Hilarity N. Suze, if you see the beggar again, can you offer her $10 to tell you how the trick was done?
That should solve the mystery.
Nothing wrong with your math skills. What you are doing is applying the wrong formula to the situation. In this case, the chances of guessing on two discrete days is not multiplicative odds. Likewise the fact that its the same person has no bearing on the matter. I can present a very simple example similar to your coin toss(which is wrong too).
Let is assume that beggar lady is known as BL and Hilarity N. Suze is HNS.
Let us say that BL will be intuiting the contents of the pocket of HNS for three days, but in this case, HNS may only have 1-3 cents, rather than the 0-99 we have been talking about, or the heads/tails that you mentioned. BL knows this or is otherwise satisfied with restricting to this range. We also assume, as you did, that shes purely guessing.
On the first day, BL guesses… well, anything. She has a 1/3 chance of getting it right.
On the second day, we give HNS a new random amount. BL guess again, and since she guesses 1 to 3, and since HNS has only 1 to 3 cents on her, by your way of approaching the problem, she should have only 1/9 of a chance of getting it right. (1/3 times 1/3) Repeat on the third day. New coin count to HNS, new guess by BL. Does BL really only have a 0.037 (3.7%) chance of guessing correctly?
You can try the experiment with a dice. Choose 3 numbers from 1-3 and write then on paper and label them by day. Now using a dice count 1 or 2 as matching 1 cent, 3-4 as matching 2 and 5-6 as matching 3 cents.
If you are correct, it should take around 9 trials to even match the second number with the die roll. But If I am right, it should happen 1 out of three times. Same goes for ‘day 3’.
Despite there being some relationship between a days coinage and the next, any change at all resets the odds. Likewise your coin flip example will show quite long streaks of heads or tails in it. Do it even ten times and you will likely see streaks.
Here is my 10 count(and 10 times). Method: shake 5 quarters in hand, spill on floor, read left to right where they land. Repeat.
HHTHHTHTTT - 2 sequences of 2 and a sequence of 3. Totals: 5 each. Again?
THHTTTTHHT - 2 sequences of 2 and a sequence of 3 again. Totals, five each.
HHTTHTHTHH - 3 sequences of 2. Totals, 6 heads, 4 tails.
THTHHTHHTT - 3 sequences of 2. Totals, 5 each.
TTHHHTTHTT - 2 sequences of 2, and 1 of 3. Totals, 6 and 4 tails.
HTTHHTHTHT - 2 sequences of 2, totals 5 each
HTHHTHHHTT - 3 sequences of 2, totals 6 and 4
HHHHTHHHHH - WOW! - runs of 4 and five. 9 and 1 for totals.
HTHHHHHTHT - run of four! totals 6 and 4
TTHHTHHHHH - run of 5 AGAIN! totals 7 and 3.
Totals: 60 heads and 40 tails. runs: 17 of two, 2 of three, 1 of four, 2 of five.
Notice the 9th group. It has that run of four, but they are technically added from 2 runs of five, right? But thats true of the whole sequence of 100 coin tosses. If we dont bullshit and consider it one long run of 100, what he have are: 18 of two, 3 of three, 4 of four, 2 of five, 1 of six, 0 of seven, 0 of eight, …
These should not be appearing by your example. That run of six is 0.50.50.50.50.5*0.5 percent or 0.015625 chance in 100. Did I beat the odds, and did I predict it would happen before I flipped the coins? Yes.
What you are thinking about is multiplicative chances such as someone getting in a car accident and marrying later in the same year. The car accident reduces the chances of a marriage happening to that person. (0.04 chance of marrying times 0.001 chance of accident). Numbers purely out of my head.
There is no ‘roll over’ effect. If you are gambling in a casino, you are not ‘due for a win’ after a streak of losses.
Everyone that BL approaches has 0-99 cents on them. Thats 100% odds. But no matter what, the next day everyone she approaches still has 100% chance of having 0-99 cents on them. In order for her odds to grow slimmer, you have to reduce her allowable guesses, such as eliminating previous guesses.
This whole thing is an example of the Monty Hall problem - Wikipedia
Of course there was. Any number of ways really. Just ask the promoter of the street fair for starters. They couldn’t have gone far. It’s ridiculous to think that someone discovered a street performer who could pick lottery numbers (to within one digit), and believe them, and not spend a little time tracking them down for a shot a millions.
This looks like overcomplicating things. I think tdn is correct.
If the odds of BL guessing correctly each time are 1 in 100, then the chances of her guessing correctly three times in just three attempts are surely 1 in a million.
