Poll

8 votes (47.05%)
6 votes (35.29%)
3 votes (17.64%)
2 votes (11.76%)
6 votes (35.29%)
2 votes (11.76%)
3 votes (17.64%)
2 votes (11.76%)
8 votes (47.05%)
6 votes (35.29%)

17 members have voted

gordonm888
gordonm888
Joined: Feb 18, 2015
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October 18th, 2020 at 4:25:22 PM permalink
Quote: Gialmere



I see that this week's FiveThirtyEight column containing the election puzzle also has a similar Price is Right puzzle. Note the slight rule changes...


Officials answers to be released next week.



DUPLICATE


The item is "a randomly selected value between 0 and 100." Thus if we limit ourselves to two digits beyond the decimal point, there are a total of 99.99 distinct possible choices.

If the contestant A picks 33.33 and the contestant B picks 66.67 then the contestant C has these choices:

1. pick 0.00 or 0.01 and win 33.32 percent of the time; A wins 33.34% of time, B wins 33.33
2. pick 33.32 and win 33.32 percent of the time A wins 33.34, B wins 33.33
3. pick 33.34 and win 33.33 percent of the time: A wins 33.33, B wins 33.33 (optimal for C)
4. pick 66.68 and win 33.32 percent of the time A wins 66.66; B wins 0.001

If the 1st contestant picks 33.33 and the 2nd contestant picks 66.66 then the third contestant has these choices:

1. pick 0.00 or 0.01 and win 33.32 percent of the time (because the price cannot be zero); A wins 33.33% of time, B wins 33.33
2. pick 33.32 and win 33.32 percent of the time: A wins 33.33; B wins 33.34
3. pick 33.34 and win 33.32 percent of the time: A wins 33.33; B wins 33.34
4. pick 66.67 and win 33.33 percent of the time; A wins 66.5 and B wins 0.01 (optimal for C)

If A chooses 33.32 and B chooses 66.65 then C chooses 66.66 and A wins 66.64; B wins 0.01; C wins 33.34
If A chooses 33.32 and B chooses 66.66 then C chooses 33.33 and A wins 33.32; B wins 33.34; C wins 33.33
if A chooses 33.34 and B chooses 66.67 then C would choose either 0 or 33.33: A wins 33.33, B wins 33.33, C wins 33.33
if A chooses 33.34 and B chooses 33.33 then C would choose 33.35: A wins 0.01, B wins 33.33, C wins 66.65


IT is OPTIMAL for A to chooses 33.33; it is OPTIMAL for B to then choose either 66.67 or 66.66. In both these cases, A will win 33.33/99.99 or 1/3 of the time.
So many better men, a few of them friends, were dead. And a thousand thousand slimy things lived on, and so did I.
unJon
unJon
Joined: Jul 1, 2018
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October 18th, 2020 at 5:15:05 PM permalink
Quote: gordonm888

Quote: Gialmere



I see that this week's FiveThirtyEight column containing the election puzzle also has a similar Price is Right puzzle. Note the slight rule changes...


Officials answers to be released next week.



DUPLICATE


The item is "a randomly selected value between 0 and 100." Thus if we limit ourselves to two digits beyond the decimal point, there are a total of 99.99 distinct possible choices.

If the contestant A picks 33.33 and the contestant B picks 66.67 then the contestant C has these choices:

1. pick 0.00 or 0.01 and win 33.32 percent of the time; A wins 33.34% of time, B wins 33.33
2. pick 33.32 and win 33.32 percent of the time A wins 33.34, B wins 33.33
3. pick 33.34 and win 33.33 percent of the time: A wins 33.33, B wins 33.33 (optimal for C)
4. pick 66.68 and win 33.32 percent of the time A wins 66.66; B wins 0.001

If the 1st contestant picks 33.33 and the 2nd contestant picks 66.66 then the third contestant has these choices:

