[Prof. Templeton]

Bonanova, Prime, and Octopuppy were sitting around a table discussing the expected probablity of a 6 to come up on a fair six-sided die. Bonanova then pulls such a die from his pocket and puts forth the following game. Starting with Bonanova, then Prime, then Octopuppy they will each take a turn rolling the die. When someone rolls a 6 they are declared the winner. If no six is rolled the first time around the table, then they continue in order until a six is rolled.

What is the probablity that Bonanova will win this game?

If they each put in a dollar to start, and they each put in another after every win and if after each win Bonanova will always be player number one, should Bonanova expect to make money?

Should he expect to make money if he isn't always player number one?

[Guest]

Bonanova, Prime, and Octopuppy were sitting around a table discussing the expected probablity of a 6 to come up on a fair six-sided die. Bonanova then pulls such a die from his pocket and puts forth the following game. Starting with Bonanova, then Prime, then Octopuppy they will each take a turn rolling the die. When someone rolls a 6 they are declared the winner. If no six is rolled the first time around the table, then they continue in order until a six is rolled.

What is the probablity that Bonanova will win this game?

If they each put in a dollar to start, and they each put in another after every win and if after each win Bonanova will always be player number one, should Bonanova expect to make money?

Should he expect to make money if he isn't always player number one?

If I recall my engineering stochastics class properly...

The odds of winning, losing, and drawing are the following (for bonanove in the first position):

W: 1/6

L: 55/216

D: 125/216

Therefore the expected value for Bonanove is: 1/6*(2)+55/216*(-1) = 17/216, which is a positive number. Bonanove should expect to make about 7.8 cents per dollar bet.

If Bonanova was either second or third in line, his expected value would be -.004 and -.07 respectively. Therefore it would be expected that he lose ~.004 dollars and ~.07 dollars for each dollar bet. he should not play.

Edited October 31, 2008 by Nyquist

[Guest]

I think it works like this

For any given turn (as a whole)

Chance of Prime winning = 5/6 x 1/6 = 5/36 = 30/216

Chance of Octopuppy winning = 5/6 x 5/6 x 1/6 =25/216

The probability of Bonanova winning on any given turn is 216-25-30 = 161/216

Pretty high it seems.

[Guest]

Let's start from the beginning and go through a few rolls to see everyone's (cumulative) probability of winning on that roll:

B: 1/6

P: 5/6 * 1/6

O: 5/6 * 5/6 * 1/6

B: (5/6)³ * 1/6

P: (5/6)⁴ * 1/6

etc.

So in total:

P(B winning)=Sum(i=0,inf)[(5/6)³ⁱ*1/6]=Sum(i=0,inf)[(125/216)ⁱ*1/6

P(B winning)=(216/91)*1/6=36/91=approx. .396

[Guest]

if we are assuming experimental probability, then the probability of rolling a 6 each time is 1/6 because the previous roll is an independent event to the second. Therefore, the probability of anyone rolling a 6 is 1/6. So Bonanova rolls and the probability of rolling a 6 is 1/6, then Prime's probability is 1/6 and Octopuppy's probability 1/6. I can only deduce that Bonanova's probability of winning is 1/6. Bonanova should not expect to make money because he has the same chance of rolling a 6 as the others, regardless of whether he goes first or not. This is what I learned in maths so feel free to correct me if I have made a mistake or said something wrong

. Unless Bonanova is feeling lucking and he has his lucky charm, I wouldn't expect him to have any more of a chance of winning that anyone else. Just because there is three of them doesn't change the probability of each roll either. It is still 1/6.

Edited November 1, 2008 by runder_1111

[Guest]

if we are assuming experimental probability, then the probability of rolling a 6 each time is 1/6 because the previous roll is an independent event to the second. Therefore, the probability of anyone rolling a 6 is 1/6. So Bonanova rolls and the probability of rolling a 6 is 1/6, then Prime's probability is 1/6 and Octopuppy's probability 1/6. I can only deduce that Bonanova's probability of winning is 1/6. Bonanova should not expect to make money because he has the same chance of rolling a 6 as the others, regardless of whether he goes first or not. This is what I learned in maths so feel free to correct me if I have made a mistake or said something wrong

. Unless Bonanova is feeling lucking and he has his lucky charm, I wouldn't expect him to have any more of a chance of winning that anyone else. Just because there is three of them doesn't change the probability of each roll either. It is still 1/6.

I found faulty work in your post.

In order for the second person to win, it is required that the first person NOT win. Therefore, the probability of the second person is dependent on the outcome of the first person's role. I don't have anything more than that since I'd rather not be doing any kind of math work outside of my probability class <_<

[Guest]

I found faulty work in your post.

In order for the second person to win, it is required that the first person NOT win. Therefore, the probability of the second person is dependent on the outcome of the first person's role. I don't have anything more than that since I'd rather not be doing any kind of math work outside of my probability class <_<

Thanks for pointing that out Sparanda. Now to rethink my probability....so for the second to win it would be 5/36 and the third 25/216. So for it to continue it would be 5^(n-1) / 6^n. But still, Prime doesn't have any less of a chance of rolling a 6 than Bonanova. Because just say Bonanova rolls a 3, that doesn't affect Prime's chance of rolling a 6. Bonanova's roll, however, does determine whether Prime gets a go or not and Prime of Octopuppy, as Sparanda said. I am.....confused now. I need to think about this some more....

