6

A money pouch contains a certain number of cents and only one dime. You and your friend are playing a game: They alternate turns and pick one coin at a time, which they put in their pockets. Whoever picks the dime wins. You are trying to decide whether it is better to play first or second: Initially you think that if you pick the dime in your first draw, you immediately win, while if you opt to play second, your probability to pick the dime increases. Which of the two options is more favorable for you? We assume that you can’t see the coins before taking them out of the pouch and also that you can’t tell from the size which coin is which.

My thought: Let's say we have $n$ coins in total. If I play first, I have probability of picking the dime: $\frac {1}{n}+\frac {1}{n-2}+\frac {1}{n-4}...$ while if I play second, I have $\frac {1}{n-1}+\frac {1}{n-3}+\frac {1}{n-5}...$. Clearly the second option gives a larger number. Is this correct?

Thank you.

Reyansh Laghari
  • 395
  • 1
    Pick a values for n. What results do you get? If the n is odd, you want to go first because you get an extra draw. But what about n= 2,4,6? It is equally likely that the dime comes on the any of n draws. – user317176 Feb 01 '22 at 20:10
  • Your formula is wrong. Add up the probability that you win plus the probability that your opponent wins give more than 100% when $n>1$. Your first term when you go first is the only accurate term. – SlipEternal Feb 01 '22 at 20:16
  • "I have probability of picking the dime: $\frac{1}{n}+\frac{1}{n-2}+\frac{1}{n-4}\dots$" By your attempt, for $n=10$ we would have a "probability" of $\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}\approx 1.1417$... That is larger than one and can not be a probability. – JMoravitz Feb 01 '22 at 20:16
  • 2
    For each of these terms in the sum, you should have conditioned on the event that the dime had not yet been found. The probability of finding the dime in the third draw given that the first two draws did not reveal the dime is indeed $\frac{1}{n-2}$ however we need to multiply this by the probability of us actually being in the case of the first two draws not revealing the dime... $\frac{n-1}{n}\cdot\frac{n-2}{n-1}\cdot \frac{1}{n-2}$ which simplifies to just $\frac{1}{n}$. See https://math.stackexchange.com/questions/1287393/ – JMoravitz Feb 01 '22 at 20:19
  • OK so if n is odd, if I play first, I have $(\frac {n-1}{2}+1) \cdot \frac {1}{n}$ probability to win, and if I play second, $(\frac {n}{2}) \cdot \frac {1}{n}$. Is it correct? – Reyansh Laghari Feb 01 '22 at 20:37
  • Yes, that is correct. – Gregory Feb 01 '22 at 20:41
  • 1
    Close. If $n$ is odd and you play second, you have probability $$\left(\dfrac{n-1}{2}\right)\left(\dfrac{1}{n}\right)$$ to win. If $n$ is even, you have probability $\tfrac{1}{2}$ no matter if you go first or second. – SlipEternal Feb 01 '22 at 20:54

2 Answers2

4

The first essential point is that the dime has no preference for position, thus its Pr of being at any position is $\frac{1}{n}$

The second essential point is that we don't know whether $n$ is odd or even, and if it is odd, we get one extra try if we go first

Thus it is always preferable to go first

true blue anil
  • 41,295
2

Suppose we extend the game as follows. We start off the same where we draw coins one at a time, but we number the draws and mark down which coin was drawn on which numbered draw. Once we reach the dime, we record the winner as usual, but we also continue drawing the remaining coins and marking the draw number of each.

In the end, we wind up with $n$ coins drawn, and the draw number for the dime is equiprobable among all $n$ draws. We mark the first player a winner if the dime was drawn in an odd-numbered draw, and we mark the second player the winner if the dime was drawn in an even numbered draw.

We know that this will yield the same results.

Next, all of the coins that were drawn on odd-numbered draws are distributed to the first player and all the coins that were drawn on even numbered draws are distributed to the second player. If the dime appears in the player's pile, we mark them as a winner again.

Again, we know that this will yield the same results.

Finally, we start over and mix up all the coins randomly and assign them into two piles. If $n$ is even, the two piles each have $\frac{n}{2}$ coins. If $n$ is odd, one pile has $\frac{n+1}{2}$ coins and the other has $\frac{n-1}{2}$ coins. We give one pile to one player and the other pile to the other. If the pile they are given contains a dime, we mark down the win.

But, we continue and allow them to take turns drawing coins from their own pile. If one pile more coins than the other, the first draw must be from the bigger pile. We mark whichever player draws the dime as the winner.

Again this yield the same results as the original game. We have just demonstrated different games, each with an injective set of outcomes. This demonstrates there exists a bijection between the sets of outcomes. Any probabilities of winning calculated for any of these three games applies to all three.

The easiest is of course the last game where you just split up the piles evenly and whoever has the dime wins.

I realize this was a lot of words, and it wasn't necessary to go through all of these steps. I explained it this way in hopes that it shows why it works.

If $n$ is even, the first player wins half the time and second player wins half the time. If $n$ is odd, the first player wins $\dfrac{n+1}{2n}$ while the second player wins $\dfrac{n-1}{2n}$ of the time.

Not knowing if $n$ is odd or even, there is a slight advantage in choosing to go first all the time, just in case $n$ is odd.

SlipEternal
  • 10,555
  • 1
  • 17
  • 38