1. pick 0.00 or 0.01 and win 33.32 percent of the time (because the price cannot be zero); A wins 33.33% of time, B wins 33.33
2. pick 33.32 and win 33.32 percent of the time: A wins 33.33; B wins 33.34
3. pick 33.34 and win 33.32 percent of the time: A wins 33.33; B wins 33.34
4. pick 66.67 and win 33.33 percent of the time; A wins 66.5 and B wins 0.01 (optimal for C)

If A chooses 33.32 and B chooses 66.65 then C chooses 66.66 and A wins 66.64; B wins 0.01; C wins 33.34
If A chooses 33.32 and B chooses 66.66 then C chooses 33.33 and A wins 33.32; B wins 33.34; C wins 33.33
if A chooses 33.34 and B chooses 66.67 then C would choose either 0 or 33.33: A wins 33.33, B wins 33.33, C wins 33.33
if A chooses 33.34 and B chooses 33.33 then C would choose 33.35: A wins 0.01, B wins 33.33, C wins 66.65


IT is OPTIMAL for A to chooses 33.33; it is OPTIMAL for B to then choose either 66.67 or 66.66. In both these cases, A will win 33.33/99.99 or 1/3 of the time.



I donít think that works:

If A picks 33.33 then B can guarantee 33.32 by picking 0 and C will pick 33.34. Leaving A with 0.00.

I think A wants to pick 66.68.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
ThatDonGuy
ThatDonGuy 
Joined: Jun 22, 2011
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Thanks for this post from:
gordonm888
October 18th, 2020 at 5:59:28 PM permalink
Quote: unJon

Quote: gordonm888

Quote: Gialmere



I see that this week's FiveThirtyEight column containing the election puzzle also has a similar Price is Right puzzle. Note the slight rule changes...


Officials answers to be released next week.



DUPLICATE


The item is "a randomly selected value between 0 and 100." Thus if we limit ourselves to two digits beyond the decimal point, there are a total of 99.99 distinct possible choices.

If the contestant A picks 33.33 and the contestant B picks 66.67 then the contestant C has these choices:

1. pick 0.00 or 0.01 and win 33.32 percent of the time; A wins 33.34% of time, B wins 33.33
2. pick 33.32 and win 33.32 percent of the time A wins 33.34, B wins 33.33
3. pick 33.34 and win 33.33 percent of the time: A wins 33.33, B wins 33.33 (optimal for C)
4. pick 66.68 and win 33.32 percent of the time A wins 66.66; B wins 0.001

If the 1st contestant picks 33.33 and the 2nd contestant picks 66.66 then the third contestant has these choices:

1. pick 0.00 or 0.01 and win 33.32 percent of the time (because the price cannot be zero); A wins 33.33% of time, B wins 33.33
2. pick 33.32 and win 33.32 percent of the time: A wins 33.33; B wins 33.34
3. pick 33.34 and win 33.32 percent of the time: A wins 33.33; B wins 33.34
4. pick 66.67 and win 33.33 percent of the time; A wins 66.5 and B wins 0.01 (optimal for C)

If A chooses 33.32 and B chooses 66.65 then C chooses 66.66 and A wins 66.64; B wins 0.01; C wins 33.34
If A chooses 33.32 and B chooses 66.66 then C chooses 33.33 and A wins 33.32; B wins 33.34; C wins 33.33
if A chooses 33.34 and B chooses 66.67 then C would choose either 0 or 33.33: A wins 33.33, B wins 33.33, C wins 33.33
if A chooses 33.34 and B chooses 33.33 then C would choose 33.35: A wins 0.01, B wins 33.33, C wins 66.65


IT is OPTIMAL for A to chooses 33.33; it is OPTIMAL for B to then choose either 66.67 or 66.66. In both these cases, A will win 33.33/99.99 or 1/3 of the time.



I donít think that works:

If A picks 33.33 then B can guarantee 33.32 by picking 0 and C will pick 33.34. Leaving A with 0.00.

