Picking first or picking last

Question

We've had quite a debate in our family regarding this one.

Five guys want to share a house with five bedrooms. Room 5 is smaller than the rest and no one wants to pick it. To decide which room they get, they put five pieces of paper in a bag, labelled rooms 1 to 5.

One at a time, each guy puts a hand in the bag and pulls one piece of paper. They reveal which room they got after each pull from the bag. The last guy to pull gets the last sheet of paper.

The debate in our family is around the odds of picking room 5. We all agree, at the beginning of the selection process, that it is 20% (1 in 5). But after the first sheet is pulled and it is not room 5, the remaining odds for picking room 5 rise to 1 in 4. Then if room 5 is not selected again, the odds rise to 1 in 3 then the 4th person has odds of 1 in 2 = 50%.

One member of our family insisted that the strategy should be to pick first (or earliest possible) to keep the odds lowest that room 5 would be picked. We contrasted this with a scenario where each guy pulled their sheet but no one revealed until all sheets were picked. Clearly the odds would be 20% for each person.

Meanwhile, another argument presented in our family is that if you pick first there is a 20% chance of picking room 5 and an 80% chance of picking any of the other rooms. However if you get the last slip of paper there is also an 80% chance that it will have already been picked, thereby giving the last person a 20% chance of getting room 5 as well. So the question to this community is: when the picks are revealed as they occur, is there any benefit or advantage to picking first or last? And, does revealing affect the overall outcome.

There are these new fangled things called paragraphs that prevent great big walls of text from chasing readers away. — Graham Kemp, Aug 29 '17 at 23:15
Questions like this come up a lot (because a lot of people wonder about the same things as you and your family members). For example, just a few hours ago: https://math.stackexchange.com/questions/2409496/is-scrabbles-method-of-determining-turn-order-fair — David K, Aug 30 '17 at 00:02
Very relevant, possible duplicate: https://math.stackexchange.com/questions/1875045/is-this-lot-drawing-fair/1875055#1875055 — Ethan Bolker, Aug 30 '17 at 00:07

score 5 · Answer 1 · answered Aug 29 '17 at 23:57

If you think the guy who picks last—let me call him Lester—if you think Lester has a greater than 20% chance of picking no. 5, then I guess you also must think he has a greater than 20% chance of picking no. 1? (If not, then I'd be interested in your explanation of the difference between slip no. 1 and slip no.5.) Likewise, Lester has a greater than 20% chance of picking no. 2, and no. 3, and no. 4? So his chances add up to more than 100%? That doesn't sound right to me. My guess is, the order of picking doesn't make any difference.

score 3 · Answer 2 · answered Aug 29 '17 at 23:24

There is no benefit to going first, middle, or last, as you have proven by the "don't look until everyone has picked" thought experiment.

As long as each paper in the hat has the same chance to be drawn among however many are there (static doesn't cling room 5 to the side of the hat, it is not larger, or anything), then it is the same principle as shuffling and dealing a deck of five distinct cards, one card to each of five players. Every player has the same probability for getting any particular card reguardless of order of dealing.

score 3 · Answer 3 · answered Aug 29 '17 at 23:40

The first argument is flawed. Yes, if the first guy does not pick room 5, then the second guy has a $25$% chance of picking room 5, but it is also true that if the first guy does pick room 5, then the second guy has a $0$% chance of picking room 5. So, what is the actual chance of the second guy picking room 5? Well, that is the chance of the first guy not picking room 5 and the second guy picking room 5, which is $0.8\cdot 0.25=0.2$, i.e $20$%

Indeed, if you do the proper math, all guys turn out to have a $20$% chance of picking room 5

In fact, the second argument is correct: There is an $80$% chance that any of the first 4 guys pick room 5, so the last guy has a $20$% chance of picking room 5.

Kudos to the person in your family who made the second argument!

Revealing or not revealing your pick makes of course not one bit of a difference as far as the probability of picking room 5, but it does conceptually help you to think about the problem.

score 2 · Answer 4 · answered Aug 29 '17 at 23:36

If no one looks at their own slip of paper, then it doesn't matter what order the slips were drawn because nobody reveals their knowledge until the end, anyway. So you may as well have a mediator hand everyone their slip personally, and in this case, the odds are clearly $1$ in $5$.