[bonanova]

If it's your turn to roll the die, your probability of winning is p0 [independent of preceding rolls].

The probabilities for the first three players are:

p1 = p0

p2 = (5/6) p0

p3 = (5/6)(5/6) p0

Since one of these three players will win, p1 + p2 + p3 = (36/36 + 30/36 + 25/36) p0 = 1.

p1 = 36/91 = .3956 = p0

p2 = 30/91 = .3297

p3 = 25/91 = .2747

Multiply by the $3 pot, the expected winnings are

E1 = .3956 x $3 = $1.1868. Net winnings +$0.1868

E2 = .3297 x $3 = $0.9891. Net winnings - $0.0109

E3 = .2747 x $3 = $0.8241. Net winnings - $0.1759

Answers to OP questions:

1. Probability bona will win the game is .3926
   
2. Should bona expect to make money on repeated games, always playing first? Yes.
   About $0.19 each game.
   
3. Should bona [or any player] expect to make money if [say] the starting player is randomly chosen? No.
   Players should expect to break even in a zero-sum game.
   

[Prime]

This is my favorite game. It is called Russian Roulette and is played with a six-shooter -- not a die.

I agree with CR’s reasoning and with Bonanova’s answer. (I did not quite understand Bonanova’s reasoning.)

The probability for each is a geometric series. That is a sum of probabilities to win on each next turn, given everyone misfired up to then.

Thus for each next turn after the first round there must be 3 misfires at the probability (5/6)³ followed by a shot at the probability 1/6. Thus common ratio "r" of the series is (5/6)³ for either player. Whereas scale factor "a" (which is a probability of shooting onerself in the first round in this setup is 1/6 for Bonanova, (5/6)*1/6 for Prime, and (5/6)² * 1/6 for Octopuppy.

Applying formula for geometric series S = a*(1 - rⁿ)/(1 - r) with n tending to infinity, we get:

Boanova: 36/91;

Prime: 30/91;

Octopuppy: 25/91

[bonanova]

I did not quite understand Bonanova’s reasoning.

Mac The Cat had it almost. Just he didn't normalize the odds.

But I do like how mine turned out.

The probability that the i^th player gets a turn is (5/6)^i-1.

The winning probabilities are simply these numbers multiplied by p₁=the probability the first person wins.

Set that sum to 1 [certainty that one of the players will win], and you get p₁.

The die has no memory: whenever it's your turn you have a probability p₁ of winning.

It's all about the probability of getting a turn.

Edit:

Edited November 1, 2008 by bonanova
Comment on MTC's solution

[Prof. Templeton]

Congrats to Chuck, Bona, and Prime. I have 36/91 for Player One. And, of course, the last scenario is a zero sum game.

[Prime]

Mac The Cat had it almost. Just he didn't normalize the odds.

But I do like how mine turned out.

The probability that the i^th player gets a turn is (5/6)^i-1.

The winning probabilities are simply these numbers multiplied by p₁=the probability the first person wins.

Set that sum to 1 [certainty that one of the players will win], and you get p₁.

The die has no memory: whenever it's your turn you have a probability p₁ of winning.

It's all about the probability of getting a turn.

Edit:

I still don't get it.

The probability of a win once you get a turn is exactly 1/6. Your formula yields 36/91. Infinite series seem so much simpler an explanation to me.

[Guest]

I still don't get it.

The probability of a win once you get a turn is exactly 1/6. Your formula yields 36/91. Infinite series seem so much simpler an explanation to me.

I don't know if either way is simpler, but here's the logic:

No matter what position you're in, and what has happened before, once the dice get to you your probability of winning must be the same. Call this probability p0. If you're going first, your probability of winning is p0. If the dice come back to you 3 rounds from now, your probability of winning is again p0. The fact that p0 can be directly calculated from an infinite series is cool, but it's not necessary.

You know that P(player 1 winning) + P(player 2 winning) + P(player 3 winning)=1

P(1 winning)=p0

The probability of the dice getting to player 2 is 5/6. Once player 2 has the dice, it's exactly like they started with the dice.

P(2 winning)=5/6*p0

The same logic gives

P(3 winning)=5/6*5/6*p0

You can then do the algebra to find p0.

[Prime]

I don't know if either way is simpler, but here's the logic:

No matter what position you're in, and what has happened before, once the dice get to you your probability of winning must be the same. Call this probability p0. If you're going first, your probability of winning is p0. If the dice come back to you 3 rounds from now, your probability of winning is again p0. The fact that p0 can be directly calculated from an infinite series is cool, but it's not necessary.

You know that P(player 1 winning) + P(player 2 winning) + P(player 3 winning)=1

P(1 winning)=p0

The probability of the dice getting to player 2 is 5/6. Once player 2 has the dice, it's exactly like they started with the dice.

P(2 winning)=5/6*p0

The same logic gives

P(3 winning)=5/6*5/6*p0

You can then do the algebra to find p0.