I think A wants to pick 66.68.



If B picks 0, then B will get only 33.32 after C picks 33.34.
On the other hand, if B picks 66.66, then B can get at least 33.33.
Remember that all three players are playing to maximize their own chance of winning.

unJon
unJon
Joined: Jul 1, 2018
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October 18th, 2020 at 6:09:10 PM permalink
Quote: ThatDonGuy

Quote: unJon

Quote: gordonm888

Quote: Gialmere



I see that this week's FiveThirtyEight column containing the election puzzle also has a similar Price is Right puzzle. Note the slight rule changes...


Officials answers to be released next week.



DUPLICATE


The item is "a randomly selected value between 0 and 100." Thus if we limit ourselves to two digits beyond the decimal point, there are a total of 99.99 distinct possible choices.

If the contestant A picks 33.33 and the contestant B picks 66.67 then the contestant C has these choices:

1. pick 0.00 or 0.01 and win 33.32 percent of the time; A wins 33.34% of time, B wins 33.33
2. pick 33.32 and win 33.32 percent of the time A wins 33.34, B wins 33.33
3. pick 33.34 and win 33.33 percent of the time: A wins 33.33, B wins 33.33 (optimal for C)
4. pick 66.68 and win 33.32 percent of the time A wins 66.66; B wins 0.001

If the 1st contestant picks 33.33 and the 2nd contestant picks 66.66 then the third contestant has these choices:

1. pick 0.00 or 0.01 and win 33.32 percent of the time (because the price cannot be zero); A wins 33.33% of time, B wins 33.33
2. pick 33.32 and win 33.32 percent of the time: A wins 33.33; B wins 33.34
3. pick 33.34 and win 33.32 percent of the time: A wins 33.33; B wins 33.34
4. pick 66.67 and win 33.33 percent of the time; A wins 66.5 and B wins 0.01 (optimal for C)

If A chooses 33.32 and B chooses 66.65 then C chooses 66.66 and A wins 66.64; B wins 0.01; C wins 33.34
If A chooses 33.32 and B chooses 66.66 then C chooses 33.33 and A wins 33.32; B wins 33.34; C wins 33.33
if A chooses 33.34 and B chooses 66.67 then C would choose either 0 or 33.33: A wins 33.33, B wins 33.33, C wins 33.33
if A chooses 33.34 and B chooses 33.33 then C would choose 33.35: A wins 0.01, B wins 33.33, C wins 66.65


IT is OPTIMAL for A to chooses 33.33; it is OPTIMAL for B to then choose either 66.67 or 66.66. In both these cases, A will win 33.33/99.99 or 1/3 of the time.



I donít think that works:

If A picks 33.33 then B can guarantee 33.32 by picking 0 and C will pick 33.34. Leaving A with 0.00.

I think A wants to pick 66.68.



If B picks 0, then B will get only 33.32 after C picks 33.34.
On the other hand, if B picks 66.66, then B can get at least 33.33.
Remember that all three players are playing to maximize their own chance of winning.



But
If A picks 33.33 and B picks 66.66, then how is B guaranteed anything? C picks 66.7, locking in 33.33 and B is hosed.

Maybe Iím thinking about it wrong.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
gordonm888
gordonm888
Joined: Feb 18, 2015
  • Threads: 38
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October 18th, 2020 at 6:28:06 PM permalink
Quote: unJon

Quote: ThatDonGuy

Quote: unJon

Quote: gordonm888

Quote: Gialmere



I see that this week's FiveThirtyEight column containing the election puzzle also has a similar Price is Right puzzle. Note the slight rule changes...


Officials answers to be released next week.



DUPLICATE


The item is "a randomly selected value between 0 and 100." Thus if we limit ourselves to two digits beyond the decimal point, there are a total of 99.99 distinct possible choices.

