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# Waiting, again

## Question

You're at a carnival and two people offer you money for getting six fair dice collectively to show all six numbers. Each charges you \$1 per roll. Peter pays you \$20 when you succeed, while Paul will pay you \$50. There's another difference. Paul lets you roll all six dice each time, but Peter makes you roll just one die each time. Do you stop and play? If so, with whom?

Edited by bonanova
Clarify wining condition

## 8 answers to this question

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Peter's game:

Once I reworded the question to "how many dice are expected to get all results from 1 to 6," I realised this is basically the Coupon Collector's Problem with dice instead of coupons.
https://en.wikipedia.org/wiki/Coupon_collector's_problem
n, in this case, is 6, which is already known to equal 14.7.  So this game is in your favour.

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I want to be sure I’ve got this right. There are six (presumably distinguishable) dice. I want to demonstrate that each is capable of showing all six faces.

Paul lets me roll all six (and tabulate which individual dice showed which numbers) charges \$1 for the combined roll, and pays \$50 when the tabulations show each die has shown all six faces.

Peter has me roll one die at a time, charges \$1 for the individual roll, tabulates the results, and pays \$20 when each die has shown all six faces.

OR...

my Termination condition is seeing all six numbers on the table at once. Paul has me roll all 6 dice each time, and only pays me if I roll a full straight (123456). Peter lets me roll one die at a time, once I’ve rolled all six dice, he lets me improve my hand by rolling a single die that duplicates another one, and pays once all six numbers are showing.

Edited by CaptainEd

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Paul's game:

The probability of winning Paul's game with a single roll is 0.01543209876.  In 50 rolls, if my understanding of basic probability is correct (it probably isn't), you have ~45.95% chance of winning.  That's still better than roulette, and tons of people play that.  Still, I wouldn't.

Peter's game:
Clarifying non-spoiler question: Does Peter let you roll all 6 dice for free (or \$1) the first roll?  Or must you pay \$6 to roll all 6 dice the first time?

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There are six dice. They are marked A. B, C, D, E and F in some order. Initially no dice are on the table.

Peter's game: You pay \$1 and roll die A or die B or die C or die D or die E or die F. You get to choose which die to roll.

Spoiler

Peter's game is equivalent to the problem of rolling a single die until all six faces have shown at least once. Employing six dice, and choosing on each turn which one to roll, simply eliminates the need to record e.g. with pencil and paper which numbers have already been rolled.

Paul's game: You pay \$1 and roll die A and die B and die C and die D and die E and die F. You must roll all of them.

Winning condition: All dice are on the table and collectively they show 1, 2, 3, 4, 5 and 6 dots.

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Thank you Bonanova, sorry to be so dense

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On 1/13/2018 at 11:25 PM, CaptainEd said:

Thank you Bonanova, sorry to be so dense

@CaptainEd - OMG no.

Awhile ago I next-to-worshiped Martin Gardner (who wrote the math games column in Sci American for so many years) because he worded his puzzles perfectly, simply and clearly. His, unlike mine, (try tho I may) never needed editing. When I wrap prose around mine to make them perhaps interesting or, sometimes, to camouflage the solution, stuff gets added that has often has to be clarified later.

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Gardner sets high standard in many ways. I was a child reading Childrens Activities and a few years later I was enjoying hexaflexagons and later mathematical games. I was kneeling behind you in worship.

i enjoy the puzzles here, and sometimes I don’t understand something that is obvious to anyone else. I think I may have a touch of ambiguity flu. Keep on puzzling, Bonanova!

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And Children's Activities had some cool features on the last page - cartoon, riddle or puzzle - as I recall.

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