# geometric series with probability

Archeological evidence indicates that the ancient (pre-Roman) Etruscans played dice using a dodecahedral die having 12 pentagonal faces numbered 1 through 12 (figure above). One could simulate such a die by drawing a random card from a deck of cards numbered 1 through k. for your own personal value of k, begin with the largest digit in the sum of the digits in your student ID number. This is your value of k unless this digit is less than 5, in which case subtract it from 10 to get your own value of k. a.) John and Mary draw alternately from a deck of shuffled k cards. The first one to draw an ace-the card numbered one- wins. Assume that John draws first. Use the formula for the sum of a geometric series to calculate (both a rational number and a four place decimal) the probability J that John wins, and the probability M that Mary wins. Check that M+J=1. b.) Now John , Mary, Paul draw alternately from a deck of k cards. Calculate separately their respective probabilities of winning, given that John draws first and Mary draw second. Check that J+M+P=1.

My student no is 10-0244 in that case my largest digit is 4 so I subtracted it from 10 and I get the difference of 6 then k=6.

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I'm probably being dense, but what does this have to do with either geometric series or differential geometry? – Martin Dec 28 '12 at 2:33
Dear Jhanette, telling the moderators "needed to submit in january.. please help" is not an appropriate use of a moderator flag. – Zev Chonoles Dec 28 '12 at 2:47
Are the draws with or without replacement? (That is, if a player doesn't draw the ace, does he put his card back and reshuffle?) (The reference to geometric series suggests it's with replacement, as does the dice analogy.) – mjqxxxx Dec 28 '12 at 4:10
The sum of the digits of your student number is 11, and its largest digit is 1, so you should actually solve it for k=9. – Jaime Dec 28 '12 at 5:49
thank you for answering my question. you gave me a big favor. :) – Jhanette Querimit Mangubat Aug 5 '13 at 2:15

We look at the three-person game, since it is more interesting. We also assume that the deck is shuffled between draws, in order to simulate tossing a fair $k$-sided die. You will find it easy to adapt the idea to the simpler $2$-person game, and if you wish, to a $d$-person game.

Let our players be A, B, and C. We suppose A tosses first, then (if necessary) B, then, if necessary, C, then, if necessary A, and so on.

To make the notation simpler, let $p=\frac{1}{k}$.

Player A can win on the first draw, the fourth, the seventh, the tenth, and so on.

The probability she wins on the first draw is $p$.

In order for A to win on the fourth draw, A, B, and C must all fail to win on their first three draws, and then a must win. The probability of this is $(1-p)^3p$.

In order to win on the seventh draw, A, B, C must all fail twice, and then A must win. This has probability $(1-p)^6p$.

Similarly, the probability A achieves her win on the tenth draw is $(1-p)^9p$. The probability A achieves her win on the thirteenth draw is $(1-p)^{12}p$. And so on.

So the probability that A wins is $$p+(1-p)^3p+(1-p)^6p +(1-p)^9 p+ (1-p)^{12}p+\cdots.$$ The above is an infinite geometric series. Recall that the infinite geometric series $a+ar+ar^2+ar^3+\cdots$ has sum $\frac{a}{1-r}$ (if $|r|\lt 1$). In our case, $a=p$ and $r=(1-p)^3$, so the probability A wins is $$\frac{p}{1-(1-p)^3}.$$ This can be "simplified" to the less attractive expression $\dfrac{1}{3-3p+p^2}$.

We could go through a very similar calculation for the probability that B wins. However, there is a shortcut. Suppose that A fails to win on her first throw (probability $1-p$). Then effectively B is now first, so has probability of winning $\dfrac{p}{1-(1-p)^3}$. So the probability B wins is $$\frac{(1-p)p}{1-(1-p)^3}.$$ A similar argument shows that the probability C wins is $$\frac{(1-p)^2 p}{1-(1-p)^3}.$$

Another way: We can avoid summing an infinite series. Let $a$ be the probability that A ultimately wins. As discussed earlier, the way for B to win is for A to fail on her first draw. Then effectively B is the first player. So the probability B is the ultimate winner is $(1-p)a$. Similarly, the probability C is the ultimate winner is $(1-p)^2a$.

But it is (almost)clear that someone must ultimately win, the probability the game goes on forever is $0$. It follows that $$a+(1-p)a+(1-p)^2 a=1,$$ and therefore $$a=\frac{1}{1+(1-p)+(1-p)^2}.$$ A little calculation shows that this is the same answer as the one obtained earlier.

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thanks andre :) – Jhanette Querimit Mangubat Jan 7 '13 at 10:48
Very well explained, thank you! At first I didn't understand why other players have a probability $(1-p)^{n-1}$, but now I realize, it's because when we're looking at Player A's rolls, we need $P($A rolls a win aka low chance) $= 1/6$, but when looking at other players, we're looking for $P($Other player fails aka probability that they don't roll a win, and because $P($win$)=1/6$ $-->$ high probability) $= 5/6.$ – Alex G May 24 at 18:07