Deal or No Deal - mathematical question

[earendel]

Those who follow DoND are aware that the show is on a "Million Dollar Mission" in which each time a contestant fails to win the $1M, another $1M is put into one of the 26 cases. Last night the contestant had 10 chances to win the million but ended up with around $275K (her case had $400 in it).

This leads to my question: How many $1M cases does it take to give the contestant the best chance of actually winning the million?

Under normal circumstances the odds start out at 1 in 26 and shift each round as fewer cases remain. But the more $1M cases there are, the more likely the contestant is to pick them - last night of the first 6 cases the contestant chose, four had $1M in them. On the last show I saw the contestant got down to where there were 2 $1M cases and two others, and she took the deal rather than the gamble (her case, IIRC, had $1M in it).

[MarleysGh0st]

This leads to my question: How many $1M cases does it take to give the contestant the best chance of actually winning the million?

That's easy: 26 cases.

[Rexer25]

I wonder how far the producers of the show would go. If no one wins $1MM, do they fill all 26 cases with that? I would imagine the odd are in favor of a big winner before they millionize all 26 cases, but just in case, would they?

[earendel]

This leads to my question: How many $1M cases does it take to give the contestant the best chance of actually winning the million?

That's easy: 26 cases.

OK, maybe I should have been more specific. What I want to know is the maximum number of cases needed to give the contestant a chance at the million without just handing her the money; or perhaps to put it another way, at what point are the odds roughly equal between the contestant and the show.

[MarleysGh0st]

OK, maybe I should have been more specific. What I want to know is the maximum number of cases needed to give the contestant a chance at the million without just handing her the money; or perhaps to put it another way, at what point are the odds roughly equal between the contestant and the show.

Well, the $1 million cases are replacing the other high level cases, which leaves the left-side cases filled with, essentially, nothing. Contestants don't want to gamble when there's only one $1 million case and one low case, so that they've got to at least get to the situation of having two $1 million cases and one low one. They could already do that now, with a little bit of luck.

Of course, if the contestant isn't a gambler, then they'd want to see three $1 million cases left before they proceed...

[andrewjackson]

This leads to my question: How many $1M cases does it take to give the contestant the best chance of actually winning the million?

That's easy: 26 cases.

OK, maybe I should have been more specific. What I want to know is the maximum number of cases needed to give the contestant a chance at the million without just handing her the money; or perhaps to put it another way, at what point are the odds roughly equal between the contestant and the show.

At the beginning of the game? 13 cases with $1 million would make it even money that if a contestant picked a case and held it to the end that person would win $1 million.

Things are going to change as they eliminate cases. It would depend on what numerical amounts got eliminated. If they eliminate all 13 cases that have $1 million then their chances are zero.

[jaybee]

Pretty soon the odds will get them that One million dollar winner. If they keep adding cases they will eventually get to ao point where there are only two cases left (or maybe three or four) - and both are for a million. As it stands now the smart ones will play it like the mom of five last night - go as long as there is one 1 million left on the board then take the deal.

As long as you can keep two or more of the 1 million up there, you'll be looking at deals in the $200,000 to $350,000 range, not a bad fallback place to be so it makes sense to keep trying for the million.

It could happen at any time but I'm going to predict that they start with 15 - one million dollar cases for the first winner.

Somewhat reminds me of when BAM went through it's millionaire dry spell and added in the 10K bonus each run.

Jaybee

[MarleysGh0st]

Somewhat reminds me of when BAM went through it's millionaire dry spell and added in the 10K bonus each run.

Except that WWTBAM didn't give contestants an expected value of $300K...$400K...$500K just for showing up. You still had to answer those 15 questions of increasing difficulty to earn that jackpot!