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A man went to Monte Carlo to try and make his fortune. Whilst he was there he had an opportunity to bet on the outcome of rolling dice. He was offered the same odds for each of the following outcomes: At least 1 six with 6 dice. At least 2 sixes with 12 dice. At least 3 sixes with 18 dice.

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Marbles and Bags

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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

Stage: 4 Challenge Level: Challenge Level:1

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 Head, where could the other Head be? Three places

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


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