# Does selecting more boxes initially change the math of “Deal or no Deal”?

Apologies if this seems to duplicate questions 835 and 886, but I think it's a different question...

In Deal or No Deal the contestant selects one of 26 boxes and then chooses boxes in groups of 6, 5, 4, 3, 2, and then one at a time until only their selected case remains. After each group of selections the "banker" offers to "buy" their selected case for a sum of money roughly based on the remaining amounts that haven't been "knocked" out by selection.

If the contestant chooses the three or four highest amounts in their first group of six they are offered very little by the "banker" and they face a tough fight to grind the offers higher.

Question: If the contestant mentally selected say 4 boxes instead of just the one initially does this change the math of the game at all?

For example, if they select one case they initially they had a one in 26 chance of getting the million dollar case guaranteeing that they couldn't knock it out, and thereby keeping the offers quite high (although the perceived risk would vary depending on which other cases are knocked out when). But if they "mentally selected" say 4 boxes and said to themselves, I won't select anyone of those cases until I'm down to the final 4... does this change the math at all? They now have a 4/26 chance of selecting the million dollars initially and are effectively choosing 6/5/4/3/2 cases from 22 cases instead of 26...

I kind of know intellectually that it doesn't change the math, but it feels like it should... can anyone explain why it doesn't matter how many cases are picked initially?

-

As you wrote, if you first choose $4$ boxes, you have a $4$ in $26$ chance that the million dollars are in one of them. If you now select a box to be knocked out, the chance that you're knocking out the million dollar box is $4/26\cdot0+22/26\cdot1/22=22/26\cdot1/22=1/26$, since the chance is $0$ if you reserved the box and $1$ in $22$ if you didn't. This is, as you guessed, the same as the $1$ in $26$ that you have if you just directly pick a box. The detour via the preselection in your mind has just inserted factors of $22$ in the numerator and denominator that cancel out.
No. Assuming the positions of the different amounts have been properly randomized, what order you choose the boxes makes absolutely no difference to the probabilities of the outcomes. You might as well choose them in order $1, 2, 3, \ldots$.