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View Full Version : Probability Quiz: EV is getting you only so far...

HesseJam
12-23-2004, 05:43 AM
In decision theory one of the most basic concepts is the Expected Value (EV) concept. If your are facing a decision to choose between alternatives, you should choose the alternative with the highest EV. But there are situations where this concept is impractical:

A casino with unlimited ressources proposes the following game to the public which will auction off the game to the highest bidder (the game will only be held once):

a) You need to pick heads or tails on a coin flip.
c) In stage 1, if you guessed correctly the casino will pay you \$1, which is your profit. You also get a freeroll for stage 2. If you missed, the game is over.
c) In stage 2, the pay out will be doubled. If you guessed correctly the casino will pay you \$2, which is your profit for now a total profit of \$3. You also get a freeroll for stage 3. If you missed, the game is over and you total profit remains the 1\$ from the previous stage.
d) If you make it to stage 3, the casino will again double the pay out to \$4.
e) You stay in the game until your first miss. The pay outs will always be doubled in each subsequent stake.

The auctions starts. How high would you go with your bid? Try to apply the EV concept. Obviously, according to that concept you can increase your bid as long as it stays below the EV. Would you take your decision according to that concept in this case?

HesseJam
12-23-2004, 09:22 AM
Just a clarification. As long as you hit your guess you go to the next stage with double pay outs. If you miss you still keep your earnings from the previous stages.

Skjonne
12-23-2004, 09:29 AM
When does the Casino close?

HesseJam
12-23-2004, 12:36 PM
I was still a bit unclear. The game has no limit in stages. As long as you hit you will be promoted to the next stage with the double payout of the previous stage. If you hit 10 times in a row and then fail then the game had 11 stages and you won (2 exp 11) - 1 = \$2047.
(1+2+4+8+16+32+64+128+256+512+1024)

Skjonne
12-23-2004, 12:48 PM
[ QUOTE ]
I was still a bit unclear.

[/ QUOTE ]

Not at all. In my crappy try-to-be-funny way, I was just stating that this bet has infinite EV hence should cost an infinite amount of \$. But Hey, I won't pay that.

The problem is interesting yet I really don't think you have the answer. Problems with unrealistic assumptions (infinte time) do not need to have (realistic) solutions

Paul2432
12-23-2004, 01:18 PM
I would bid upto EV=0.

Paul

HesseJam
12-23-2004, 04:15 PM
Yes, EV is infinite

1/2 * 1 + 1/2*1/2*4 + 1/2*1/2*1/2*8 ....
= 0.5 + 0.5 + 0.5 + 0.5 .......

I have an answer, maybe not THE answer since this depends on each individual.

One thing seems clear: nobody will pay an infinite amount of money nor can anybody. Nobody will probably sell his home, cash out his bank account and put his whole fortune into the game even if this would be clearly +EV .

Why? Because your own risk-utility-function is not properly reflected. Even If you put in only \$1000 you will fail about to 99,8% to get your money back (you need to guess right 9 times in a row). But the 0.2 % who get there suddenly face a fantastic game. 0.1 % of them will go and win a Million bucks and can now play for two Million and so on and on.
Now, the problem here is that the EV concepts treats each dollar with the same value. While at large each dollar has of course the same value (one dollar /images/graemlins/crazy.gif ) not each dollar has the same value to you. Let's say you win one Million, which might be a life changing event for some people. Will it be the same life changing event if they win a second Million? Probably not. Now, this someone who had won the Million is proposed an even odds bet that he could triple his Million if he makes good. Otherwise he loses one Million which is all his money. Most people would not do this because the value of the second and third Million combined is much lower to them than the first Million.

So, in the coin flip proposition you would bid as much as you personally think you could afford to loose for that small chance to win big. It's like the lottery, the difference is that it is infinite + in EV whereas the lotteries are negative. The result (how much to bid for the game)will be different for each person.

Skjonne
12-23-2004, 06:36 PM
Oh, that were where you were heading /images/graemlins/smile.gif

Yes, that is of course correct all of it. Allthough not that applicable to poker?

But now that the bottle is opened:

Many people don't understand why a guy like me, with a degree in economics and matematics play the lottery. Utility-function! If I were to win 2M bucks, my life would change. My Utility function wrt money steepens (sp?) dramatically at the point where I can quit my 9-5 job!

Some years ago, I had a colleague that had a problem. He had appr. \$600 to spend on his summer holiday. All his friends were going three weeks to Spain and he really wanted to go along. He very very seriously considered going to the casino and put \$600 on red (or black, I don't remember). \$1,200 would make him a holiday, \$0 - could he care?

Summary: blah blah blah (Quiz time: Am I drunk?)

MortalWombatDotCom
12-23-2004, 07:55 PM
i would collude with every potential bidder to play for one penny and split the proceeds. we would randomly select which of us would have to cough up the penny. infinite expectation, fraction of a penny risk. woot.

12-23-2004, 08:03 PM
That's a really interesting concept, I like it.

RocketManJames
12-23-2004, 08:25 PM
Right... I have even heard of a company (maybe an Urban Legend) that basically got to a point where it had X in the bank. With X, it had basically 0% chance to make it, but with 2X it had boosted its chances to say 80%. So, the founders put X on an even money bet.

Also, how does this relate to poker? I think it might be a stretch, but in a tournament, isn't the utility function based on the player's strength vs the field, the bet sizes at the moment and in the future, and also the distribution of the chips amongst all the contenders?

-RMJ

HesseJam
12-25-2004, 03:05 AM
In poker I can see one relation: Playing marginal situations (slightly +EV but high variance) when not properly bankrolled can let you bust out (=death) earlier. For me this is the exact opposite to the lottery. In poker playing high variance / small +EV situations gives you only a small profit (compared to your bankroll) with a high risk. In the lottery, the loss is small but if you hit it ...
So, it can be perfectly rational to choose a playing style that is a bit more risk averse than, let's say TAG, provided it still shows profit and it actually has a lower variance.

irchans
12-26-2004, 12:29 PM
One way to answer this question is to apply the logarithmic utility function as in Kelly criterion or Certainty Equivalent. (See Thorp (http://www.bjmath.com/bjmath/thorp/paper.htm), this link (http://www.investopedia.com/articles/trading/04/091504.asp), or Wong (http://www.bjmath.com/bjmath/ror/ce.htm).)

<font class="small">Code:</font><hr /><pre>
Bankroll Game Worth
\$1000 \$4.97
\$10,000 \$6.62
\$100,000 \$8.28
\$1,000,000 \$9.94
\$10,000,000 \$11.60
\$100,000,000 \$13.26
\$1,000,000,000 \$14.92
\$10,000,000,000 \$16.50
</pre><hr />

I used the following Mathematica code to generate the table.

<font class="small">Code:</font><hr /><pre>

eExpectedLog = Sum[
1/2.^i*Log[ B + 2^(i - 1) - 1],
{i, 1000}];

For[i = 3, i &lt;= 10, i++,
B = 10^i;
Print[{B, Exp[ eExpectedLog ] - B}]
]

</pre><hr />

SeanSkill
12-28-2004, 08:23 AM
If you give me one rebuy I will bid .50 LOL