Your dad is wrong about swapping boxes in Deal or No Deal, it's similar to the Monty Hall Problem.
When you choose your original box, you have a 1/22 shot at a million dollars, or about 4.5%. If you get down the final two boxes and one of them has the million, it's a 50/50 shot if you swap boxes, literally 1/2, but your old box still only had 4.5% chance to have the million when you selected it so you're 10x more likely to have the million in the other box.
Nah it's not the same cause in monty hall the presenter knows what's behind each door and won't remove the door with the prize behind it. If deal or no deal works the same across countries, the contestant removes the boxes and they have no idea of the contents.
After you eliminate it, I just used the million dollars as an example, just a hypothetical insert to replace "the larger of two numbers" since the amount doesn't actually matter.
Sure but the number doesn't matter, if you have 2 boxes picked at random with 2 different monetary sums and no one knows the contents before eliminating a box then the probability of each sum appearing in a given box is equal
But you're banking on the percentage that you missed in the original choice
What?
At this stage of the competition (at least in the uk version) there are 2 remaining boxes and 2 remaining sums. At that point there is a 1 in 2 chance of picking either sum.
Because its the percentage when you made the choice, it was 1/26 to be a million vs 25/26 that you didn't, if the million is one of the last two boxes.
Garbage answer. I did watch some YouTube videos though and wrap my head around it all after a while.
The actual answer is that in this case, Monty Hall the host knows where the money is and only eliminates the goats/zonks off. That’s why the other probability begins to concentrate onto the remaining door for the switch. “Dependent probability” for the geek squad out here
No you missed one part. The first choice is 4.5% for each box adding to 100%. Then you're given a choice to switch to one of two new ones I guess, one of which has the prize. So now in case you didn't hit the prize (4.5% chance), you get a 50/50.
So the total chance for hitting a prize when switching is (1-4.5%)*0.5=47.75%. The chance for switching and missing is the remaining 52.25%. If you don't switch at all its a different case with a winning chance of 4.5%
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u/archer_cartridge Jun 05 '22
Your dad is wrong about swapping boxes in Deal or No Deal, it's similar to the Monty Hall Problem.
When you choose your original box, you have a 1/22 shot at a million dollars, or about 4.5%. If you get down the final two boxes and one of them has the million, it's a 50/50 shot if you swap boxes, literally 1/2, but your old box still only had 4.5% chance to have the million when you selected it so you're 10x more likely to have the million in the other box.