I would modify this experiment somewhat. The woman might be reluctant to approach Hilarity repeatedly or may act differently towards a repeat mark. Send different people into the lot at different times. Some should exit their car and then immediately walk a path that goes directly past this woman. They should note the amount guessed.
Others should go to the coffee shop and buy a drink, noting the cost of the drink. Then they should sit down outside with the drink very visible. If the woman approaches note the amount she guesses. We’re looking for any relationship between the cost of the drink and the amount she guesses.
In addition to sending single individuals also send pairs of people on each type of mission and see if she consistently offers to guess the combined amount for pairs.
davidm, thanks for explaining. I get it now. It still doesn’t explain how she made such accurate guesses, though. We can agree that it’s not psychic powers. Can we agree that it’s not just lucky guesses? That leaves some sort of magic trick. And if street magicians regularly do a very similar trick, I think it’s a safe assumption Lady Beggar, somewhere along the way, learned the trick.
I’m pretty sure my math is sound. Prior successes and failures have nothing to do with it. It’s just straight probability.
Let’s say that I flip a coin and I “win” if heads comes up. My chances are 1/2, as there are only two possible outcomes. Let’s say I do it again. Once again my chances are 1/2. But to have it happen twice in a row? If I predict that I’m going to flip heads twice, then there are 4 possible outcomes – HH, HT, TH, TT. That’s a 1/4 chance. If I predict three heads in a row, there are 8 possible combinations. So for each iteration, multiply by the denominator one more time. 1/21/21/2=1/8. 1/1001/1001/100=1/1000000.
Remeber, it’s not that she has to guess right every now and then. To get the results she got, she had to guess a very specific and pre-determined sequence of numbers.
If that doesn’t convince you, try this: I just wrote down three numbers between 0 and 99. Can you guess all three? In sequence? What do you think your odds are?
tdn,
Yes, we can agree that it’s probably not just lucky guesses. I added “probably” because it’s not totally beyond the realm of possibility that she simply got lucky, and it’s not impossible that Hilarity is inadvertently leaving out some vital piece of information. (It’s also not impossible that the whole story is manufactured, but I’m trusting the OP and discounting that possibility.)
I think that the weight of evidence (her successful guesses and the nature of those guesses) leans heavily towards the idea that there is a trick involved.
tdn,
I should have added that I don’t think that it’s the same as the street magician’s trick being discussed. In the first place that trick as I understand it, involves the magician pulling a handful a change from his pocket, having the spectator guess how much is in the hand, and then revealing that the spectator guessed correctly. That’s a different trick and could likely be accomplished through some slight of hand of one kind or another. Also, what the lady appears to be doing seems to rely on hanging out for hours in a specific location watching for known situations where the trick will work. That won’t work for a street magician. He needs to be able to perform more or less on cue.
I think the coffee shop incident is easily explainable as being based on the cost of the drinks Hilarity and friend had in front of them.
The 83 cents in the parking lot is less easy to explain and may have been a different trick altogether.
I think she’s using two different tricks. When the change for a person’s or group’s purchases can be calculated, she guesses based on that. Otherwise, when there are no clues available, she guesses an amount that statistically will pay off the highest amount over time.
But I’m not convinced of the correctness of the second part. It leaves an element of chance and she did guess correctly, at least in Hilarity’s case.
What’s that joke about the physics professor at MIT who comes into the classroom and says “I just saw a car with the license plate ‘AGB492’ - what are the odds of THAT!!??”
I counted my change this morning and I have $15.33 - 6 twoonies, a looney, 5 quarters, 8 dimes, 5 nickels and 3 pennies. The damned parking lot machine gave me $12 in twoonies change. OH, plus an American quarter I should set aside for future travel… Yep, she better not try that trick in Canada.
I’m not sure what all is involved here but he’s playing the odds by saying “within a day”. He triples his chances of being correct on each guess that way. I’m curious, can you tell us the days of the month that you and your wife were born on?
Something similar was going on with the lottery number. Think about it. For one digit, If you choose any digit other than 0 or 9 then your guess has a three in ten chance of being within one of the correct number. That doesn’t completely explain it, if I’m doing the math right there’s only a 1 in 81 chance that he’d get all four correct within 1, so there may be something more going on there.
