First, the answer to the previous Math Question, as Pax and Evan answered, is 50%. Here's why:
The probability of the team of at least one run is 100% minus the probability of the team scoring no runs. There is a number of ways in which the team can score no runs. Let O be a strikeout, and H be a hit.
- The team can get three strikeouts (OOO). Probability = (1/2)(1/2)(1/2) = (1/8)
- The team can get three strikeouts and one hit in three ways:
HOOO, OHOO, OOHO. Probability of each: (1/2)(1/2)(1/2)(1/2) = 1/16. Probability of one hit and three strikeouts: 3 * (1/16) = (3/16).
- The team can get three strikeouts and two hits in six ways:
HHOOO, HOHOO, HOOHO, OHHOO, OHOHO, OOHHO. Probability of each: (1/2)(1/2)(1/2)(1/2)(1/2) = 1/32. Probability of one hit and three strikeouts = 6 * 1/32 = 6/32
Probability of no hits: 1 - ( (1/8) + (3/16) + (6/32) ) = 1/2 = 50%.
Now, here's two more math questions:
Math Question #2: What's the most amount of money that a contestant can win on a single show of Jeopardy!? Assume that you can place the Daily Doubles wherever you want.
Math Question #3: Can you guarantee a tie in Tic-Tac-Toe for X? For O? How is it done?
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Sorry. I changed both my snwers and then I ended up being at like 9 comments so I figured it would be easier to just erase the ones I wrote and repost.
Jeopardy question: $566,400
For max jeopardy money the daily double is in a $200 spot in the first round and is picked as the last question in the round. There are six categories. So in the first round you get
[(200*5)+(400*6)+(600*6)+(800*6)+(1000*6)]=$17,800
The daily double gives you $35,600 after round 1.
In round 2 there are 2 daily doubles, put them in the $400 spots and pick them last. So you get
[(400*4)+(800*6)+(1200*6)+(1600*6)+(2000*6)]=$35,200
add the amount from round one
35,600+35,200= $70,800
then daily double #1
70,800*2 = 141,600
daily double #2
141,600*2 = 283,200
Then final jeopardy
283,200*2 = $566,400
Question #3 - in tic tac toe (which some of my students ubelievably refer to as "x-o-x") the entire game hinges on the second move.
Assuming x is always first and o is always second, by the way.
x should never lose, but can guarantee victory if o makes the wrong second move.
o should tie every time.
I'm not really sure how to phrase this as a proof, but see if you can follow even though I'll probably make this more confusing than it needs to be.
Assuming x is trying to win and not tie, in the first move x should pick any corner. To guarantee a tie, o should then pick the middle square. After that the game is just a series of blocks to a tie. This is the ideal game.
if x picks a corner first and o picks any square other than the middle, then x is guaranteed to win. in the 3rd move x should pick a square where o has to block and is not in line with the original o. After o blocks, x can pick a square that gives two paths to victory. o will block one, but x will then win on the next move.
This doesn't really go towards actually proving anything because there are more possible starting moves and such, but this should be the ideal game.
In the ideal game, x should pick a corner, o should pick the middle, x should pick the opposite corner, and o should pick a non-corner. Then blocks to a tie.
In this game, x actually has 2 chances for o to make a mistake. x can win as described in the previous comment. Also x will win if in the 4th move o chooses a corner rather than a non-corner.
If the game goes x-corner, o-center, x-opposite corner, o-any remaining corner, x-the one remaining corner. At this point x has two paths to victory and will definitely win.
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