Depends upon what you are putting odds on -
Each Guess
or
Three Guesses in a row
The odds for each are different, as explained above.
This was in pre-internet days, when information of this kind was a bit harder to track down than it is now. Sure, we could have done it, but it might have taken a lot more legwork than meets the eye.
Besides, you have to consider the ridiculous side to it. Suppose we had tracked down the guy. What would we have said to him? “We’re totally convinced you’re psychic, and we want you to give us the lottery number for tomorrow… and the next day… and the next day… etc.”
And finally, your saying we “believed” the guy is off the mark. We didn’t, really. It’s more easy for me to believe that the guy has some secret intuition for guessing people’s birthdays within a day than it is to believe that he has the power of predicting lottery numbers within a digit. We obviously “believed” to the extent of risking a dollar on the lottery – but to make it my life’s work to try and exploit this guy to become a lottery millionaire? I don’t think so.
On the first day, she guessed correctly, a 1/100 chance. But after that’s already happened, the chance of it *having happened *is.. wait for it… 1/1. Outcomes we already know *cannot influence future outcomes *unless they in some way change the odds (e.g., if **Hilarity **carried a bag of 100 tokens, each with a different integer from 1-100, and correct guesses were removed from the bag). You’re also only looking at a snippet of her potential interactions: you didn’t see the hundred times earlier and later in the day when she bombed out.
The problem is you’re looking at this after the fact and trying to connect three separate incidents that represent a tiny sample size.
Hell with that–I can offer you **triple **chances! For some *very *special buyers, I might even be persuaded to share my time-tested scientific method for increasing your odds tenfold.
I was running the scam in this thread first.
FIND YOUR OWN THREAD!! The marks are catching on.
Heck, I can tell you how to win 100% of the time.[sup]*[/sup]
[sup]*[/sup]Net profit not guaranteed
Yes, but we’re not talking about rolling each individual die until the correct number shows up. We’re essentially talking about rolling all three at once. and getting a specific result from each die. Even if rolled individually, what are the odds of all three coming up right on the first try?
That is what we’re talking about here. Suppose that BL has the task of guessing right with three people in a row. If the three people have, say, 38 cents, 2 cents, and 67 cents, she doesn’t have to guess any one of those right to win. She has to guess 38, 2, and 67 IN THAT ORDER. One misstep along the way, and she’s lost.
That’s how combination locks work. You can’t correctly guess 2 of the 3 numbers on a lock and get the lock 2/3rds of the way open.
Now one can argue (and some have) that HNS only witnessed the hits and not the many misses that occured in the parking lot at other times of day. Fair enough. What are the odds that she only witnessed the hits?
But to do so is meaningless, because at the time each roll happens, the previous ones have already taken place, so their probability is no longer 1/6 but 1/1. Let’s say that a SDMB poster meets the same stranger Monday, Tuesday, and Wednesday, and the scenario involves a die.
1.) Monday. Stranger: “I bet if I roll this die, it will come up on a six.” She rolls a 6. This was a 1/6 chance.
2.) Tuesday. The stranger has already rolled a 6. There is a 1/1 chance that this has happened. Stranger: “Today, I bet on three.” She rolls a 3. This was a 1/6 chance. It did not become a 1/36 chance just because she’d guessed correctly yesterday.
3.) Wednesday. The stranger has already rolled a 6 and a 3. There is a 1/1 chance that this has happened. Stranger: “Today, I bet on four.” She rolls a 4. This was a 1/6 chance. It did not become a 1/216 chance just because she rolled a 6 on Monday and a 3 on Tuesday.
Now, if the SDMB member were to meet the stranger again…
4.) Thursday. The stranger has already rolled a 6, a 3, and a 4. Stranger: “Today, I have three dice. I bet that they will be a two, a three, and a six.” She rolls a 2, a 3, and a 6. *This *was a 1/216 chance. It did not become a 1/46,656 chance because of the three previous correct rolls.
Isn’t that pretty much what she did? I can see your point, but I guess it depends on how we look at it. If she started her day thinking that she’s going to do her trick three times and hope to succeed at each individual attempt, then the odds are higher. I she decided that she was going to do three times, all with success, then her odds are much lower. That’s pretty much saying that her odds are determined by what she considers to be a success. Her actions don’t change one whit.
Pretty cool, but it’s given me a headache.
No, that’s not what she did. Three sequential predictions are absolutely different from three simultaneous predictions. Treating the former like the latter is absolutely wrong, and as useless as, say, discussing bumblebees when explaining how the wing of an airplane works.
