So, watching Deal or No Deal the other evening, and I get to wondering … so you pick a case that has a dollar value inside of it. 1 in 26 has $1-million. You then start choosing other cases to see what’s inside them. At one point in the proceedings it came down to a 50/50 shot. They were telling the contestant that they had a 50% chance of being a millionaire. Question: isn’t their chance still only 1 in 26? Yes, there’s a 50% chance that that case has a million dollars in it, but the original odds of 1 in 26 never changed, did they?
Sequential doors There is a generalization of the original problem to n doors. In the first step, the player chooses a door. The game host then opens some other door that is a loser. If desired, the player may then switch to another door. The game host then opens another as-yet-unopened losing door, different from the player’s current choice. Then the player may switch again, and so on. This continues until there are only two unopened doors left: the player’s current choice and another one. How many times should the player switch, and when, if at all? One possible strategy is to stick with the first choice all the way through but then switch at the very end. With four doors, this strategy can be proven optimal; it has been asserted that with n doors, this strategy is also optimal and gives a probability of winning equal to (n−1)/n (Bapeswara Rao and Rao 1992). This problem appears similar to the television show Deal or No Deal, which typically begins with 26 boxes. The player selects one to keep, and then randomly picks boxes to open from amongst the rest. In this game, even until the end, the box the player initially selects and all boxes left unrevealed are equally likely to be the winner. The distinction is that any box the player picks to open might reveal the grand prize, thereby eliminating it from contention. Monty on the other hand, knows the contents and is forbidden from revealing the winner. Because the Deal or No Deal player is just as likely to open the winning box as a losing one, the Monty Hall advantage is lost. Assuming the grand prize is still left with two boxes remaining, the player has a 50/50 chance that the initially selected box contains the grand prize.
In that scenario the player gets to change their “door” each round, thereby giving them the actual probability of the remaining chances. In Deal or No Deal the contestant is stuck with their very first choice of cases from the beginning to the bitter end. Their probability never changes from 1 in 26, methinks …
So… if it gets down to two and the original is 1/26, does that mean the other one is 25/26?
Of course the probability changes, and it changes because it was never a set objective property of the case in the first place - it was just an expression of your best information from your perspective. Once you learn things (opening the other cases) your information changes so the probability changes too.
No offense, but this really is kind of a dumb question. If you were playing Russian Roulette, and you pulled the trigger five times in a row, and five times you survived, do you honestly think the odds of blowing your brains out on the sixth attempt is still one-in-six?
Not a good analogy for the system, Kevin. To try to make your analogy work, if you chose 1 chamber – and that’s the chamber you were able to keep as your head-shot – and then got to fire each other chamber one at a time into the air … I assert that with that set-up your chances of blowing your brains out are 1-in-6 UNTIL the shot goes into the air.
Originally Posted by Kevin KView Post No offense, but this really is kind of a dumb question. If you were playing Russian Roulette, and you pulled the trigger five times in a row, and five times you survived, do you honestly think the odds of blowing your brains out on the sixth attempt is still one-in-six?
Yes, it is. The odds don’t change. It’s like getting heads on a coin flip ten times in a row. Each time, it’s still 50-50. The coin doesn’t know how many heads have come up.
Now, the odds of getting heads ten times in a row may be different, but for each individual toss, it’s 50-50.
And he’s obviously talking about a Russian Roulette version where the chamber is spun once, and then each chamber fired one after the other with no further spins of the chamber. Completely different probability system. To change the Deal or No Deal system to match his analogy, you’d choose a case as YOURS, open IT, and then pick another one.
For Deal or No Deal, I’d say you have a 1-26 chance of picking the million dollar suitcase. After that, the odds that you HAVE the million dollar suitcase go up, but the odds that you picked it are still 1-26.
I think that distills it, Poxy. And James Kimbell made a good argument, too. And Raspberry Leper – I think the odds are 100% that all of the women are foxy.
Originally Posted by Richard DicksonView Post Yes, it is. The odds don’t change. It’s like getting heads on a coin flip ten times in a row. Each time, it’s still 50-50. The coin doesn’t know how many heads have come up. Now, the odds of getting heads ten times in a row may be different, but for each individual toss, it’s 50-50.
Wow, you guys really don’t understand probability at all.
Each coin flip is independent of each other coin flip. So yes, if you get heads ten times in a row, the eleventh flip is still 50-50 because the flips are independent of one another.
In my Russian Roullette example, each event is dependent on the previous event. Each time we pull the trigger, we reveal the state of one of the chambers. If we pull the trigger five times and all five chambers are empty, by definition (not to mention, common fucking sense) the last chamber contains the bullet. It is 100%, unquestionably true, and we don’t need to pull the trigger to find out. If you think otherwise, please feel free to track down a revolver and find out for yourself.
The cases on Deal or No Deal are not independent of one another. If they were, we’d see episodes where multiple cases had the million dollars, or none at all. Each case is not an independent random event, we have an ordered sample of items which makes up a set of values. Each time we eliminate a value, our set shrinks, and the odds change.
Quote:
Originally Posted by Richard DicksonView Post For Deal or No Deal, I’d say you have a 1-26 chance of picking the million dollar suitcase. After that, the odds that you HAVE the million dollar suitcase go up, but the odds that you picked it are still 1-26.
That makes no sense. The odds that you picked the million and the odds that you have the million are the exact same thing.
Originally Posted by Kevin KView Post …an ordered sample of items which makes up a set of values. Each time we eliminate a value, our set shrinks, and the odds change…
That’s a good point.
And, no, I don’t know anything about probability – hence my query.
No worries, you had an honest question. Dickson’s the one telling me I’m wrong without knowing what the hell he’s talking about.
The Russian Roulette analogy is correct though. Let’s say we were determining the odds that the last chamber had the bullet. At first it’d be 1-in-6. Then when we learn that the first chamber is empty, the odds change: Now the odds that the final chamber has the bullet is 1-in-5. And if the second chamber is empty, the odds that the last chamber has the bullet is 1-in-4. And so on.
Look at it this way: If one of the hotties opens the million dollar case, are the odds that you’re a millionaire still 1-in-26? Or course not. Same goes for the opposite.
I don’t watch the show, but- how often do they get down to exactly two, with one having the million and one not? If it happens a lot, it should become obvious that the original pick doesn’t lose 96% of the time.
Originally Posted by Kevin KView Post That makes no sense. The odds that you picked the million and the odds that you have the million are the exact same thing.
Independent of getting in on this argument, it actually does make sense, since we’re talking about two different events. When you made your original choice, the odds of selecting the million-dollar briefcase were 1 in 26. As that event concludes the moment you make your choice, the probability of picking the million dollar briefcase on your initial pick will remain 1 in 26.
The next set of odds have to do with the result of opening the briefcase. When you have more information, the odds that your briefcase contains a particular value (not necessarily the million) increases because there are fewer possibilities available.
Kevin, the problem with your Russian Roulette example is that you’re not playing it correctly. As Blofeld already pointed out, you’re supposed to spin the chamber before each shot. Consequently, you’re never eliminating a chance for the bullet to come up. Theoretically, a game of Russian Roulette could go on indefinitely, assuming that the spin results in sufficient randomness.
What Gred David said. Kevin, as you spin the chamber you find out what the odds are of guessing where the bullet is. But the chances of getting your head blown off are still the same.
If you eliminate the spinning, Kevin’s example is spot on (and I’m not 100% on Hoyle’s rules for russian roulette, but I’m pretty sure spinning the chamber after every attempt is one variation of the game and not necessary).