This is where the reasoning breaks down for me:

If P0 is probability of Player1 winning, then why would probability of Player2 be P0*5/6? 5/6 is the probability of a player missing his shot. We multiply dependent events. So that product is like saying: "First, Player1 must win, then Player1 must not win." Also, the probability P0 cannot be tied up to any specific round, but must represent the sum total of all rounds.

Let me take a shot at explaining Bonanova’s solution the way I could understand it.

Let's call the three turns of rolling dice (Bonanova, Prime, Octopuppy) a round.

Before any given round the probability for Bonanova to win in that round is B=1/6; for Prime -- P=(5/6)*1/6; for Octopuppy -- (5/6)²*1/6. The total probability for the 3 men to win in any particular round "n" is probability that they get to that round "r_n" times the sum of their probabilities for a single round: r_n*(B+P+O) where (B+P+O)=(1/6 + 5/36 + 25/216)=91/216.

The overall probability for someone winning in some round over an infinite number of rounds is: (B+P+O) + r₂*(B+P+O) + r₃*(B+P+O) + ... to infinity.

Which we can rewrite as (B+P+O)*(1+r₂+r₃+r₄+...).

Let's designate the infinite sum (1+r₂+r₃+r₄+...) with the letter "S". Knowing that the total probability of someone rolling 6 ever is 1, we have an equation: (B+P+O)*S = 1; or S*91/216 = 1. From which we find: S=216/91. Consequently:

B = 1/6 * 216/91 = 36/91;

P = 5/36 * 216/91 = 30/91;

O = 25/216 * 216/91 = 25/91.

So I don't see that Bonanova's solution managed to avoid infinite series. It just does not mention it. The only difference is that Bonanova finds the value for the infinite sum from an algebraic equation using the fact that overall probability must equal 1. Whereas, I found the value for geometric series using its well-known formula. Either way, it's the infinite sum.

[Guest]

This is where the reasoning breaks down for me:

If P0 is probability of Player1 winning, then why would probability of Player2 be P0*5/6? 5/6 is the probability of a player missing his shot. We multiply dependent events. So that product is like saying: "First, Player1 must win, then Player1 must not win." Also, the probability P0 cannot be tied up to any specific round, but must represent the sum total of all rounds.

Let me take a shot at explaining Bonanova’s solution the way I could understand it.

Let's call the three turns of rolling dice (Bonanova, Prime, Octopuppy) a round.

Before any given round the probability for Bonanova to win in that round is B=1/6; for Prime -- P=(5/6)*1/6; for Octopuppy -- (5/6)²*1/6. The total probability for the 3 men to win in any particular round "n" is probability that they get to that round "r_n" times the sum of their probabilities for a single round: r_n*(B+P+O) where (B+P+O)=(1/6 + 5/36 + 25/216)=91/216.

The overall probability for someone winning in some round over an infinite number of rounds is: (B+P+O) + r₂*(B+P+O) + r₃*(B+P+O) + ... to infinity.

Which we can rewrite as (B+P+O)*(1+r₂+r₃+r₄+...).

Let's designate the infinite sum (1+r₂+r₃+r₄+...) with the letter "S". Knowing that the total probability of someone rolling 6 ever is 1, we have an equation: (B+P+O)*S = 1; or S*91/216 = 1. From which we find: S=216/91. Consequently:

B = 1/6 * 216/91 = 36/91;

P = 5/36 * 216/91 = 30/91;

O = 25/216 * 216/91 = 25/91.

So I don't see that Bonanova's solution managed to avoid infinite series. It just does not mention it. The only difference is that Bonanova finds the value for the infinite sum from an algebraic equation using the fact that overall probability must equal 1. Whereas, I found the value for geometric series using its well-known formula. Either way, it's the infinite sum.

I do not think I used an infinite series approach to arrive at the same answer, but let me know if you think I did:

I just considered the first round of rolling:

B = 1/6

P = 5/6 * 1/6 = 5/36

O = 5/6 * 5/6 * 1/6 = 25/216

The relative ratios of these 3 probabilities will always remain the same so:

B = 1/6 / (1/6 + 5/36 + 25/216) = .3956

P = 5/36 / (1/6 + 5/36 + 25/216) = .3297

O = 25/216 / (1/6 + 5/36 + 25/216) = .2747

Thanks.

[Guest]

This is where the reasoning breaks down for me:

If P0 is probability of Player1 winning, then why would probability of Player2 be P0*5/6? 5/6 is the probability of a player missing his shot. We multiply dependent events. So that product is like saying: "First, Player1 must win, then Player1 must not win." Also, the probability P0 cannot be tied up to any specific round, but must represent the sum total of all rounds.

P0 is the probability of any player winning when they have the dice in their hands. There is a 5/6 chance that player 2 will get the dice; once player 2 has the dice, they have a p0 chance of winning.

[Prime]

P0 is the probability of any player winning when they have the dice in their hands. There is a 5/6 chance that player 2 will get the dice; once player 2 has the dice, they have a p0 chance of winning.

Have it your way. You've killed the Infinite Series in less than six shots.