A little math quiz
September 21st, 2012 at 6:58:11 AM
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Here is a little quiz question to exercise your brain...
Three people play three rounds of a betting game.
The person who comes in last must double the money of the other two.
After three rounds, each person has been last only once and each has $24.
What are the three starting bankrolls?
Place your answer in spoiler tags.
Source: MENSA Puzzle Clandar, 2010, May 1
Three people play three rounds of a betting game.
The person who comes in last must double the money of the other two.
After three rounds, each person has been last only once and each has $24.
What are the three starting bankrolls?
Place your answer in spoiler tags.
this one is easy.
Source: MENSA Puzzle Clandar, 2010, May 1
In a bet, there is a fool and a thief.
- Proverb.
September 21st, 2012 at 7:43:55 AM
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Quote: WongBoWhat are the three starting bankrolls?
$39, $21, $12
September 21st, 2012 at 8:00:13 AM
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Working backwards,
Just before the last round (12, 12, 48), the guy with 48 having lost. This must be the case, because the two 12's were doubled.
Just before that (6, 42, 24). This must be the case because of the symmetry, the 6 and 24 are doubled, the remainder must belong to the middle player.
Just before that (39, 21, 12). This must be the case because 42 and 24 must be the doubled amounts, the remainder must belong to the first player.
So, (39, 21, 12).
Now working forwards to double check: (39,21,12) -> (6,42,24) -> (12,12,48) -> (24,24,24).
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September 21st, 2012 at 8:17:43 AM
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Pacomartin is correct.
There was no stipulation about how the results were resolved.
Thank you, teliot for posting the work! Saved me some typing.
There was no stipulation about how the results were resolved.
Thank you, teliot for posting the work! Saved me some typing.
In a bet, there is a fool and a thief.
- Proverb.
September 21st, 2012 at 9:20:56 AM
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FOLLOW UP
A community group has 500 people. At the dance, new members pay $20 for tickets whereas longtime members pay $14. The planner assumes that all the longtime members will come, but 30% of the new members won't attend. How much ticket revenue does he think will be collected?"
A community group has 500 people. At the dance, new members pay $20 for tickets whereas longtime members pay $14. The planner assumes that all the longtime members will come, but 30% of the new members won't attend. How much ticket revenue does he think will be collected?"
September 21st, 2012 at 9:24:21 AM
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Quote:FOLLOW UP
A community group has 500 people. At the dance, new members pay $20 for tickets whereas longtime members pay $14. The planner assumes that all the longtime members will come, but 30% of the new members won't attend. How much ticket revenue does he think will be collected?"
Cute.
Revenue = $7000
n = new
o = long time
r = revenue
n + o = 500
r = 0.70*n*$20 + o*$14
= $14*(500 -o) + o*$14
= $14*500 - $14*o + o*$14
= $7000
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September 28th, 2012 at 7:54:33 AM
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Ok, here is another one...
Two players. A and B take turns drawing one ball at a time
From a bag containing four red and two black balls.
If Player A goes first, what is the probability
That Player A will draw a red ball before Player B?
Source MENSA calendar 2012, Jan.7
Two players. A and B take turns drawing one ball at a time
From a bag containing four red and two black balls.
If Player A goes first, what is the probability
That Player A will draw a red ball before Player B?
there are only two cases that lead to this outcome.
Source MENSA calendar 2012, Jan.7
In a bet, there is a fool and a thief.
- Proverb.
September 28th, 2012 at 8:10:16 AM
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Quote: WongBoOk, here is another one...
Two players. A and B take turns drawing one ball at a time
From a bag containing four red and two black balls.
If Player A goes first, what is the probability
That Player A will draw a red ball before Player B?there are only two cases that lead to this outcome.
Source MENSA calendar 2012, Jan.7
There is a 2/3 chance Player A will pick the red before the black on the first try. If he doesn't there is 1/5 chance Player B will pick the remaining black ball on his try, thus assuring that Player A will pick red.
Therefore -
2/3 + 1/3(1/5) = 11/15 chance of Player A drawing a red ball before Player B
Therefore -
2/3 + 1/3(1/5) = 11/15 chance of Player A drawing a red ball before Player B
September 28th, 2012 at 8:12:30 AM
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Could you use spoiler tags please!
Correct answer.
Correct answer.
In a bet, there is a fool and a thief.
- Proverb.
September 28th, 2012 at 8:14:10 AM
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Quote: WongBoOk, here is another one...
Two players. A and B take turns drawing one ball at a time
From a bag containing four red and two black balls.
If Player A goes first, what is the probability
That Player A will draw a red ball before Player B?there are only two cases that lead to this outcome.
Source MENSA calendar 2012, Jan.7
p(A picks red 1st) = 4/6
p(A picks black, then B picks black, then A picks red) = (2/6)*(1/5)*(4/4) = 2/30
sum = 22/30
Double-check: since A goes first, the only way A doesn't draw red before B is to pick black first, then have B pick red next, because there aren't enough black balls in the urn to last for 4 turns.
p(A black, B red) = (2/6)*(4/5) = 8/30
8/30 + 22/30 = 1. Verified.
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice."
-- Girolamo Cardano, 1563
September 28th, 2012 at 8:17:20 AM
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Quote: WongBoCould you use spoiler tags please!
My apologies, never used these before, but will going forward.
September 28th, 2012 at 8:19:55 AM
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If you hit the quote button on my earlier post you can see what to type.
Or click formatting codes, at the bottom right of the screen when writing a post.
Or click formatting codes, at the bottom right of the screen when writing a post.
In a bet, there is a fool and a thief.
- Proverb.
