I'm preparing a 4 phase prediction effect. The spectator chooses from 4 cards in the first phase. To meet that prediction is a 1 in 4, or 25% chance. Now phase two has the spectator choose from 4 cards, and singly this would be another 25% chance. But to combine it with the first phase, what percentage of a chance would it be to get both the first and second predictions correct? Then the third and fourth phases are the same. So what would be the percentage that they can get all 3 predictions correct and then all 4 predictions correct consecutively? Astronomical? lol

Your help will be greatly appreciated!

1st phase
1/4 = 25%

2nd phase
1/4 * 1/4 = 1/16 = 6.3%

3rd phase
1/4 * 1/4 *1/4 = 1/64 = 1.7%

4th phase
1/4 * 1/4 * 1/4 * 1/4 = 1/256 = 0.4%

I Think its been a while scince I did probability

Mark

I agree. But it might be that when the second card is chosen, it's chosen from the remaining three cards, and the third card likewise being chosen from the remaining two cards.

And in that case, probability of first one being correct is 1/4,
probability of first two being correct is 1/4 * 1/3 = 1/12,
probability of first three being correct is 1/4 * 1/3 * 1/2 = 1/24,
probability of all being correct is also 1/24.

It all depends on the definition of the problem.

Dave

Mr. Le Fevre is, of course, correct. Are you predicting a free choice out of four cards each time (in math-geek, four independent random events, each with a probability of 1/4)? Or are you predicting the arrangement of four cards (in math-geek, a permutation of four unique objects)?

It is a prediction of four cards each time. Thanks for your help so far!

10cardsdown,

If I understand your question, the chance of predicting one of four cards correctly four times is(as previously stated)1/4 x 1/4 x 1/4 x 1/4 (or one over four to the fourth power) which calculates to 0.0039. This, of course, is less than four tenths of a percent chance of success through luck alone. It might be more impressive to express the chance of failure as 99.6%

Sid

Okay, you probability geeks. I'm working on Gaeton Blooms "Quarte" effect.

In this one, you have a four digit number apparently predicted in advance. First spectator chooses a card from a group of 9 cards which are simply the digits 1-9. This is removed. Second spectator chooses a card from the remaining eight. Third Spectator chooses a card from the remaining seven. And the last spectator choose a card from the remaining six. This creates a four digit number and your prediction is correct.

It seems to me that the odds are: 1/9 x 1/8 x 1/7 x 1/6. Am I right?

Dennis Loomis
http://www.loomismagic.com

From a true proability standpoint, you are correct. The odds of you getting it right by chance are 3023 to 1, or 1 in 3024.

From a showmanship standpoint, I'd mention that there are 9,000 four digit numbers (1,000-9,999), so you can simply claim that the odds are 1 in 9,000 for your audience!

Thanks, Scott. I like your thinking. Why bother them with the truth? We are, after all, not DISillusionists.

Denis Loomis
http://www.loomismagic.com

Ahh....its' 50/50 each and every time. You're either right or you're wrong...50/50...who needs statistics?

Proving once again the figures don't lie, but liars figure...

Aren't' probabilities, combinations, and permutations fun...yes, this much ---> n! where n=52 <---for you math types

Who needs statistics? Well, we in Nevada do. No, not just the casinos for their obvious need to properly study probabilities, but everyone who lives in the state of Nevada.

As long as the casinos keep figuring probabilities properly, they keep raking in money. And as long as the casinos keep raking in money, Nevada doesn't have to turn to individual citizens for a state income tax!

You guys and gals probably don't care, but I'm going to tell you a little story anyway. It came back to me when reading S. Steele's post above, and seeing the “!” factorial sign. I was a math major in college until coming up against the factorial on a number theory midterm (worth 50% of the class grade). The question, which I remember to this day 25 years later, was, "How many numbers between 100! and 1000! are evenly divisible by 7?" I had no idea, and I realized sitting there contemplating the sure “F” I was going to get in that class that I really didn't care what the answer was. It was the only time in my education I was forced to write my name on a blank exam, turn it in, walk out, and never look back. (I definitely did get an F in that class.) I'm now in medicine, not math. But I still enjoy math-based effects, and have always wanted to know the answer to the above factorial question. Do any of you have any ideas? I suppose I could work it into a magical effect if I just knew the answer.
Bob

When I first ran into factorials, I thought the “!” meant you had to shout the number: ONE HUNDRED! ONE THOUSAND!



0pus

Every number between 100! and 1000! is divisible by 7. Think about it . . . they all have a factor of 7 because by definition all those numbers begin 1x2x3x4x5x6x7 . . .

Rest easy,
Jack Shalom

I had to take Statistics in college as a requirement. It was difficult!! And not my favorite class. I'm more of an English, writing, reading literature kind of person. Nonetheless, I enjoyed reading the probability posts above. It is interesting stuff!

The initial problem clearly wanted to know how many of those numbers were EVENLY divisible by 7. Not all of those numbers are.

For instance, while 100! might be evenly divisible by 7 (it does indeed have 7 as a factor), the number 100!+1 could not be. And I'm sorry to say, 100!+1 lies between 100! and 1000!.

Now, had the question been how many of the numbers 100!, 101!, 102!, 103!...1000! are evenly divisible by 7, then you would be correct Jack.

Jason

If we count 100! and 1000! there are (1000!-100!)/7+1 numbers divisible by 7. This number is extremely large so I can at least give an approximation. The extra 1 and 100! are small in this context so 1000!/7 is a very good approximation. The Stirling Approximation says that ln(n!) is approx nln(n)-n, where ln is the natural logarithm (I'll call the decimal logarithm "log" later). So ln(1000!/7)=ln(1000!)-ln7 which is approx 1000ln(1000)-1000-ln7. Now to get this into base 10 I'll multiply by log(e) and get about 2565, so the number we seek is of the order

10^2565

and I'll applaud anyone that can write down the exact number.

/Tomas

Right, Jason. I misread the problem!

Jack