How much do you recall about that lottery? Was there some restriction on what numbers could be chosen? Was a player choosing four different independent digits or were they choosing a four digit number within some range?
How long ago did you first see her? I had assumed that the second occurrence had happened recently, but now it sounds like it was quite some time ago.
My guess is that it’s a trick and the passage of time has made you remember the result (that she “guessed” the amount) but you’ve forgotten exactly what happened. It’s unlikely that she stated the amount cold; there was probably a prelude that you’ve forgotten. The incident popped back into your head recently, your memory is that she guessed the amount cold, and now it’s remarkable and worth posting about. If it was really so remarkable, why didn’t you post back when it happened?
Anyone who can predict lottery numbers and be off by one digit per number is wasting his time working a street fair. I’m assuming the sequence was correct, right?
I’m the 2nd, she’s the 26th. Can’t remember the number he guessed, but it was within 1.
Statistically, I think you’re right, but this was – if you will – a different activity, in that he was predicting the future, so to speak, rather than guessing at a fact that was hidden to him.
It’s pretty simple – the TV presenter (Dawn Hayes?) pulls four ping-pong balls, one after the other, out of the machine. Each ball has a digit on it, from zero to nine. Any four-digit number can be produced, in other words.
Statistically the two are equivalent. Guessing a future fact is guessing a hidden fact. It’s just that it’s “hidden” by time rather than by walls or distance.
You could be right. It hasn’t been that long really–back in the spring. I don’t know why the incident popped back in my head.
It certainly hasn’t been long enough that I could think I’m never going to see this lady beggar again, but for some reason I don’t think I am.
But as to forgetting details, I could have done that by the next day, you know? I probably would have forgotten the whole incident if a similar one involving the same lady hadn’t happened soon after.
Sequence was correct, just every digit was off by one. As to the observation that if you can pull lottery numbers out of the ether, you don’t need to be working street fairs – well, the same argument is true of the beggar in the OP.
The whole thing remains mysterious to me. If the guy was purely guessing, we can say to a certainty the statistical likelihood of his being right. But it’s hard for me to see how it could be a trick, in the stage-magician sense of the word.
It’s still a conundrum, the same as in the OP – to which we still don’t have a fully satisfactory explanation, if you ask me.
I imagine the change trick has something to do with the fact that one would stop hunting in their purse once you found the “correct” amount. That’s a problem with doing inventory. If you tell someone there are 11 widgets and they find 10, they will keep on looking for the extra one someplace. If there were really 12 and they found 11, then they might just stop looking,
I would like that theory, except for the second time - where she guessed the total amount of change between the two of them. If I were looking through a purse for change I don’t think I’d be comparing notes with my companion during the process.
Really? I see it going like this" “I have 11 cents, how much do you have”? “10. That makes 21 cents”. “Whoops, here is another penny. That’s 22 cents, amazing”!!
I work in retail, I see it every day. Someone starts digging around for change, comes up short then digs deeper in the bag. Still short? Look in another pocket of the bag. Repeat til they find enough or the person behind them kills them.
Sorry if I seem a bit skeptical but if the above is true why didn’t you track the street fair guy the next day, ask for the next numbers, buy a ticket for every combination of ‘one number off’ wait for the ping pong balls to drop and count your millions?
Yes, really. When this happens at work, don’t they ever overshoot the mark? Don’t they ever go looking for $1.50 and stop with $1.55 or $1.60?
“I need 10 cents more.”
“Here. I found a quarter.”
Think about it. If you were depending on people stopping when they find the correct amount, which would you do:
[ol]
[li]Offer to guess the combined total of their change.[/li][li]Offer to guess their change individually.[/li][/ol]
Assuming that you’re depending on them stopping when they get the correct amount, which of those two would you think is more likely to give at least one hit?
Come to think of it, that argument also applies to a single individual. What are the odds that they’ll hit the exact amount while randomly digging for change?
The lady guesses 83 cents. You have more than that in your purse but it’s mixed in with non-change stuff. You start digging. You reach 80 cents. What are the chances that the next three coins you find will all be pennies? If she guesses higher than you actually have then it seems to me that the odds are you will overshoot the mark before stopping.