## Coin Toss

Jake says to his buddy Chirag, "I’ll bet you 1/2 the money in my wallet on the toss of a penny- heads I win, tails I lose."

The penny was tossed and the money handed over. Jake repeated the offer again and again, each time betting 1/2 the money in his wallet. We are not told how long the game went on, or how many times the penny was tossed, but this we know: that the number of times that Chirag lost was exactly equal to the number of times that he won.

Now, did Jake gain or lose by this little venture? Or is it impossible to determine?

*Feel free to enter your answer below; submissions will be unmasked Thursday.*

Margot| Profile April 16th, 2008 - 1:09 amJake would walk away with a loss. The longer they play, the larger his loss.

zxo| Profile April 16th, 2008 - 8:10 amHe loses money.

The simple explanation is that after a loss, he has 1/2 as much money as he did before, but after a win, he does not double his money but only wins 1.5 times his starting fund.

For each W-L pair then, he ends up with X * 1.5 * 1/2 or .75X

where X is the amount of money he started with before the W-L pair.

misha| Profile April 16th, 2008 - 8:27 amJake lost, and probably a lot. Let’s assume that Jake started with $100 in his wallet. After two iterations of the game, where there was one heads and one tails, Jake ends up with $75. After four iterations, with 2 heads and 2 tails, Jake has $56.25.

This is because, if he starts with x dollars, after one win and one loss he has

x(1.5)(.5) = 0.75x, or 75% of what he started with.

Anastasia11| Profile April 16th, 2008 - 10:37 amJake will always lose. And the more times he plays the less he will end up with.

So if he plays twice he will end up with 3/4 the amount he started with. If he plays 4 times he ends up with 9/16.

So if x is the starting amount, and n the number of games (which has to be even) his final amount is x*(3/4)^n/2

ikickass| Profile April 16th, 2008 - 11:06 amI think Jake will have less money than he began with.

Shawn| PUZZLE GRANDMASTER | Profile April 16th, 2008 - 11:16 amEach win multiplies Jake’s wealth by 1.5.

Each loss multiplies Jake’s wealth by 0.5.

For “Z” number of bets, Jake’s wealth increases by [(1.5)*(0.5)]^Z/2, because half of the bets are winners, half are losers.

But (1.5)*(0.5)=0.75 which is less than 1.0. Anytime a number between 0 and 1 is raised to a power greater than 1, the result is a lower number, so this is a losing proposition. In fact, the more times the penny is flipped, the more money Jake loses.

To break even, Jake needs to set the multiplier if he wins to be the inverse of the multiplier if he loses.

Diego| Profile April 16th, 2008 - 11:33 amIf he places his won money on his wallet every time, then he ends up losing, maybe something to do with the more he wins the more he has to bet, and the more he loses the less he can win.

If this happens to be a trick, say he has $100 in his wallet and bets $50, say he wins, now he totals $150 ($100 dollars in his wallet and $50 in his hand) but he keeps betting “half of what he has in his wallet”, he will bet again $50 instead of $75. This way he will end up even. Unless he loses the first games and runs out of money hehe

tad| Profile April 16th, 2008 - 11:53 amJake lost money.

hex| PUZZLE MASTER | Profile April 16th, 2008 - 1:19 pmIf we start with X dollars in Jake’s wallet, he:

wins -> remains 3x/2

loses -> remains x/2

So the cumulative result of several tosses is a multiplication (which can be done in any order).

Consequently, after n (even number) turns (n/2 wins and n/2 losses), what will remain in Jake’s wallet is: x(3/4)^(n/2) where ^ symbol denotes “raised to the power”

Essentially, this means that the outcome is predetermined only by the number of tosses, and sadly Jake will lose at then end of such a game as 3/4 < 1 and thus raising it to a positive power will only make it smaller.

If he plays an almost infinite number of turns, he will have almost no dollars left

jasc| Profile April 16th, 2008 - 2:13 pmJake always loses money.

We know he wins as many times as he loses.

To break even after a loss, you’d have to bet and win the very same amount (but he can only possibly win half). Loses money.