If the contestant A picks 33.33 and the contestant B picks 66.67 then the contestant C has these choices:

1. pick 0.00 or 0.01 and win 33.32 percent of the time; A wins 33.34% of time, B wins 33.33
2. pick 33.32 and win 33.32 percent of the time A wins 33.34, B wins 33.33
3. pick 33.34 and win 33.33 percent of the time: A wins 33.33, B wins 33.33 (optimal for C)
4. pick 66.68 and win 33.32 percent of the time A wins 66.66; B wins 0.001

If the 1st contestant picks 33.33 and the 2nd contestant picks 66.66 then the third contestant has these choices:

1. pick 0.00 or 0.01 and win 33.32 percent of the time (because the price cannot be zero); A wins 33.33% of time, B wins 33.33
2. pick 33.32 and win 33.32 percent of the time: A wins 33.33; B wins 33.34
3. pick 33.34 and win 33.32 percent of the time: A wins 33.33; B wins 33.34
4. pick 66.67 and win 33.33 percent of the time; A wins 66.5 and B wins 0.01 (optimal for C)

If A chooses 33.32 and B chooses 66.65 then C chooses 66.66 and A wins 66.64; B wins 0.01; C wins 33.34
If A chooses 33.32 and B chooses 66.66 then C chooses 33.33 and A wins 33.32; B wins 33.34; C wins 33.33
if A chooses 33.34 and B chooses 66.67 then C would choose either 0 or 33.33: A wins 33.33, B wins 33.33, C wins 33.33
if A chooses 33.34 and B chooses 33.33 then C would choose 33.35: A wins 0.01, B wins 33.33, C wins 66.65


IT is OPTIMAL for A to chooses 33.33; it is OPTIMAL for B to then choose either 66.67 or 66.66. In both these cases, A will win 33.33/99.99 or 1/3 of the time.



I donít think that works:

If A picks 33.33 then B can guarantee 33.32 by picking 0 and C will pick 33.34. Leaving A with 0.00.

I think A wants to pick 66.68.



If B picks 0, then B will get only 33.32 after C picks 33.34.
On the other hand, if B picks 66.66, then B can get at least 33.33.
Remember that all three players are playing to maximize their own chance of winning.



But
If A picks 33.33 and B picks 66.66, then how is B guaranteed anything? C picks 66.7, locking in 33.33 and B is hosed.

Maybe Iím thinking about it wrong.



UnJon, I agree with your objection. That branch of my solution should be eliminated.
So many better men, a few of them friends, were dead. And a thousand thousand slimy things lived on, and so did I.
Gialmere
Gialmere
Joined: Nov 26, 2018
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October 19th, 2020 at 8:13:32 AM permalink


A, B, C, and D are all different positive integers.

A < B < C < D

Find the largest D so that

1/A + 1/B + 1/C + 1/D = 1
Have you tried 22 tonight? I said 22.
Wizard
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Wizard
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Gialmere
October 19th, 2020 at 9:31:40 AM permalink

I get 42.

1/2 + 1/3 + 1/7 + 1/42 = 1

I won't give a solution unless the answer is confirmed, but it was pretty much trial and error.
It's not whether you win or lose; it's whether or not you had a good bet.
charliepatrick
charliepatrick
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Gialmere
October 19th, 2020 at 11:26:47 AM permalink
Quote: Wizard


I get 42.

1/2 + 1/3 + 1/7 + 1/42 = 1

I won't give a solution unless the answer is confirmed, but it was pretty much trial and error.

I think you can prove it.

(i) A=2.
If A wasn't 2 then the first largest value of the first three would be 1/3+1/4+1/5 = 47/60 leaving 13/60 for 1/D. This is greater than 1/C (1/5) so this cannot be a solution.

(ii) B=3 or 4.
Similar logic that 1/2+1/5 = 3/10. As 1/C+1/D=3/10, 1/C>3/20. The only value left is 1/6 (1/7<3/20) and this doesn't create a solution.
Also 1/2+1/6 leaves 1/3, so 1/C >1/6 which contadicts 1/6 being 1/B.

