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Puzzles with numbers and things
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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.
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Originally Posted by elvaagoRe: 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. :-))
The expected number is:
Sum over each outcome j of (j times the probability that the outcome is j)
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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. :-))
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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.
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Originally Posted by April madnessif 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
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if I change my nickname to Ape madness, can I get a point on this thread?
sorry for off-topic
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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
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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%.
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Originally Posted by snookersfundid you see that one above for 6?
2-2^-40??
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did you see that one above for 6?
2-2^-40??
well, I tried to compensate before for the 1 for the first round....
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Originally Posted by snookersfunor 6) better: 2-2^-39Are you "doing a chasmmi" and solving the problem perfectly but with the wrong numbers?
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Originally Posted by snookersfun1) 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
2) is not
3), 4) and 5) are correct
6) Well, yes, nearly 2 - but what exactly?
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Originally Posted by chasmmiok im a bit confused but...
1) 1
2) 3
3) 1
4) 1.5
5) 2
6) 2
???
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