what, the 30th time we have seen the Monty Hall problem posted this week? Change your pick, always change your pick.

This game is definitely worth playing as long as you always change your pick.

SpaceAce

12-23-2004, 09:00 PM
[ QUOTE ]
This is, what, the 30th time we have seen the Monty Hall problem posted this week? Change your pick, always change your pick.

This game is definitely worth playing as long as you always change your pick.

SpaceAce

[/ QUOTE ]

Well, technically, you need not change your pick every time to make the game worth playing. I don't wanna do the calculations, but I'm sure it's still worth playing you can change your pick 99 times out of a 100.

/images/graemlins/laugh.gif

dogsballs
12-27-2004, 08:38 PM
Here's a link. It's called the Monty Hall Paradox (http://math.ucsd.edu/~crypto/Monty/montybg.html) and a version seems to get posted here every other week or something like that.

JimDoc
12-28-2004, 12:15 AM
I'm playing........but I'm an idiot who just loves the action