The problem is that the odds can change once something is a known quantity–and in the case of these guesses, as soon as they are made, they are known quantities with a 1/1 chance of happening. Did you check out the link to the Monty Hall problem up above? If not, you really should. What you’re talking about right now is the same as insisting that if Monty is standing in front of an open door with a goat inside it that that door still only has a 1/3 chance to contain a goat.
Did you start each shopping excursion with no change? If so, she could have listened to or looked at the amount as you were checking out, and say it was $25.17, she’d subtract 17 from 100 and you’d have 83 cents.
No, he’s not. He’s looking at the problem explaining the possibility that both times the beggar got it right. The analogy with the Monty Hall problem would be guessing once, being shown a door, and then closing it and having the choices rerandomized. Because, as you point out, what the lady does last time has no influence on the next–exactly UNLIKE the Monty Hall problem.
Come on. I flip a coin 3 times, and I want all three times to give me heads. Of course the chance is 1 in 8, no matter how long in between I wait to make the flips. Yes, if I make more flips, that messes with the odds, but if you assume I’m picking three of those flips to watch at random, then the chances are the same, as there are eight possibilities, and only one gives the desired result.
It doesn’t matter that I’ve already made one flip, and now the chances are 1:1 for which ever side I got. That doesn’t change the fact that, the chances for the system as a whole remains 1 in 8. Don’t believe me? Try it yourself.
And seeing how the incredulity in the OP is not that it merely happened once, that’s obviously the approach we should be taking.
Assuming the method is as one would suppose, (and that change is the modulus between your current monetary amount and a dollar), tdn is perfectly correct. The beggar has a 1:1000000 chance of being correct at any three random times.
The only factor that comes into play is that we have no idea how many times the beggar has been wrong, nor if she has some method that gives her choices better than chance alone would dictate.
I think you are misunderstanding the monty haul problem.
BL is not looking for 3 heads in a row. Shes just looking for a match on that given instant. That is 1/2 if its coins, or 1/100 if its pennies.
The observers, you, me and HNS are watching for a pattern. HHH in your example, but we would be equally impressed if it were TTT. So that right there doubles the likelihood you suggest.
Now if we were ancient Greeks or Romans we might be hopeful of a Venus throw which is all combinations different. In terms of throwing three coins that might be HTH or THT
so
HHH = 1/6
TTT = 1/6
HTH = 1/6 - coin venus #1
THT = 1/6 - coin venus #2
and that leaves only two possibilities:
TTH
HHT
each of which are 1/6. All probabilities must total to 1.0. To reiterate: you have a 1/6 chance of getting any of those. The pattern is simply a human and cultural invention. You will throw, you will get something, and then you will decide if its a* remarkable* pattern.
In monty haul, your first choice is a non choice as you dont lose based on it. Its like a warm up shot in hockey. Nobody is counting the points, but you cant lose either. In monty haul its the second choice that counts. And what you are doing is deciding “did I make my choice in the past, or am I making it now?”. Do I make it then at 1/3, or now at 1/2?
We most certainly would not. That is like saying if HNS, you , and I had $0.01, 0.02, and 0.03 in our pockets respectively, we would be impressed if BL guessed $0.64, 0.65, and 0.66. Or that if she guessed that HNS had 0.02, you had 0.03 and I had 0.01 would somehow count. It wouldn’t. I fail to see how the Monty Hall problem applies at all. The guesses are completely independent of each other.
I agree with tdn and BigT that her chances of guessing the change in one, two, and three people’s pockets is 1/100, 1/10,000 and 1/1,000,000 respectively. If you disagree, please give a numerical answer, without any coin toss or car analogies.
No, not at all. In this case you decide the remarkable pattern before any throws are made at all. And it doesn’t matter if all three coins or dice are tossed at the same time or are spaced apart by a year. Sure, once that first toss is successful, the odds of the next two coming out right goes up, but that doesn’t change the odds of the original prediction.
Wow you Canadians have weird money. 
Why can’t you guys (and most of the world) stick with bills that are all the same size, color, and texture, and quarters that have the same face and 50+ different backs? I guess we do have those $1 coins like you; but don’t forget our awesome rarely-seen half-dollars and $2 bills that some people think are fakes. (Do you have any idea how hard it is to teach money to kids these days?)
A few years ago I got back into snorkelling for the first time since I was a kid. On my first outing, I found treasure in the briny deep. It was 3 quarters, a dime, and a looney.
I still have the looney.