On the other hand, if you are drawing slips in sequence and then revealing your slip to yourself and others, then whoever goes first has a $1$ in $5$ chance of drawing the bad room. Clearly whoever goes first can do no better than $1$ in $5$ without cheating.

Now $4$ in $5$ times, the second person has to draw, and they draw with odds $1$ in $4$ of getting the bad room. The total odds that they get the bad room is the product: $$ \frac{4}{5}\times\frac{1}{4} = \frac{1}{5}, $$ so they are also not getting off any better.

Now, $3$ in $5$ times, the person who draws third gets the bad slip with odds $1$ in $3$, so their total odds of getting the bad slip are again the product, which is $\dfrac{1}{5}$.

Similarly, $2$ in $5$ times, the person who draws fourth gets the bad slip with odds $1$ in $2$, so their total odds of getting the bad slip are again the product, which is $\dfrac{1}{5}$.

Ultimately, there is a $1$ in $5$ chance that the person who draws last gets the bad slip, and at that point, they know they are going to get it because nobody before them got it. Thus they get the bad slip $1$ in $5$ times.

This shows us that in either case, there is not an advantage no matter which order you draw the slips, or if you do it by revealing or not.

The Entramanure · Answer 5 · 2017-08-29T23:55:49.630

1

The probabilities are the same.

Person 1: 20% chance $\times$ 100% chance it has not yet been picked = 20%
Person 2: 25% chance $\times$ 80% chance it has not yet been picked = 20%
Person 3: 33% chance $\times$ 60% chance it has not yet been picked = 20%
Person 4: 50% chance $\times$ 40% chance it has not yet been picked = 20%
Person 5: 100% chance $\times$ 20% chance it has not yet been picked = 20%

edited Aug 29 '17 at 23:55

answered Aug 29 '17 at 23:28

The Entramanure

21

you are good with the first 4 lines. Player 5 has a 100% chance of getting room 5 if it has not already been picked (20%) – Doug M Aug 29 '17 at 23:32
that is true, but I am assuming slips are not revealed. – The Entramanure Aug 29 '17 at 23:33
Then tell me how you came up with 80%. it does not follow from the lines before it. – Doug M Aug 29 '17 at 23:37
This is true, but then the last player becomes a different probability problem. If you roll a dice and get 6, you do not have a 1/36 chance of getting another 6. – The Entramanure Aug 29 '17 at 23:38
And I got 20% for all the lines. – The Entramanure Aug 29 '17 at 23:38
Indeed if you roll dice, dice do not have memory. If you draw without replacement, the results of the first draw indeed do influence the others.
You should get 20% for all of the lines. But, I still say that what shows on line 5 does not follow from what is on line 4.
– Doug M Aug 29 '17 at 23:44
You know what, you're right. the sequence is 1/5, 1/4, 1/3, 1/2, 1/1., and then 1/1, 4/5, 3/5, 2/5, 1/5. – The Entramanure Aug 29 '17 at 23:55
Let us continue this discussion in chat. – The Entramanure Aug 29 '17 at 23:56
Thank you all for the insightful and well written analysis of the problem and explaining the reasoning behind equal probability result. Cheers – Mike Lapenna Aug 30 '17 at 00:11

Doug M · Answer 6 · 2017-08-29T23:47:26.010

Pick first pick last, it doesn't matter. The lottery doesn't know the difference.

Suppose everyone picks, and keeps there picks secret. Then one by one, they reveal what their pick shows. There is no way that the information of the reveal changes what is already written on a slip of paper and is in my hand.

So how does this change if the reveal happens simultaneously, sequentially, or before the next draw. 20% chance regardless of what order you pick.

But we can look at this with a probability tree.

Suppose we consider the guy picking 2nd.

There is a 20% change that the guy picking 1st already has the bad room (and an 80% chance that he doesn't)

The chance that player 2 gets the bad room then is $0.80\cdot 0.25 = 0.20$

Player 3? there is a 20% chance that player 1 has the bad room, 20% chance player 2 has it, 60% chance that the lot with that number is still in the bag and a 1/3 chance of pulling that lot = 20%

etc.

Picking first or picking last

6 Answers6