## 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**

If B doesn't stop A making RF (because B can't) then A wins, so B has to pick something like Four Aces thru Jacks. Then A makes 10-high Str Flush which can't be beaten.

Quote:charliepatrickNot sure about this but it seems to work

A's first move has to stop B making Royal Flush. If A picks cards all higher than T's, then B can pick Four Tens and cannot be stopped from making 10-high Str Flush or winning with the four 10s. So A should pick four 10s and any other card.

If B doesn't stop A making RF (because B can't) then A wins, so B has to pick something like Four Aces thru Jacks. Then A makes 10-high Str Flush which can't be beaten.

I believe you have provided the correct answer, Charlie.

Quote:GialmereWhat about the joker in PGP?

Since this problem is getting stale, I'll dispense with the spoiler tags.

Let's rephrase the question to:

"Two numbers are chosen between 0 and 1 along the uniform distribution. That will divide the segment into three pieces. What is the probability the longest piece is less than than 0.5?"

The answer would be the same if the first number were drawn between 0 and 0.5, since it must be in one half or the other. May as well be the half between 0 and 0.5. Let's call that number x.

Let the second number be y. The only range for y, where the longest segment is less than 0.5, is from 0.5 to 0.5+x. This distance equals x.

The range of values of x and y where the longest segment is under 0.5 = ...

Integral from 0 to 0.5 of x dx =

x^2/2 from 0 to 0.5

= 0.5^2/2 = 1/8.

However, we must double this, because this only counts the values of x<0.5.

So, the answer is 2/8 = 1/4.

Quote:charliepatrickNot sure about this but it seems to work

A's first move has to stop B making Royal Flush. If A picks cards all higher than T's, then B can pick Four Tens and cannot be stopped from making 10-high Str Flush or winning with the four 10s. So A should pick four 10s and any other card.

If B doesn't stop A making RF (because B can't) then A wins, so B has to pick something like Four Aces thru Jacks. Then A makes 10-high Str Flush which can't be beaten.

Correct!

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My wife says she is going to divorce me because I love poker more then her.

I think she’s bluffing.

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...

Quote:The RiddlerThis week, we return to the brilliant and ageless game show, “The Price is Right.” In a modified version of the bidding round, you and two (not three) other contestants must guess the price of an item, one at a time.

Assume the true price of this item is a randomly selected value between 0 and 100. (Note: The value is a real number and does not have to be an integer.) Among the three contestants, the winner is whoever guesses the closest price without going over. For example, if the true price is 29 and I guess 30, while another contestant guesses 20, then they would be the winner even though my guess was technically closer.

In the event all three guesses exceed the actual price, the contestant who made the lowest (and therefore closest) guess is declared the winner. I mean, someone has to win, right?

If you are the first to guess, and all contestants play optimally (taking full advantage of the guesses of those who went before them), what are your chances of winning?

Officials answers to be released next week.

I would pick two hairs over 200/3 = let's say 66.666668. The second player would pick a hair over 100/3, say 33.333334. The last player would pick 0.

This way you have essentially a 1/3 chance of winning, despite the positional disadvantage.

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.

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.