## 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**

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.

Quote:gordonm888Quote: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 don’t think that works:

I think A wants to pick 66.68.

Quote:unJonQuote:gordonm888Quote: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 don’t 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.

Quote:ThatDonGuyQuote:unJonQuote:gordonm888Quote: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 don’t 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

Maybe I’m thinking about it wrong.

Quote:unJonQuote:ThatDonGuyQuote:unJonQuote:gordonm888Quote: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 don’t 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.

ButIf 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 I’m thinking about it wrong.

UnJon, I agree with your objection. That branch of my solution should be eliminated.

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

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.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) 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.

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:charliepatrickI 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!

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