Here was my opening question (based on a similar problem by MIT's Richard Stanley):
You meet a billionaire with a well shuffled pack of cards. There are 52 cards in total - 26 have a £ on them and 26 have a zero on them. The billionaire is going to slowly turn the cards face up, one by one. You can raise your hand at any point — either just before he turns over the 1st card, or the 2nd, or the 3rd, etc, - and the moment you raise your hand, you win £1 million pounds if the next card he turns over has a £ on it, and you win nothing if the next card he turns over has a zero on it.
Prior to the game starting, what’s your best strategy for maximising your chances of winning £1 million pounds?
There were quite a few profferings of strategies, such as "wait until 51 cards have been flipped", and "wait until you've had more zeros than £s", and then I gave the answer:
James Knight
The answer is, prior to the game starting, your strategy is inconsequential - you have a 50% chance whatever you decide. You can plan to raise your hand the first time you’ve seen more zeros than £s, or the first time you’ve seen 1, 2, 3, 20 or whatever zeros, but your chance to win will still be 50%.
Some people were not convinced - they insisted that they had control over the probability. Here's how the conversation ensued:
Alex Schamenek
James, if what you say is true then card counting wouldn't work. It does. You are wrong.
James Knight
Alex, I used to be a professional gambler. Card counting does work by increasing the probability, but my OP challenge is not the same as card counting. What you've argued is essentially the same as arguing that because you can't run a car on orange juice you can't enjoy it as a health drink!
Louis le Hutin
James, I am interested in the proof of your statement. It is trivially clear for the first round, but it is not so clear in the next ones.
Johnny Coroama
The best strategy is to -Assign the deck values of +1 or -1 (+1 for each 0, and -1 for each £) - after a running count of +1 or -1 - wait for a statistical disparity -and choose the opposite of the count, if the deck was truly shuffled the higher the deck count is in either direction the greater the chance that the next card will be the opposite, this obviously doesn't work in large models, but its perfect for smaller models of guessing.
James Knight
Louis, the strategy must be decided prior to beginning, which means any strategy you decide prior to starting is not going to increase your chances one jot. After play has begun things change, but they can equally change in your favour or against you, and you won't know which beforehand.
Louis le Hutin
Johnny, I agree... to a degree. It might be a better strategy to wait until the difference is significant (let's say, 60-40),
James Knight
Louis, you can't say 'it might be better to wait' - it's a 50:50 whether it's better to wait, which is why I asked what’s your best strategy for maximising your chances - there is no strategy for maximising your chances.
Johnny Coroama
Your count will give you a good indication, I'd say around +6 or -6, your looking at .60 percent chance the next card is the opposite, if indeed the deck was shuffled
James Knight
Johnny - see my last post.
Johnny Coroama
Picking toward the beginning is bad, picking toward the end is bad
James Knight
It makes no difference Johnny.
Johnny Coroama
If you want a statistical advantage it sure does
Louis le Hutin
James, I agree, if you are talking about the FIRST move. Things change once you get past the first move.
James Knight
Louis, Johnny - there are no advantages because the odds of having an advantage at any one point are the same - 50:50 each game. In other words, you don't have any idea whether the game in which you partake is going to facilitate the run you want or not - it's 50-50.
Johnny Coroama
You are dealing with 52 cards, if the first 26 cards are 0's you don't think you have an advantage guessing what the last 26 are going to be ?
Louis le Hutin
I disagree, James, perhaps because I am a Bayesian. Past information DOES give you an advantage. Come on, if the 26 0 cards have appeared you have a sure bet.
James Knight
Yup guys. And if you choose the winning lottery number, you have a 100% chance of winning the jackpot. Do you want to conclude that you’ll win the jackpot 100% of the time?You're both confused. J
Johnny Coroama
Exactly, so deduce from the sure bet, statistical disparity, look for the disparity by assigning values, and then act when the disparity increases the odds of your guess
James Knight
Your 26 card point is irrelevant to strategy - it would be a run that bore no correlation to strategy. The odds of the first 26 cards being £ is the same as the odds of the first 26 cards being zero. But in one case after the 26 cards are shown you have a 100% of winning and in the other case after the 26 cards are shown you have a 0% chance of winning. The average of 0 and 100 is 50, so it's still a 50:50, even if you could guarantee that you had a 1/2 shot at your desired run.
James Knight
Johnny >>Exactly, so deduce from the sure bet, statistical disparity,
look for the disparity by assigning values, and then act when the disparity
increases the odds of your guess<<
This won't work for exactly the same reason that planning to go and pick the winning lottery numbers at the weekend won't work.
Johnny Coroama
Lol no James, you are wrong. The number of cards remaining is not 50:50. not in small models - pointless to argue with you any further - its a rather simple concept
James Knight
Johnny you just don't understand - anyone knows that if you reduce the number of zeros then you have a better chance of winning. But everyone also ought to know that there is no strategy for reducing the number of zeros!!
Johnny Coroama
James you are confusing conditional probability / Heads and Tales - with propositional logic
James Knight
I'm not, I'm really not!
Johnny Coroama
Since your scenario deals with the proposition that 52 cards are in a deck, half of which are 0's and the other half £ - statistical disparity gives someone better then 50:50 chance
James Knight
You're wrong. After the hand is raised, you might as well just pull the last card of the deck as these events are statistically identical. I think you'll have no trouble seeing that one cannot get better odds than 50-50 on the last card, so one cannot get better odds with any strategy.
Louis le Hutin
Indeed, if you can stay your hand until you have a good chance of winning, and you have no adverse consequences if you stay your hand, I don't see how not staying your hand until you have good odds is a losing strategy
James Knight
Because, Louis, the odds of your game being favourable or non-favourable are 50:50, so there is no strategy to aid you in this.
There were no further replies to my final post, which hopefully means that Louis and Johnny came to realise the error in their thinking. As it happens, here is an illustration that I wrote but ended up not needing to share in the philosophy forum - one which gives even clearer exhibition to how there is no prior strategy better than the 50:50 probability.
Suppose we are now going to start one game, with two players - Louis and Johnny - each trying to win £1 million pounds from the billionaire, who will give them £1 million each if they both get a £ result. Louis's strategy is to take the last card (which is the same as taking the first card). The odds of Louis winning are 50/50. Johnny thinks he's cuter - his strategy is to wait for a favourable situation where there are more £ cards left than zero cards. Then he pounces!
Suppose Johnny gets his good run (which is itself a 50:50, as if there were a second game he has 50% chance of not getting a favourable run). When Johnny raises his hand, the probability of the next card being a winner is the same as the last card being a winner, because the deck is shuffled, so it's still the same odds for Louis and Johnny at any stage of the game. Johnny, despite all his plotting, will never be better off than Louis, who agreed from the very beginning to take the last card with 50/50 odds.
Johnny thinks he has an advantage because he thinks that he might get a favourable run of cards that increases his chance of winning above 50%. This is true, but if that were to happen, Louis's chances have increased by the exact same amount too, proving that Johnny's strategy confers no advantage over Louis.
This shows beyond any doubt that there is no possible prior strategy that you can use to increase your chances of winning.