(iii) 1/2 + 1/4 (do this one first for interest).
Now find 1/C + 1/D which add up to 1/4. The values for 1/C can be between 1/5 and 1/7 (as 1/8 leads to 1/D=1/8).
1/5 gives : 1/2 1/4 1/5 1/20 (10 5 4 1 / 20)
1/6 gives : 1/2 1/4 1/6 1/12 (6 3 2 1 / 12)
1/7 doesn't work

(iv) 1/2 + 1/3 .
Now find 1/C + 1/D which add up to 1/6. The values for 1/C can be between 1/7 and 1/11 (as 1/12 leads to 1/D=1/12).
1/7 gives : 1/2 1/3 1/7 1/42 (21 14 6 1 / 42)
1/8 gives : 1/2 1/3 1/8 1/24 (12 8 3 1 / 24)
1/9 gives : 1/2 1/3 1/9 1/18 (9 6 2 1 / 18)
1/10 and 1/11 don't work

Also note that 1/N - 1/(N+1) = 1/(N (N+1)) so 1/N = 1/(N+1) + 1/(N (N+1)). In this case N=6 gives 1/6 = 1/7+1/42.
Gialmere
Gialmere
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October 19th, 2020 at 4:27:15 PM permalink
Quote: Wizard


I get 42.

1/2 + 1/3 + 1/7 + 1/42 = 1

I won't give a solution unless the answer is confirmed, but it was pretty much trial and error.


Quote: charliepatrick

I think you can prove it.


(i) A=2.
If A wasn't 2 then the first largest value of the first three would be 1/3+1/4+1/5 = 47/60 leaving 13/60 for 1/D. This is greater than 1/C (1/5) so this cannot be a solution.

(ii) B=3 or 4.
Similar logic that 1/2+1/5 = 3/10. As 1/C+1/D=3/10, 1/C>3/20. The only value left is 1/6 (1/7<3/20) and this doesn't create a solution.
Also 1/2+1/6 leaves 1/3, so 1/C >1/6 which contadicts 1/6 being 1/B.

(iii) 1/2 + 1/4 (do this one first for interest).
Now find 1/C + 1/D which add up to 1/4. The values for 1/C can be between 1/5 and 1/7 (as 1/8 leads to 1/D=1/8).
1/5 gives : 1/2 1/4 1/5 1/20 (10 5 4 1 / 20)
1/6 gives : 1/2 1/4 1/6 1/12 (6 3 2 1 / 12)
1/7 doesn't work

(iv) 1/2 + 1/3 .
Now find 1/C + 1/D which add up to 1/6. The values for 1/C can be between 1/7 and 1/11 (as 1/12 leads to 1/D=1/12).
1/7 gives : 1/2 1/3 1/7 1/42 (21 14 6 1 / 42)
1/8 gives : 1/2 1/3 1/8 1/24 (12 8 3 1 / 24)
1/9 gives : 1/2 1/3 1/9 1/18 (9 6 2 1 / 18)
1/10 and 1/11 don't work

Also note that 1/N - 1/(N+1) = 1/(N (N+1)) so 1/N = 1/(N+1) + 1/(N (N+1)). In this case N=6 gives 1/6 = 1/7+1/42.


Correct!

The fun little twist in this one is that it first asks for the largest number and then quickly turns it on its head by shoving it into a fraction and asking for the smallest value. Clever tweak.

----------------------------------------

Have you tried 22 tonight? I said 22.
Wizard
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Wizard
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October 19th, 2020 at 4:36:17 PM permalink
If you ask this again, I think you should specify D can't be infinity. Because 1/2 + 1/3 + 1/6 + 1/infinity = 1.
It's not whether you win or lose; it's whether or not you had a good bet.

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