### Gambling at Monte Carlo

A man went to Monte Carlo to try and make his fortune. Is his strategy a winning one?

### Marbles and Bags

Two bags contain different numbers of red and blue marbles. A marble is removed from one of the bags. The marble is blue. What is the probability that it was removed from bag A?

### Coin Tossing Games

You and I play a game involving successive throws of a fair coin. Suppose I pick HH and you pick TH. The coin is thrown repeatedly until we see either two heads in a row (I win) or a tail followed by a head (you win). What is the probability that you win?

# The Better Bet

##### Age 14 to 16 Challenge Level:

Well done Clare from Arundel, you have used a very clear systematic method and explained it well too!

#### For the 4 Coins

One possibility is H H T T and the chance of that is $\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}$ = $\frac{1}{16}$

But H T H T is also a possibility and that too has a $\frac{1}{16}$ chance of happening.

So I'll need to know how many ways there are that I could get two heads plus two tails, and win.

To find that I'll make a list systematically :

Start with a Tail, where could the other Tail be? Three places

H H T T
H T H T
H T T H
T T H H
T H T H
T H H T

So there are six ways to win and they all have the same chance $\frac{1}{16}$, which means I have a $\frac{6}{16}$ chance of winning with this game.

What does a chance of $\frac{6}{16}$ mean? Obviously not that I will win $6$ games in $16$ every time.

But, if I was guessing my average wins out of $16$, it would be better to guess $6$ than to guess $5$ or $7$ for example.

I get $£3$ when I win, so $6$ wins is $£18$.

I think my most reasonable guess is that on average I'll get $£18$ for a cost of $£16$.

I'll guess my average is $£2$ profit in $£16$ staked, or $£1$ in $£8$.

#### Now for the dice.

I'll imagine that they are different colours (red green blue) so I can know them apart.

I could get no sixes at all : $\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6} = \frac{125}{216}$

Or just one six on the red : $\frac{1}{6}\times\frac{5}{6}\times\frac{5}{6} = \frac{25}{216}$

Or just one six on the green : $\frac{5}{6}\times\frac{1}{6}\times\frac{5}{6}$ = $\frac{25}{216}$

Or just one six on the blue : $\frac{5}{6}\times\frac{1}{6}\times\frac{1}{6}$ = $\frac{25}{216}$

I could get two sixes

Not the red : $\frac{5}{6}\times\frac{1}{6}\times\frac{1}{6}$ = $\frac{5}{216}$

Not the green : $\frac{1}{6}\times\frac{5}{6}\times\frac{1}{6}$ = $\frac{5}{216}$

Not the blue : $\frac{1}{6}\times\frac{1}{6}\times\frac{5}{6}$ = $\frac{5}{216}$

Or I could get all three to be six : $\frac{1}{6}\times\frac{1}{6}\times\frac{1}{6}$ = $\frac{1}{216}$

Now I'll use my logic about averages again.

My most reasonable guess for what happens on average per 216 plays of the game is to say that :

125 times I'll get nothing,
75 times I'll get one six and win £2,
15 times I'll get two sixes and win £4,
and just once I'll get all three sixes and win £6
216 plays will have cost me £216 and I'll have won £150 + £60 + £6 = £216

So my most reasonable guess for an average would be no overall profit or loss.

That means that the better bet is the tossed coin not the dice.

Thank-you Clare