Deal Or No Deal!?

[stevie555]

Anyone ever see that 'deal or no deal show', hosted by howie mendel? My question is how to determine if it is mathematically correct to take the 'deal' offered by the 'banker.'

Lets say you are a contestant, and you are down to the final 4 breifcases, one of which is in your possesion and three which are out on the stage. The 4 cases have totals as follows: .01, 500, 75,000, and 400,000.

Therefore, your EV is .01*.25 + 500*.25 + 75,000*.25 + 400,000*.25 = 118,875 right?

And as long as the banker's offer is less than or equal to the calculated EV, you should always say "NO DEAL, HOWIE!" right?

In the above example, the deal offered by the banker was 70,000. I don't know if the values in the cases were correct, but I do know for sure that there was a 400,000 case still in play. Therefore, even if the remaining cases were .01, 1, 10, and 400,000, your offer should always be about 100,000 (25% of 400,000) to be accepted, right?

[stevie555]

If you've never seen the show, it works like this: A contestant begins by picking a briefcase out of 20 possible off of the stage which is then put into her possesion, but the value remains unknown.

Then, she picks cases off cases from the stage one by one and the value inside the cases, ranging from .01 to 1,000,000 are revealed.

Based on the values in the remaining cases, 'the banker' makes her an offer that she can either take right then or she can choose another case and recieve another offer.
If she chooses to, the contestant can decline every deal offered and take whatever is inside the breifcase in her possesion.

Obviously, picking low values from the stage is ideal because it increases the probability that she has a high valued case in her possesion, thus increasing the banker's offer.

[Alan3]

If the banker does not have any extra information, then you should keep playing as long as the banker's offer is less than the average of the remaining briefcases.

But what if the banker knows the contents of the first briefcase you chose? Then it becomes a game where the banker is trying to manipulate you into choosing a smaller payoff than you would get from the final briefcase. In that case I think you are just screwed.

[metsandfinsfan]

-EV can sometimes make sense because of variance

let's say there are 3 cases left.
.01
$1
$1,000,000

They offer you $200,000 which is definitely -EV (Ev would be about $333,333)

You would be a moron not to take the deal anyway

[hero]

lottery larry in the other gambooling forum has been discussing this, check his post out

[CallMeIshmael]

[ QUOTE ]
If the banker does not have any extra information, then you should keep playing as long as the banker's offer is less than the average of the remaining briefcases.

[/ QUOTE ]

This is only correct when the contestent is incredibly wealthy (which is never the case)

Since utility isnt a linear function of money, it is possible (and often the case in the show) that a -EV $ play is +EV utility.

[Alan3]

[ QUOTE ]
This is only correct when the contestent is incredibly wealthy (which is never the case)

[/ QUOTE ]

I agree.

[stevie555]

Thanks for putting it into prospective. Sounds like it would only be the right choice of strategy if you were offered to participate in the gameshow several times, in which case the particular variance in a single round would not be as significant. Reminds me of the beerzman post where people said they would risk their entire bankroll on a 4:1 shot.