Probability of 7 players picking their own name from a hat

[zcaspar]

I recently played a poker tournament which, before beginning the game, required each player to pick another player's name from a hat. A player would later be awarded extra points if they later knock out this player from the tournament. Our mistake was that someone kept selecting their own name. This happened four times before we gave up on the idea. My question is what is the probability, given a perfectly random selection process, of this name picking process failing (i.e. at least one person picking their own name) four times in a row.

Here is what I believe to be the answer, I would be grateful if someone could confirm / correct me:

Event A is the probability that any one player picks their own name.

$$P(A) = 1/7, P(A') = 6/7$$
Given there are 7 players, the probability of one or more of the seven selecting their own name is 1 - probability of no players selecting their own name:

$$= 1 - (6/7)^7 = 0.6601$$
The probability of this happening four times in a row is $$0.6601^4 = 0.1898$$

 

[pka]

I recently played a poker tournament which, before beginning the game, required each player to pick another player's name from a hat. A player would later be awarded extra points if they later knock out this player from the tournament. Our mistake was that someone kept selecting their own name. This happened four times before we gave up on the idea. My question is what is the probability, given a perfectly random selection process, of this name picking process failing (i.e. at least one person picking their own name) four times in a row.

This requires an application of the DERANGEMENT principle. Say that we have $N$ hats is a hatcheck room. The meeting is over and the hatcheck-monkey gives the hats out randomly. There are $\mathcal{D}(N)=N!\sum\limits_{k = 0}^N {\frac{{{{( - 1)}^k}}}{{k!}}}$, for every one to get the wrong hat. That is a total rearrangement of the hats.

 

[zcaspar]

This requires an application of the DERANGEMENT principle. Say that we have $N$ hats is a hatcheck room. The meeting is over and the hatcheck-monkey gives the hats out randomly. There are $\mathcal{D}(N)=N!\sum\limits_{k = 0}^N {\frac{{{{( - 1)}^k}}}{{k!}}}$, for every one to get the wrong hat. That is a total rearrangement of the hats.

Thanks, using the derangement principle, I get 1854 derangements. There are 7! arrangements = 5040. So the probability of a derangement is 1854/5040 = 0.3679. Therefore getting 4 arrangements in a row is $$(1-(0.3679))^4 = 0.1597$$

 

[zcaspar]

I recently played a poker tournament which, before beginning the game, required each player to pick another player's name from a hat. A player would later be awarded extra points if they later knock out this player from the tournament. Our mistake was that someone kept selecting their own name. This happened four times before we gave up on the idea. My question is what is the probability, given a perfectly random selection process, of this name picking process failing (i.e. at least one person picking their own name) four times in a row.

Here is what I believe to be the answer, I would be grateful if someone could confirm / correct me:

Event A is the probability that any one player picks their own name.

$$P(A) = 1/7, P(A') = 6/7$$
Given there are 7 players, the probability of one or more of the seven selecting their own name is 1 - probability of no players selecting their own name:

$$= 1 - (6/7)^7 = 0.6601$$
The probability of this happening four times in a row is $$0.6601^4 = 0.1898$$

Would be interested to know why this answer is wrong however

 

[pka]

Would be interested to know why this answer is wrong however

In a collection of seven the probability of a total derangement is $\dfrac{\mathcal{D}(7)}{7!}=\dfrac{1854}{7!}=0.367857142$.
Now $1-0.367857142$ is the probability that at least one get her/hisown.
Beyond that I fail to follow your point.

 

[zcaspar]

In a collection of seven the probability of a total derangement is $\dfrac{\mathcal{D}(7)}{7!}=\dfrac{1854}{7!}=0.367857142$.
Now $1-0.367857142$ is the probability that at least one get her/hisown.
Beyond that I fail to follow your point.

Thank you for your help!

 

[Dr.Peterson]

Would be interested to know why this answer is wrong however

You're asking about your original answer, right?

Event A is the probability that any one player picks their own name.

$$P(A) = 1/7, P(A') = 6/7$$
Given there are 7 players, the probability of one or more of the seven selecting their own name is 1 - probability of no players selecting their own name:

$$= 1 - (6/7)^7 = 0.6601$$
The probability of this happening four times in a row is $$0.6601^4 = 0.1898$$

The problem here is that the players' choices are not independent. You can't assume that.

 

[zcaspar]

You're asking about your original answer, right?

The problem here is that the players' choices are not independent. You can't assume that.

Yes, the original one. I understand why @pka 's answer is correct...OK, yes, I see that now, subsequent choices depend on what has gone before - that helps thanks.