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View Full Version : Who has the edge, and how would you play this...?

RocketManJames
05-23-2003, 06:10 PM
Here is a dumb game. I am confused by who has the edge and how you would play this.

Person A has 2 marbles. He secretly puts both marbles in his left pocket, both marbles in his right pocket, or he puts one in each. Person B pays \$1 (or X) to guess if it is left, right, or both.

Payout:
Choosing Left or Right correctly: +\$1.50 (or 1.5X)
Choosing a Tie correctly: +\$6 (or 6X)

Now, given this, how should Player A play this? What percentage of the time should he split the marbles? He probably doesn't want to do it too often, as he has to pay a lot more if Player A chooses the split. But, if he never does, Player B will soon figure that out, and never choose "both" thereby winning easily since he's getting paid 3:2 on an even bet (if Player A never splits the marbles).

So, *without any a priori knowledge* of the probabilities that Player B will choose Left, Right or Split... How should player A play this? What about for player B? How do (or should) their respective strategies change as the game goes along?

Excuse me if this is a well-known type of problem that many people have solved over and over on this Forum. I was just thinking about this today and I can't really figure it out.

-RMJ

RocketManJames
05-23-2003, 07:08 PM
Also, say it wasn't such a large payout for the "split"... what if the pay-out was \$2 or \$3?

How does this affect the edge? How can edge be calculated assuming reasonable participants and allowance of strategy modification and response to strategy modification by the opposition?

-RMJ

DPCondit
05-23-2003, 07:46 PM
Player B 'PAYS' \$1.00 each time? Then he only nets .50 when he is right and loses 1.00 when he is wrong, for an EV of -.25 each time, assuming player A uses an optimum strategy of randomly placing all the marbles in one pocket or another, and player B guesses one or another (no split guesses). Player B can guess one pocket or another all he wants, he is going to keep losing at .25 a pop, and more than that if he guesses splits.

Now if Player B "BETS" (not "pays") \$1 each time and gets his dollar back along with the \$1.50 payout, then he makes \$1.50 when he is right and loses only 1.00 when he is wrong, for an EV of .25 (positive this time), when both sides use the "one side or another" approach. However, I just don't see how A can improve his position by throwing in any splits with the heavily stacked payout for splits.

Player B, in the first example, has no winning strategy against A playing optimum strategy of always going to one side or another. Anytime B deviates from always calling one side or another he loses 1.00.

In the second example, there is nothing that A can do to have a positive expectation if B plays optimally. B can choose to randomly call a split with .20 probability and randomly choose one side or the other the other times with .80 probability and will have at least a breakeven expectation against A, no matter what A does. If A deviates from always choosing one side or another, he loses .20 expectation every time he deviates (against player B's optimum strategy). If B notices that A NEVER chooses splits, he may be able to squeeze some extra value by lowering the probable number of times that he calls for a split.

Of course, it should go without saying that A and B must both make sure their "guesses" are truly random and unpredictable to their opponent.

Don

DPCondit
05-23-2003, 08:33 PM
My answer was in response to the first post. the second, follow-up post was not up when I started putting my thoughts together for my answer. (and I don't have time to try and figure out another one of these right now).

Don

irchans
05-24-2003, 12:18 AM
It seems to me that the correct strategy for person A is to put marbles in both pockets 1/9 of the time and randomly choose one pocket 8/9 of the time. I suspect that when person A follows that strategy, then person B's expectation is minimized. Person B has the same expectation if he picks both or either pocket. (This is a standard game theory type question.)

If you change the payoff for choosing a both correctly to P. Then the correct strategy for player A is to choose both with percent = 3/(3+4 P).

When P is 3, choose both 3/15=1/5 of the time.
When P is 4, choose both 3/19 of the time.
When P is 6, choose both 3/27=1/9 of the time.

I got the equation

percent = 3/(3+4 P)

by equating the payoffs for B's choices.
<pre><font class="small">code:</font><hr>
Payoff for B choose both= Payoff for B choose Right
percent * P = (1-percent) * 1/2 * \$1.50
percent * P = \$0.75 - percent * \$0.75
percent * (P+ \$0.75) = \$0.75
percent = \$0.75 / (P+ \$0.75)
percent = 3 / (4 P+ 3)
</pre><hr>

cheers,
irchans

DPCondit
05-24-2003, 01:01 AM
Doggone it!

Yep that's right, that's the saddle point. Oh well, irchans is correct, that is a better strategy for A. I messed up that part of my answer.

Don

RocketManJames
05-28-2003, 10:09 PM
Thanks irchans and DPCondit. I do not think I would have figured this out on my own. I guess I'll need to pick up some elementary game theory book.

-RMJ