Interesting Puzzle

50/50

Only if they are equal number of red vs green balls in the urn... and to be like some of the members that ANALize everything.

As long as there are 50 reds and 50 greens that are mixed. and not that the green balls were put in first and are on the bottom of the urn and the red balls that were put in last are on the top. and they weren't mixed up.

All urns used in mathematical puzzles must be certified to allow equal access to all balls contained therein. The 100 balls in the urn can be from all green to some mixture to all red. When you take the one sample, it will be any one of the 100.

but if its 60 reds and 40 blues.. then odds change.

but I had the flu (due to circumstances) when I had statistics back in college.

And at other times, I get bored and stir things up...

There won't be <...equal number of red vs green balls in the urn...> either [at the start] or [after one red ball has been removed]; as both these conditions cannot be true.

One could always pour the contents of the <...urn...> into a cast iron bathtub for a closer look, perhaps.

If the balls are truly randomly mixed the odds or getting 2 same color balls is 50/50 before you pull out the first ball.After that,the odds of pulling the same color ball next diminish because you now have less balls (49) the same color as the one you pulled,so the odds are weighted in favor of a different color ball.

(If someone kicks the balls,however,they will always be blue.)

You are correct. If there are an odd number of balls left after pulling the first ball out, it can't be 50-50, and pulling the second ball out has to be based on the new odds.

One cannot tell the numeric probability for the next ball color picked without knowing the previous ball distribution. An incremental formula can be made with an incremental unknown variable. However, since the question is "...will it be more likely to be red or green (or are the two colors equally likely)?" The answer is YES it will be more likely that a red or green ball will be pulled because an odd number (99) of balls are in the urn. An equally likely distribution can only happen with an even number of two types of balls. I have no idea if red or green will be more likely but one of them will be more likely.

GA

But, you have "lost" your /99 in your calculation

0/99 + 1/99 + 2/99 .........+ 99/99 = 50

Now you need to divide by 100 to get ½

The odds are even.

You can check this easily with an earn with 3 balls. After you removed the first red; there are three equal possibilities: two greens; a red and a green, and two reds. The probabilities of another red are now 0/2, 1/2 and 2/2.

Add these up 1½ and divide by 3 = ½

The trick to understanding this is that by drawing the first red you have eliminated the possibility of all greens; the other possibilities are equal. You don't need to start trying to work out how much more likely it is that the earn contained more reds because you have drawn one.

Randall,

I struggled with the divide by 99 or by 100 as I was working this out. I believe you are correct in your 50% probability as the answer.

--JMM

GA, you nailed the logic - #10

The sum of 0 to 100 is 5050. Half is 2525. The sum of 0 to 70 is 2485. The sum of 0 to 71 is 2556. So, pulling a red ball puts the initial distribution at 50% 71 or more; 50% 70 or less. Therefore the next ball also being red is about 69.57% likely.

I decided to write a simulator program to investigate this puzzle, and found that if the inspected ball was red, the urn had more red balls approximately 75% of the time. If the inspected ball was green, the urn had more green balls approximately 75% of the time.

There are 101 different amounts of red balls (0-100) in the urn and 100 different balls that may be chosen, for a total of 101 x 100 = 10100 possibilities, all equally likely. (It doesn't make any difference what order the balls are in, we are only interested in which color has the majority.)

It only takes the computer a couple of seconds to tabulate the urn ball color distribution into 4 groups:

• red ball selected - red majority in urn - 3624 - (74.73%)
• green ball selected - red majority in urn - 1225 - (25.27%)
• green ball selected - green majority in urn - 3775 - (74.75%)
• red ball selected - green majority in urn - 1275 - (25.25%)

The percentages come out an even 25% & 75% if you put the inspected ball back in before looking at the majority color in the urn (ignoring ties).

Another perverse way to look at this problem, is to first realise that once the urn is in place with the randomly selected ratio of green and red balls, then the probability of getting either a red or green ball on the first drawing is equal. But, suppose that you remove that first ball without looking at it, then, draw the second ball: isn't that second ball now just a randomly selected ball from the original configuration of the urn, so the chances of it being red or green have to be equal.

In fact because the first choice was random, it doesn't matter whether you look at it or not. Because all of the probabilities apply in exactly the same way to both red and green balls: if the first drawing is red OR green then the chances of the second drawing are even.

You can keep drawing balls 'til the urn is empty, without knowing what the original configuration was the chances are always even.