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by henktijms
February 26th, 2013, 7:15 am
Forum: Brainteaser Forum
Topic: Good old die
Replies: 11
Views: 11342

Good old die

<t>Wileysw, you give very clever counterexamples! The conditions I suggested for the underlying distribution of the loaded die are not strong enough to ensure that the induced Markov chain is aperiodic. Take your example with p_1=p_6=0.5 and r=5. Then the one-step transition matrix of the Markov cha...
by henktijms
February 25th, 2013, 2:51 pm
Forum: Brainteaser Forum
Topic: Mary Ann
Replies: 1
Views: 9484

Mary Ann

<t>An isolated island is ruled by a dictator. Every family on the island has two children. Each child is equally likely a boy or a girl. The dictator has decreed that each first-born girl (if any) in the family should bear the name Mary Ann (the name of the beloved mother-in-law of the dictator). Yo...
by henktijms
February 25th, 2013, 12:57 pm
Forum: Brainteaser Forum
Topic: Good old die
Replies: 11
Views: 11342

Good old die

<t>Thanks. The simplest way to answer the question is to use the Markov chain approach outlined by Vanubis1. Let $S_n$ be the total score of the first $n$ rolls of a fair die. For fixed $r$, define the random variable $X_n$ as $S_n$ (modulo $r$). Then $S_n$ is divisible by $r$ only if $X_n=0$. The $...
by henktijms
January 26th, 2013, 6:58 am
Forum: Brainteaser Forum
Topic: Good old die
Replies: 11
Views: 11342

Good old die

What is the limiting probability that the sum of the outcomes of n rolls of a fair die is divisible by r for any r>=2 as n goes to infinity? The conjecture is that the answer is 1/r. Do you think the result is also true for a loaded die?
by henktijms
September 23rd, 2012, 8:46 am
Forum: Book And Research Paper Forum
Topic: Third edition of Understanding Probability book
Replies: 3
Views: 12879

Third edition of Understanding Probability book

<r>The third edition of my book Understanding Probability has just come out. The largely revised third edition contains new sections on Kelly betting, hitting probabilities in random walks and Brownian motion, Markov chain Monte Carlo simulation, and Bayesian probability among others, seehttp://<URL...
by henktijms
October 8th, 2011, 11:19 am
Forum: Brainteaser Forum
Topic: carnival game
Replies: 3
Views: 24464

carnival game

<t> A carnival booth offers the following game of chance. Under each six inverted cups is a colored ball, in some random order.The six balls are colored red, blue, yellow, orange, green and purple. You wager \$5 to play and you get six tokens. All you have to do is to guess the color of the ball und...
by henktijms
July 15th, 2011, 8:02 am
Forum: Brainteaser Forum
Topic: Is this a bet you can't lose?
Replies: 5
Views: 28813

Is this a bet you can't lose?

<t>QuoteOriginally posted by: phvanIf this is true and you realize that the casino is picking its numbers this way can't you just pick a distribution for your own numbers to put the odds back in your favour? Something producing the inverse of Benford's law for instance.No, the crux is that casino wi...
by henktijms
July 14th, 2011, 7:55 am
Forum: Brainteaser Forum
Topic: Is this a bet you can't lose?
Replies: 5
Views: 28813

Is this a bet you can't lose?

<t>Yes, you should read the number as the product (10^4)*(10^u). You hit the nail on the head by noting that the probability density from which the casino samples the real number x=10^u is skewed to the left. This density is nothing else than the so-called Benford density f(x)=1/[xln(10)]. The rando...
by henktijms
July 8th, 2011, 12:26 pm
Forum: Brainteaser Forum
Topic: Is this a bet you can't lose?
Replies: 5
Views: 28813

Is this a bet you can't lose?

<t>Imagine that a casino offers you the following game. The dealer hits a button and from a slot in the table comes a slip of paper with a four-digit positive integer on it that you cannot see. Next you use a keypad to choose a number of your own-- any positive integer you like with as many digits a...
by henktijms
June 22nd, 2011, 4:42 pm
Forum: Brainteaser Forum
Topic: An offer you can't refuse
Replies: 5
Views: 29485

An offer you can't refuse

<t>You give a certain good in commission to person A asking this person to sell the good to another person B. The values the persons A and B assign to the good are independent random variables that are uniformly distributed between 0 and 1. The persons A and B know this fact and know of course their...
by henktijms
June 16th, 2011, 12:21 pm
Forum: Brainteaser Forum
Topic: Wheel of fortune
Replies: 24
Views: 34511

Wheel of fortune

Yes, you are wright. In the original case of two players, the second player has a win probability of 54.6%
by henktijms
June 16th, 2011, 6:58 am
Forum: Brainteaser Forum
Topic: Wheel of fortune
Replies: 24
Views: 34511

Wheel of fortune

<t>Perhaps I have not stated clearly enough the original problem and confused some readers. Let us take the two players case . First player A spins the "continuous" wheel once or twice and next player B knowing the score of player A spins the wheel once or twice. The player having the highest score ...
by henktijms
June 15th, 2011, 2:00 pm
Forum: Brainteaser Forum
Topic: Wheel of fortune
Replies: 24
Views: 34511

Wheel of fortune

<t>MCarreira,Yes, you are absolutely right that one should carefully take into account the chance of player A losing in the second draw. This is done in my solution approach, though in a somewhat hidden way. The solutions for the cases of two players and three players have been checked by extensive ...
by henktijms
June 15th, 2011, 11:45 am
Forum: Brainteaser Forum
Topic: Wheel of fortune
Replies: 24
Views: 34511

Wheel of fortune

<t>It seems that this challenging problem is much more difficult to solve than I thought. The two-players case has been solved by rather simple arguments in my book Understanding Probability, Chance rules in everyday life (p. 358), but the three players case seems not to allow for a simple solution ...
by henktijms
June 10th, 2011, 11:55 am
Forum: Brainteaser Forum
Topic: Wheel of fortune
Replies: 24
Views: 34511

Wheel of fortune

<t>Two players $A$ and $B$ in turn draw one or two random numbers between 0 and 1. For each player, the decision whether to go for a second draw depends on the result of the first draw. The winner of the game is the player having the highest total score, from one or two draws, without going over 1. ...