Added the code to pastebin: http://pastebin.com/tGs2cBU2

Sorry for the delay

]]>Yeah. I can do that. Check back in the new year. It’s on my old laptop so I have to dig it out first.

One possible reason for the difference might be when our respective bits of code choose to ‘swap’. If I remember correctly, I had my code swap when there was only one value greater than or equal to the desired winnings, even if that was half-way through the number of boxes; the argument being it was analogous to taking the banker’s offer (although I admit the banker’s offer does not come round every turn).

Come to think of it, that almost certainly explains the difference: the higher the amount of money you want to win, the less likely you are to have a prize of that value in your box, and the more likely you are to up with a winning prize when you swap: I think the odds are stacked in your favour because statistically you’ll usually need to get rid of more lower-value boxes before most of the higher-value boxes are gone. Hmm…

]]>Would you consider posting the code ? ]]>

As for a proof: say there’s 25 boxes, then:

The probability that you pick the $1m box is 1/25. The prob. it’s in the remaining boxes is 24/25.

The prob. it’s still in the remaining boxes after the banker opens 1 box at random is 24/25 x 23/24.

The prob. it’s in the remaining boxes after the banker opens another box. is 24/25 x 23/24 x 22/23

The prob. it’s in the last box after the banker has opened 23 of the 24 boxes is 24/25 x 23/24 x 22/23 x ……….3/4 x 2/3 x 1/2 = 1/25, i.e.the same as the prob. you picked it in the first place.

So if you get down to the situation where the $1m is still out there (and 23 times out of 25 it won’t be) then it’s 50/50 between the last 2 boxes.

“I wrote a short computer program”. it must have a bug in it.

]]>Sorry for the slow response (I’ve been away) but thanks for your comment. As you can tell, you’re not the first person to say the odds are 50/50 for DOND but can you prove it? Or to put it another way: if the odds are 50/50, why does my table above show the opposite; namely that when you swap you are more likely to get your desired winnings or higher (provided it is still there for the taking)?

]]>This statement is true; when our mind, body & spirit aren’t in alignment it takes a lot of energy, everything becomes harder and the pain and stress of not surrendering to your life’s true purpose causes the day to day to become the grinding wheel of life…

The moral of the story is to always be yourself, live your own life. The fact that you have very few memorable days in your work serves as a clue that you aren’t doing your life’s work.

Success leaves clues, look at those days that really made you feel alive, that is your heart calling out, those true impulses and desires of our hearts and mind are in alignment with our purpose…

Follow them

]]>You explained it yourself at the the beginning of the article “.. in the Monty Hall problem, the presenter always knows where the prize is located.”. In DoND this isn’t the case. DoND is like the MHP where Monty has amnesia: you pick a door, Monty randomly chooses 1 of the other 2 – it’s a goat fortunately – the 2 unopened doors are now equally likely to contain the car.. Why? In the 2 out 3 times that amnesiac Monty doesn’t open the car door , 50% of the time you’ve already picked it.

It’s the same in DoND, say there’s 24 cases. You pick 1 , it’s 1/24 chance of it being $250K. There;s a 23/24 chance it’s in 1 of the other 23 cases. If you open 22 of those 23 cases at random it’s a 1/23 chance the 1 not opened is the $250K. 23/24 X 1/23 = 1/24 = same chance as you picking it to begin with.

In DoND it makes no difference if you pick 2 cases to begin with and then open the other 22, open 22 to begin with, or pick 1 then open 22 cases., you’re still left with 2 cases both equally likely to contain whatever 2 prizes happen to be left.

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