Numbers

Made by: Robby Donoho – Purdue University Senior (E-Mail: rdonoho@purdue.edu)
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You also don’t get fantasy points for a game winning goal scored in a shootout. How come this is not a game winning goal?

It is because it isn’t happening within the natural flow of the game. The game has ended, and the NHL in it’s infinite wisdom has chosen this method of breaking a tie during the regular season.

I personally like the shootout. I can’t see why you wouldn’t, but I understand opponent’s to the shootout points of view to some extent.

What would, other than a full O.T. period be an alternative solution?

I am interested in how to calculate a game with uneven odds.

For instance, given a dice game where one player has 2 dice, and the other player has 3, what are the odds of each winning / tying. (The player with the highest total value wins.)

Thanks
one additional point: im much more interested in the how than anything else, and I’m looking for a mathematical way of doing it.

Lemme see if I understand.

Player A rolls two dice.
Player B rolls three dice.
Highest total wins.

Gimme a minute, I’m working on it.

Ok, done.

Possible outcomes for 2 dice and their probabilities (sum – probability)
2 – 1/36
3 – 2/36
4 – 3/36
5 – 4/36
6 – 5/36
7 – 6/36
8 – 5/36
9 – 4/36
10 – 3/36
11 – 2/36
12 – 1/36

Possible outcomes for 3 dice (sum – probability)
3 – 1/216
4 – 3/216
5 – 6/216
6 – 10/216
7 – 15/216
8 – 21/216
9 – 25/216
10 – 27/216
11 – 27/216
12 – 25/216
13 – 21/216
14 – 15/216
15 – 10/216
16 – 6/216
17 – 3/216
18 – 1/216

So you want P(3-dice sum > 2-dice sum)

This would be the sum of the following probabilities:

P(3 dice-sum >12) OR
P(3-dice sum = 12 AND 2-dice sum <12) OR
P(3-dice sum = 11 AND 2-dice sum <11) OR
P(3-dice sum = 10 AND 2-dice sum <10) OR
P(3-dice sum = 9 AND 2-dice sum <9) OR
P(3-dice sum = 8 AND 2-dice sum <8) OR
P(3-dice sum = 7 AND 2-dice sum <7) OR
P(3-dice sum = 6 AND 2-dice sum <6) OR
P(3-dice sum = 5 AND 2-dice sum <5) OR
P(3-dice sum = 4 AND 2-dice sum <4) OR
P(3-dice sum = 3 AND 2-dice sum <3)

You do not need P(3-dice sum = 2 AND 2-dice sum <2), because the sum of the 2-dice roll cannot be less than 2.

Take ‘em one at a time:

P(3 dice-sum >12) = 21/216 + 15/216 + 10/216 + 6/216 + 3/216 + 1/216 = 56/216 = 2016/7776 ( I got a denominator of 7776 because of what happens next).

P(3-dice sum = 12 AND 2-dice sum <12) is the product of each probability, so it equals
25/216 * (1/36 + 2/36 + 3/36 + 4/36 + 5/36 + 6/36 + 5/36 + 4/36 + 3/36 + 2/36) =
25/216 * 35/36 =
875/7776

Similarly,

P(3-dice sum = 11 AND 2-dice sum <11) = 27/216 * 33/36 = 891/7776

P(3-dice sum = 10 AND 2-dice sum <10) = 27/216 * 30/36 = 810/7776

P(3-dice sum = 9 AND 2-dice sum <9) = 25/216 * 26/36 = 650/7776

P(3-dice sum = 8 AND 2-dice sum <8) = 21/216 * 21/36 = 441/7776

P(3-dice sum = 7 AND 2-dice sum <7) = 15/216 * 15/36 = 225/7776

P(3-dice sum = 6 AND 2-dice sum <6) = 10/216 * 19/36 = 100/7776

P(3-dice sum = 5 AND 2-dice sum <5) = 6/216 * 6/36 = 36/7776

P(3-dice sum = 4 AND 2-dice sum <4) = 3/216 * 3/36 = 9/7776

P(3-dice sum = 3 AND 2-dice sum <3) = 1/216 * 1/36 = 1/7776

Ok, add ‘em all up:
2016/7776
875/7776
891/7776
810/7776
650/7776
441/7776
225/7776
100/7776
36/7776
9/7776
1/7776
———
Sum = 6054/7776, or 1009/1296, or .7785, or 77.85%

Choose your answer.

I hope you actually follow the process for two reasons:
1) To learn a little bit more about how this type of problem flows, and
2) I am human. I could have very easily made a slight error somewhere.

Best bet is to do it again on your own, and see if your answers match.

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A team at our school is not that bad, honestly. But we have had bad records in the past years and nobody on the team seems to think we can win. We try, it’s not like were fooling around or not trying. We just go into a game thinking were going to lose, which I know it not a good thing. Our coach has told us we can win and we’re not going to start winning unless we can believe we can, but still nobody thinks we can. How can I get my teammates to change their minds?

Show them YOU want to win.

A good leader results in a good team.

As long as someone can step up and MAKE your team win, everyone will follow.

Nothing is more important than that person that busts their butt everyday.

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I have a hundred dollars and I’m trying to win 2000. I know the likelihood isn’t very high either way, but out of these options which one would be the best — buying 100 1$ tickets, buying 50 2$ tickets, buying 20 5$ tickets, buying 10 10$ tickets or 5 20$ tickets? I don’t know much about scratch tickets so which combination would give me the highest probability of winning?

1. Go to the website of your state lotto and find out what the odds are for each scratch and win ticket.
2. Find out how many tickets are left and how many prizes are left and what they are(the sizes of prizes left)
3. Buy the more expensive tickets, either the $10 or $20 tickets as in general, they print a less of these and the prizes are always much bigger when you do win. Also, if you think about it, what are the tickets that most people buy? You guessed it, the cheaper tickets. So, they make a lot more of them and the prizes get burned up pretty quick.

Hope that helps.

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