Dumont D.

asked • 02/21/16

Poker winning and bill paying

8 poker players play 1 game at a restaurant every night. Each player takes a turn paying the entire bill for the night. Assuming everyone has equal skills at poker, what is the chance that the person paying the bill will win the game that night?
 
The question comes from a waiter who gets higher tips when the person paying happens to be the winner.

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Daniel R. answered • 02/21/16

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use a product rule...1/8x1/8=1/16
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Dumont D.

My friends have been debating this with each other. One guy that was a statistics major said 1/64. However, I believe he's answering a DIFFERENT question: "What is the chance that Player 1 will both win AND be the one paying for dinner?"
 
But for this question, I'm thinking it's a 1/8 chance. Player 1 is paying tonight and he has a 1/8 chance of winning. Right?
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02/21/16

Daniel R.

A genetics professor can answer the question lol. There is a 1/8 chance that the player is paying and also a 1/8 chance that the player is winning. In statistics, there is a product rule that when the chance of two different situations, when multiplied together give you the probability.
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02/21/16

Dumont D.

But what we're saying is that at one person has a 1/1 chance of paying and everyone else has a 0/0 chance of paying. This is because they take turns paying each night.
 
The ultimate reason this question was asked is that someone said to the waiter "Oh you must get a big tip every 8 nights" but he said "No it's more like every 64 nights." Since they take turns paying, that paying person has a 1/8 chance of winning and therefore giving a big tip. Therefore every night there's a 1/8 chance that the person paying will be the winner. And subsequently he will get a big tip every 8 nights. Right?
 
I think the waiter observed that it's a lower probability because of a factor I excluded from the question here: the individual skills of each person.
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02/21/16

Daniel R.

well rereading the question... It's only asking for the chance of the person paying the bill to win. So there is 100% chance he's already paying the bill but a 1/8 chance he wins... Idk where the 64 is coming from
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02/21/16

Daniel R.

I did make a typo lol 1/8 X 1/8 is 1/64... I think your friend is right.
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02/21/16

Daniel R.

The chance that he wins on the day he pays is 1/64. I'm sticking to my original argument. Sorry for the confusion haha.
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02/21/16

Dumont D.

I think he would have a 1/64 chance of winning and paying on the same day - ONLY if the payer was chosen randomly. Right? However, in this case the payer is already decided at the beginning. 
 
So then the question is Does the waiter have a different probability of encountering an event in which the payer and winner are the same person? There's a 100% chance that someone will pay and a 100% chance that someone will win. But is it 1/8 that they will be the same person? Or is it 1/64?
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02/21/16

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