What are the odds?

MrGameShow · 2009-03-20 03:00:16

I'm creating a new game where there are five levels. Each level contains one dollar sign (a WIN) and the rest of the spaces are X (a STRIKE).

There are 5 levels, starting with 2 on the first level, up to 6 on the last level.

You must find the dollar sign / win on each level before you can advance, and the game is over when you have found three strikes.

I'm trying to figure out the odds of winning on each level, and the overall game.

I know combinatorics are involved, but just not sure how to figure it all out.

Any help would be appreciated!

mathsyperson · 2009-03-20 06:47:16

I would do this by making a table of level against strikes and filling it with probabilities.

| 0 | 1 | 2 |
Level 1: | 1/2 | 1/2 | 0 |
Level 2: | 1/6 | 1/3 | 1/3|
Level 3: |1/24| 1/8 |5/24|
:
:

In Level 1, you will either win immediately or use up a strike, and the probabilities of those are even.

In Level 2, you will use 0, 1 or 2 strikes, all with equal probability.
If you didn't use any in the previous level, then those are the outcomes for total strike usage.
If you used one, then you'll have either used 1, 2 or 3 overall after winning Level 2.
Combine these to get the probabilities in the Level 2 row of the table.
(There's a 1/6 chance that 3 strikes have been used by now, but we're not interested in that because if that happens then you've lost)

The Level 3 probabilities are found in the same way.

P(0 strikes) = 1/6*1/4.
P(1 strike) = 1/6*1/4 + 1/3*1/4.
P(2 strikes) = 1/6*1/4 + 1/3*1/4 + 1/3*1/4.

You can get rows for Levels 4 and 5 like this as well.

You then get the total probabilities of winning a certain level by adding up its row.
You will definitely win Level 1, you have a 5/6 chance of winning Level 2, and a 9/24 chance of winning Level 3.

By my calculations, you then have a 7/60 chance of winning Level 4, and a 1/36 chance of winning Level 5.

So if it's an all-or-nothing game, a quicker way to do it would be to get the contestant to throw two dice and give them the money if they score a 12.

MrGameShow · 2009-03-20 15:01:40

I'm not quite sure where the numbers are coming from - can you break down the Level 3 equations a little more to help me understand how to figure out the 4th and 5th?

I appreciate the solutions, but it doesn't help to be able to solve these things on my own later

mathsyperson · 2009-03-20 23:56:19

I got the probability for 0 strikes after the 3rd level by saying that that will happen only if no strikes had been used after winning Level 2 (probability 1/6), and no strikes are used on Level 3 (probability 1/4). So the probability of having used no strikes after winning Level 3 is 1/6*1/4 = 1/24.

For one strike it's a little harder, because there are two ways of that happening.
Either no strikes were used before Level 3 and one strike was used at Level 3
or one strike was used before Level 3 and no strikes are used at Level 3.
The probabilities for those two events are 1/6*1/4 and 1/3*1/4 respectively, and adding them gets the total probability.

For two strikes it's a similar thing, but now there are three ways for it to happen:
0 strikes before Level 3, 2 during Level 3
1 before, 1 during
2 before, 0 during

I found another way of doing it that I think is much better than that method though.
Every game can be represented by a 5-digit number, where each digit represents the number of strikes used on a level.

The luckiest possible game would be 00000, and the worst possible would be 12345.
There are 6! = 720 different possible numbers, and each one of them is equally likely.

A number translates to a 'win' if its digits total no more than 2, so now you just need to list them.

00000
00001
00010
00100
01000
10000
00002
00020
00200
02000
*
00011
00101
01001
10001
00110
01010
10010
01100
10100
11000

(*Note that 20000 isn't valid because you can't use 2 strikes on the first level)

So out of the 720 possible games, 20 of them are winners.
Therefore the probability of a win is 20/720 = 1/36.
In this case it was easy enough to list them all manually, but if the problem had been slightly different it could still be solved by combinatorics.

Math Is Fun Forum

#1 2009-03-20 03:00:16

What are the odds?

#2 2009-03-20 06:47:16

Re: What are the odds?

#3 2009-03-20 15:01:40

Re: What are the odds?

#4 2009-03-20 23:56:19

Re: What are the odds?

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