A win seems like a blessing, but he will end up losing 1.5 times the original bet. Loses money

He always loses more than he wins.

wally| Profile April 16th, 2008 - 3:58 pmHe will always lose. If you limit it to only 4 tosses and a $40 starting point, no matter what combination (of which there are 6) he will always end up with $22.50.

da Moose| Profile April 16th, 2008 - 4:05 pmI despised probabilities in Maths lessons. My head hurts just trying to imagine how to attempt this one. I’m well out of it. :o)

foger1979| Profile April 16th, 2008 - 4:08 pmLost.

michaelc| Profile April 16th, 2008 - 4:18 pmJake lost out to the tune of the equation below.

S * 0.75^n

where ^ is the power symbol, and n is the number of times that the coin was tossed equally, and S is the starting number of dollars.

Let’s say S is 100$ and n is 2.

Could have happened he won 2 in a row, and then lost 2 in a row.

100$

150$

225$

112.50$

56.25$

Or the maybe it was the other way around…

100$

50$

25$

37.50$

56.25$

Either way it must satisfy the formula

100 * 0.75^2

$56.25

Jake needs another hobby besides gambling!

michaelc| Profile April 16th, 2008 - 4:24 pmIf x is the starting amount,

(X-0.5X) + 0.5(X-0.5X)

or if you win first then lose,

(X + 0.5X) – 0.5(X + 0.5X)

both simplify to 0.75X for each win/lose transaction.

Thus the reason for S 0.75^n formula.

jay| Profile April 16th, 2008 - 4:38 pmJake lost money…

abcbcdcdef| Profile April 17th, 2008 - 4:03 amHe loses?

Suppose there are only two rounds,

If he wins first then loses, he loses because the money in his wallet increased after he wins therefore he loses more;

if he loses first then wins back, he still loses because the money he won back cannot compensate the loss.

Expanding this theory, as the no. of wins=no. of loss, Jake loses in whole.

joe| Profile April 17th, 2008 - 7:01 amWell whenever he loses he divides his wallet money by 2. When he wins he multiplies the amount in his wallet by 1.5

So, as long as it is played more than twice he will lose out I believe, no matter the order.

In fact if it goes on and on and the numebr of times they play tends to infinity what he has left will tend to zero – or in our realistic case the lowest denomination of his currency.

RK| Founder | Profile April 17th, 2008 - 7:57 amThis problem initially looks a lot harder than it actually is to solve.

Joe, Michaelc, Jasc, Shawn, Hex, Anastasia11, Misha, Zxo, Wally, abcbcdcdef provide some excellent explanations that are definitely worth checking out if you couldn’t quite get the correct answer.

hex| PUZZLE MASTER | Profile April 17th, 2008 - 10:58 amI noticed that almost all answers directly treated the problem as win/loss pairs, ie a win followed immediately by a loss or vice versa.

A complete proof must mention that the final outcome is a product of terms (1/2′s and 3/2′s) in any order, and since multiplication is n-way associative and commutative (can be done in any order), we can regroup it into a product of 3/4′s.

This is very important as we do not know the order of wins and losses.

Tommy| Profile April 19th, 2008 - 2:58 pmI think that most of us here should know the commutative property of multiplication. Basically, everyone here knows that 2×3=3×2.

seanRhoades| Profile April 24th, 2008 - 11:29 pmLet n be an even positive integer, and X be the original amount of money in Jacks wallet, then

Jack’s wallet = X*(sqrt(3)/2)^n.

If we graph this, Jack ends up losing all of his money to Chirag, as n goes to infinity. Here’s how I figured out the pattern,

Toss1 = X – X/2 = X/2 a loss

Toss2 = Toss1 – Toss1/2 = Toss1/2 = X/4 another loss

Toss3 = Toss2 + Toss2/2 = 3*Toss2/2 = 3X/8 a win

Toss4 = Toss3 – Toss3/2 = Toss3/2 = 3X/16 another loss

Toss5 = Toss4 + Toss4/2 = 3*Toss4/2 = 9X/32 a win

Toss6 = Toss5 + Toss5/2 = 3*Toss5/2 = 27X/64 a win

Because there must be an even number of tosses, and an equal number of wins to losses, the numerator goes up by a factor of 3^(n/2), but the denominator goes up 2^n, we can pull out the n exponent giving us (sqrt(3)/2)^n

Since (sqrt(3)/2) < 1, it must decrease as n goes to infinity.