agreeQuote:lilredroosterwhen I found out about the correct answer to that puzzle years ago I tried for a long time to figure out if there was some way to take advantage of it while betting on craps in a casino with the bets that they allow.

and I couldn't figure out any way.

it is a classic bar bet (I hear and a few friends say so)Quote:lilredroosterI don't think there is any way to use it to your advantage. if there was it would be well known. but on the rare chance that one of you guys or gals knows one please let it fly.

The reason one has the higher prob of success rolling at least one 6 and 8

is that prob of either event is 10/36 and the two 7s to get the 1st 7 is just 6/36

Sally

and you could easily expand your answer to show the average number of rolls until the game endsQuote:WizardSally, please check if you approve of the new wording.

I did just that as I wanted to know (using your data)

either wins: 7.021952479 rolls

given win by 6,8: 6.891608392 rolls

given win by 7,7: 7.178446873 rolls

for the bar bet 2 player version I get

either wins: 14.24521197 rolls

given win by 6,8: 13.65067664 rolls

given win by 7,7: 15.04916522 rolls

Sally

Is this wording acceptable:

Players A and B take turns rolling two six-sided dice. Each roll counts towards the goal of the player rolling only. Player A has a goal to roll both a total of six and eight, in either order, and duplicates are allowed. Player B has a goal to roll a total of seven twice. The dice are rolled until either event occurs. If player A rolls first, what is the probability of player A winning? What if player B rolls first?

yes, excellent (the Markov chain solution easy answers: What if player B rolls first? and I think the answer is a surprise!)Quote:WizardSo the question on the table is the same thing, except the two players take turns rolling and each roll counts for the player rolling only?

Is this wording acceptable:

Players A and B take turns rolling two six-sided dice. Each roll counts towards the goal of the player rolling only. Player A has a goal to roll both a total of six and eight, in either order, and duplicates are allowed. Player B has a goal to roll a total of seven twice. The dice are rolled until either event occurs. If player A rolls first, what is the probability of player A winning? What if player B rolls first?

I know you can do this in Excel with basic math. I got stumped, that is why I used the MarkovSally method

Sally

added:

these type of questions/bets seem at 1st so simple to figure out

but once the attempt to figure it out starts, it seems not

as simple

I always start with a simulation

so for: What if player B rolls first? He says 77 rolls B4 6and8

> print(end.time - start.time)

Time difference of 4.670225 mins

> p = count7/sims

> error = 3.29*sqrt(p*(1-p)/sims)

> error

[1] 0.0009997058

> p

[1] 0.4734789

> p-error

[1] 0.4724792

> p+error

[1] 0.4744786

> 1-p

[1] 0.5265211

> ##########################

> sims

[1] 2700000

> count68

[1] 1421607

> count7

[1] 1278393

If player A goes first, the exact answer is 3,955,200 / 6,880,129 =~ 57.487%.Quote:WizardPlayers A and B take turns rolling two six-sided dice. Each roll counts towards the goal of the player rolling only. Player A has a goal to roll both a total of six and eight, in either order, and duplicates are allowed. Player B has a goal to roll a total of seven twice. The dice are rolled until either event occurs. If player A rolls first, what is the probability of player A winning? What if player B rolls first?

You can solve it writing 8 equations...1 for each state. A, C, E and G are player A rolling, the other states are player B rolling. (x,1) means player A has made nothing while player B has made 1 seven. Solve for A.

A (x,x) = 5/18D + 13/18B

B (x,x) = 1/6E + 5/6A

C (1,x) = 5/36 + 31/36D

D (1,x) = 1/6G +5/6C

E (x,1) = 5/18H + 13/18F

F (x,1) = 5/6E

G (1,1) = 5/36 + 31/36H

H (1,1) = 5/6G

To get the probability of player B winning if he/she rolls first, just set B as the initial state instead of A, and you get a loss probability of 3,623,875 / 6,880,129, so the win probability is the complement of that: 3,256,254 / 6,880,129 =~ 47.33%.

You are given two indistinguishable envelopes, each containing money, one contains twice as much as the other. You may pick one envelope and keep the money it contains. Having chosen an envelope at will, but before inspecting it, you are given the chance to switch envelopes. Should you switch?

one way to approach: you don't know what's in the envelope you picked. but let's say it's $50. if you have the lesser envelope and you switch you will get the better envelope which has $100. so you will gain $50. If you had the better envelope and you switch you would get the lesser envelope which has $25 and you would lose $25.

so by switching you could either gain $50 or lose $25.

so it seems like the best plan is to switch envelopes. right?

it seems like some pretty good logic. if it's not right why not?

Quote:Ace2If player A goes first, the exact answer is 3,955,200 / 6,880,129 =~ 57.487%.

I agree. To prove I solved it too, here are more decimal places: 0.57487294.

If B goes first I get the probability A winning is 0.52671614.

I solved by the same method as Ace.

What's the probability A wins and what's the probability of a tie?

No one would doubt you about solving it alsoQuote:WizardI agree. To prove I solved it too, here are more decimal places: 0.57487294.

If B goes first I get the probability A winning is 0.52671614.

I solved by the same method as Ace.

You mentioned 1st how you would do it, then removed it

0.574872942

higher than the roll being for both players.

I knew that was the case from just rolling the dice.

It was harder to win trying to get 77 when player1(6&8) went first

no wonder this bar bet cashes big time for 6&8!

*****

here is my Excel matrix and recursion in Google (4 those so interested)

https://goo.gl/amR95r

the probabilities from the 8 states

from: p1,0- p2,0 = player1 with 0 collected- player2 with 0 collected

68 wins | 77 wins | from | mean # of rolls |
---|---|---|---|

0.574872942 | 0.425127058 | p1,0- p2,0 | 14.24521197 |

0.285932139 | 0.714067861 | p1,0- p2,7 | 8.854746474 |

0.741736092 | 0.258263908 | p1,6or8- p2,0 | 9.939263639 |

0.491803279 | 0.508196721 | p1,6or8- p2,7 | 6.590163934 |

0.526716142 | 0.473283858 | p2,0- p1,0 | 14.34680106 |

0.238276782 | 0.761723218 | p2,7- p1,0 | 8.378955395 |

0.700080623 | 0.299919377 | p2,0- p1,6or8 | 10.38108035 |

0.409836066 | 0.590163934 | p2,7- p1,6or8 | 6.491803279 |

Sally

yes, there are many other variations one can come up with. I have 12 of them currentlyQuote:Ace2Let's say the 2 players roll separate sets of dice simultaneously.

I did NOT think of yours... this variation has a 'feel' of being closer to fair (because of the tie)

coming soonQuote:Ace2What's the probability A wins and what's the probability of a tie?

Sally