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View Full Version : Probability question for all you math guys....

Wayfare
01-18-2005, 03:59 PM
You and your friend are playing a game. You take a standard 52-card deck and pull a card. If it is black, your friend pays you a dollar. If it is red, you pay him a dollar. You do not shuffle the deck after picking each card. You have the power to continue playing until the deck is used up, or to stop at any time.

At what point should you stop the game? What is the average EV for you playing this game?

Lost Wages
01-18-2005, 04:04 PM
You quit as soon as you are ahead. If you are ahead then there are more red cards than black cards remaining so it is -EV to continue.

Lost Wages

Wayfare
01-18-2005, 04:21 PM
Yes, but will the variance of the future trials make up for this fact?

topspin
01-18-2005, 04:47 PM
[ QUOTE ]
Yes, but will the variance of the future trials make up for this fact?

[/ QUOTE ]

This is an interesting question. It appears that you shouldn't always "quit while you're ahead"; the option value of being able to stop if you get lucky early on with drawing blacks outweights the disadvantage of being more likely to draw reds. Or, as you put it, you're being compensated by the variance.

I don't think there's an easy theoretical way to calculate the EV; you can set up a backward recursion beginning with one card to go, then two cards to go, etc. and crunch it out with a computer. I was curious enough to do it in MATLAB and the EV worked out to be about +2.6245.

If anyone's curious enough I can post the code.

QuikSand
01-18-2005, 04:49 PM
In the long run (playing this game many times), I will win an infinite amount of money.

As soon as I get \$1 ahead, I quit the game. I'm ahead \$1. That will happen a substantial share of the time -- certainly more than 50% of the time. In the event that I am never ahead by \$1, I simply play through the whole deck and break even at the end. It's clearly a +EV game, played properly or even sensibly.

There probably is an optimal strategy to maximize your EV from a single iteration of this game -- perhaps your EV is greater is you want until you are +\$2, for instance. You'll get there a smaller share of the time, but you'll win more in each win, of course. At some point, the gains from added winnings per win will be offset by the reduced share of the time that you will get that far ahead. The cutoff would probably also be dependent upon the number of cards left, as well. Stopping at +\$1 with 51 cards to go may not be optimal, but stopping at \$1 with 1 card left clearly is.

I suspect the exact calculation of such is possible... but it's beyond me at the moment. The framework, though, isn't that tough.

QuikSand
01-18-2005, 04:51 PM
topspin &gt; QuikSand

Wayfare
01-18-2005, 05:02 PM
Thanks topspin! That is great -- anyone else want to comment on that EV?

BTW, I am asking how to maximize EV, everyone knows that an infinite sequence of +EV games will result in infinate EV. If you had a finite set of trials, how to maximize your likely total +EV is the question I was trying to ask.

pzhon
01-18-2005, 05:30 PM
From an exact recursive calculation, you should quit at

26:20,
25:19,
24:18,
23:18, *5*
...
17,12,
16:12, *4*
...
11:7,
10:7, *3*
...
6:3,
5:3, *2*
...
3:1,
2:1, *1*
1:0,
0:0 *0*

Very roughly, it looks like you should quit when you are ahead about sqrt(cards remaining/2).

The value of this game is 41984711742427/15997372030584 = 2.624475...

The probability that you are never ahead is precisely 1/27: Imagine adding a red card to the top of the deck, then cutting to a red card and removing it. You can get from 27 orderings to each other, and precisely one ordering in each group of 27 is never ahead. However, you don't always quit when you are ahead. The probability you end up with nothing when you follow the proper strategy is 27340098667643/123979633237026 = 0.220521...

tylerdurden
01-18-2005, 05:51 PM
[ QUOTE ]
In the long run (playing this game many times), I will win an infinite amount of money.

As soon as I get \$1 ahead, I quit the game. I'm ahead \$1. That will happen a substantial share of the time -- certainly more than 50% of the time. In the event that I am never ahead by \$1, I simply play through the whole deck and break even at the end. It's clearly a +EV game, played properly or even sensibly.

[/ QUOTE ]

You're assuming that your opponent will always choose to continue to the end of the deck, which seems unlikely.