Probability question

Factual Questions
1

On the game show The Price is Right, there’s a game where the contestant is shown five prizes. The price of each prize is shown, with one number missing from each price. The contestant is given five blocks. Each block has a different number on it, and the contestant has to use the blocks to fill in the missing numbers in the five prizes, using one block per prize. What are the odds that a contestant will place all five blocks incorrectly? (Example: The order of the missing numbers is 2,4,5,1,8. The contestant uses the order 1,5,8,2,4)

I believe the odds of getting all five blocks placed correctly would be 1 in 120, but I don’t know how to figure getting all the blocks placed incorrectly.

2

If this was purely a game of chance, which it isn’t, it would be 1 in 5.

First pick - 4 in 5 (80%)
Second pick - 3 in 4 (75% of 80% = 60%)
Third pick - 2 in 3 (66.6% of 60% = 40%)
Fourth pick - 1 in 2 (50% of 40% = 20%)

I’m sure there is a more elegant way of calculating this but I think it’s right.

3

If the odds are actually 1 in 120 of doing it correctly, wouldn’t the odds of doing it incorrectly be 119 in 120? The probabilities have to add up to 100%, or in this case 120.

4

ski nailed it. You are correct that the odds of a random placement of 5 blocks being in the correct order are 1 in 120. The number of permutations of x blocks is x! and 5! = 120. Only one of those can be correct (if all the numbers are different). Therefore the number of incorrect placements are all the combinations that aren’t correct, or 119 in 120.

5

Yes but he was looking for the probability of getting all 5 numbers wrong. You’re forgetting about the outcomes where only one, two or three numbers are correct.

6

The chances of placing all the blocks incorrectly are:

1st pick = 1/4 (1 chance of being right, 4 of being wrong)
2nd pick = 1/3 (1 chance of being right, 3 of being wrong)
3rd pick = 1/2 (1 chance of being right, 2 of being wrong)
4th pick = 1/1 (1 chance of being right, 1 of being wrong)

Looking at this as an accumulator we have:

5/4 x 4/3 x 3/2 x 2/1 = 120/24 = 5 = 4/1.

The chances of placing all the blocks incorrectly are 4/1 or 1 in 5.

7

Which is what manwithaplan said.

8

Whoops, I just read your question again. You don’t want to know how many combinations are merely wrong, you want to know how many combinations are there where all the blocks are in the wrong positions.

This is much more interesting. Let me get back to you on that one. . . .

9

Damn, too late again. :mad:

10

jseigle, why do you want to get back to him/her? It’s been answered already, most clearly by Nostradamus.

11

Sorry - didn’t see your last post.

12

I think it works out that:

Total arrangements = 5! = 120
All correct = 1! = 1
All wrong = 4! = 24

(Which is what was said before, but I prefer to know the general idea!).

13

Well, I think the answer is 1 in 3.

There are 120 possibilities in total.

The wrong possibilities (beginning with block 2 in the 1st position) are:
2 1 5 3 4
2 1 4 5 3
2 3 4 5 1
2 3 5 1 4
2 3 1 5 4
2 4 5 1 3
2 4 5 3 1
2 4 1 5 3
2 5 1 3 4
2 5 4 1 3
2 5 4 3 1

which is 11 possibilities.
Lest you think starting with “2” is in some way special, here’s another 11 for starting with “3”:
3 1 2 5 4
3 1 5 2 4
3 1 4 5 2
3 4 1 5 2
3 4 2 5 1
3 4 5 1 2
3 4 5 2 1
3 5 1 2 4
3 5 2 1 4
3 5 4 2 1
3 5 4 1 2

2 more 11s for combinations starting with 4 and 5, makes 44 possibilities, and 44 out of 120 is around 36%, or greater than 1 in 3.

14

I haven’t looked closely at the previous answers, so I’m not sure if those were right or not.

Anyway, this is known as a derangement. As the number of objects gets larger, the probability that all of them are picked wrong gets closer to 1/e.

In this case, the probability all are wrong is

1 - 1 + 1/2 - 1/6 + 1/24 - 1/120

15

which equals 44/120 as shown above.

Thank you, Cabbage, for explaining why.

16

hibernicus has it right, manwithaplan et al. have it wrong.

Their flaw:
Start with 2 in the first position.
2 * * * *
If we pick 1 for the next position there are only two possible arrangements:
2 1 4 5 3
2 1 5 3 4
If we pick 3 for the next position there are three possible arrangements:
2 3 1 5 4
2 3 5 1 4
2 3 4 5 1
Hence, the choices are not independent when calculated this way.

Here’s a math-ish way of doing it.

We need to pick a permutation of 5 objects, etc. All permutations can be written using the parenthesis cycle notation: (1 2 3)(4 5) means 1 maps to 2 maps to 3 maps to 1 and 4 maps to 5 maps to 4. In order for all five to be out of position there can be no unit cycles- no (1). For 5 objects the only simple parenthesis forms are (* * * * ) and ( * )( *).

For the (* * * * ) group there are 5! choices but we divide by 5 since the starting point is irrelevant: 24 ways.
For the ( * )( *) group there are also 5! choices but we divide by 3 for the first group and 2 for the second group to ignore starting points: 5!/6 is 20.

20+24=44. Ta-da.

(Note that hibernicus’ blocks include 6 of the first type and 5 of the second type, hence his 11.)

And I didn’t even have to bring up Catalan numbers!

17

Knew I shouldn’t have got involved in a maths thread.:o And I thought that all that time spent in the bookies would help.

18

Point given, point taken. :slight_smile: