Puzzles with numbers and things

[davis_greatest]

1) What is the expected number of turns required to mark 1 ball? 1 turn
2) What is the expected number of turns required to mark both balls? infinity???

3) What is the expected number of balls marked after 1 turn? 1
4) What is the expected number of balls marked after 2 turns? 1.5
5) What is the expected number of balls marked after 3 turns? 1.75
6) What is the expected number of balls marked after 100 turns? 1.9999999999... would have to check that better

1) is correct
2) is not

3), 4) and 5) are correct

6) Well, yes, nearly 2 - but what exactly?

[davis_greatest]

or 6) better: 2-2^-39

Where do you get 39 from?! Are you "doing a chasmmi" and solving the problem perfectly but with the wrong numbers?

[snookersfun]

did you see that one above for 6?
2-2^-40??

well, I tried to compensate before for the 1 for the first round....

[davis_greatest]

did you see that one above for 6?
2-2^-40??

Yes, I did... but check the question... you will see no mention of 39 or 40

[snookersfun]

aaargghhh that was a Chasmmi:
2-2^-99, sooorrry

[davis_greatest]

Yes! I think we have all the answers now!


1) What is the expected number of turns required to mark 1 ball? 1
2) What is the expected number of turns required to mark both balls? 3

Questions 3 to 6 can be solved using the formula in message 905, with n=2 balls, marking m=1 ball each time.

3) What is the expected number of balls marked after 1 turn? 1
4) What is the expected number of balls marked after 2 turns? 1.5
5) What is the expected number of balls marked after 3 turns? 1.75
6) What is the expected number of balls marked after 100 turns? 2(1-(0.5)^100) = 2 - 0.5^99

So, what does this show? It shows that:
- the expected number of turns required to mark 2 balls is 3,
but
- the expected number of balls marked after 3 turns is not 2!

In fact, the expected number of balls marked after k turns only tends to 2 as k tends to infinity.

So, back to snookersfun's question in round 66 where we had a bag of 600 balls: this is the same principle as why the expected number of turns required to mark 90% of the balls is NOT the same as the number of turns after which the expected proportion of balls marked is 90%.

[davis_greatest]

Updated scoreboard

For round 69, I shall give half a point each to chasmmi and to snookersfun...

HERE IS THE SCOREBOARD AFTER ROUND 69 BUT BEFORE ROUND 68, APE BREAK MADNESS

snookersfun……………………….…..31½
abextra...............................16
davis_greatest.....................13½
Vidas..................................12½
robert602.............................6
elvaago...............................6
chasmmi..............................5½
The Statman……………………..……4
Semih_Sayginer.....................2½

[April madness]

if I change my nickname to Ape madness, can I get a point on this thread?


sorry for off-topic

[davis_greatest]

if I change my nickname to Ape madness, can I get a point on this thread?

Change it to Barry the Baboon, and we'll see

[April madness]

I can't do that, my name is Madness

[davis_greatest]

Then how about Baboon Madness?

Anyhow....

Round 70

Gordon is playing with 7 snooker balls. He arranges them on the floor to form 5 rows of 3 snooker balls in each row. Show how he can do this.

[elvaago]

Re: 921

I object to this. How can you mark half a ball or one quarter of a ball? Do you stop halfway the marking process?

I submit that the expected number of balls is either 0, 1, or 2. (And I was working on my answers in this format before the forum went off the air. :-))

[davis_greatest]

Re: 921

I object to this. How can you mark half a ball or one quarter of a ball? Do you stop halfway the marking process?

I submit that the expected number of balls is either 0, 1, or 2. (And I was working on my answers in this format before the forum went off the air. :-))

snookersfun asked me a similar question about her 600 balls question, and my answer is the same: the expected number does not need to be a whole number. If you toss a coin, the expected number of heads is one-half, not zero or one.

The expected number is:

Sum over each outcome j of (j times the probability that the outcome is j)

[elvaago]

I suppose you're right. You can expect something, but you need not be right. Just like I expect to win the lottery tomorrow, but I probably won't.

[davis_greatest]

I expect you're right there!