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Gambling at Monte Carlo

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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Balls and Bags

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

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Win or Lose?

A gambler bets half the money in his pocket on the toss of a coin, winning an equal amount for a head and losing his money if the result is a tail. After 2n plays he has won exactly n times. Has he more money than he started with?

Coin Tossing Games

Stage: 4 Challenge Level: Challenge Level:2 Challenge Level:2

The games involve tossing a fair coin until my chosen sequence or your chosen sequence occurs.

If I pick HH and you pick TH the only way that I can win is if the first two throws are HH. This occurs with probability 1/4 so P(you win) = 3/4.

What if I pick HH and you pick HT ?

If the first two throws are HH then I win, and if they are HT then you win. If we get TH, then the third throw is equally likely to be H (I win) or T (you win), so if the first pair is TH, then the probability that you win is 1/2. If the first two throws are TT let p be the probability that you win. The third throw is equally likely to be H (we have seen that the probability that you win starting from TH is 1/2 ) or T (and the probability that you win from TT is p). Thus

p = (1/2 x 1/2) + (p x 1/2) so that p = 1/2.

This leads to P(you win)}= (0 + 1 + 1/2 + 1/2 )/4 = 1/2.

What if I pick HH and you pick TT ?

If the first pair is HH then I win, and if the first pair is TT then you win. Let p and r be the probability that you win if the first pair is HT and TH respectively. By considering the outcome of the third throw, we get

p = 1/2 + (r x 1/2)

r = p x 1/2

Solving these equations, we find p=2/3 and r=1/3.
This means that P(you win) = (0 + 1 + 2/3 + 1/3)/4 = 1/2.

What is the best choice you can make?
The best choice you can make is TH.

What should you choose if I choose TT ?
By symmetry between H and T, if I choose TT then you should choose HT, and then P(you win = 3/4.

What happens if I choose HT ?

1. We have already considered HH versus HT and found that they are equally likely to win. So if I choose HT and you choose HH , then the probability that you win is 1/2.
2. Suppose you choose TH. If the first pair is HH or HT then I win, and if it is TH or TT then you win, and so P(you win)=1/2.
3. Suppose that you choose TT. We have already considered TT versus HT , and we know that the probability that HT wins is 3/4. Hence the probability that you win with TT against HT is 1/4.

Assuming that you always make a choice that maximises your chance of winning, what should I choose to maximise the probability that I win?

We still have to consider what happens if I choose TH.
By symmetry with the case where I choose HT , we see that if you choose HH then you win with probability 1/4, if you choose HT then you win with probability 1/2, and if you choose TT then you win with probability 1/2.

Using all the above results, I should choose HT or TH, because either of these choices makes my probability of winning equal to 1/2, whereas for HH and TT my winning probability is only 1/4 (assuming that you always make a choice that maximises your chance of winning).

Now suppose that we look at triples instead of pairs.
What is the probability that you win if I choose HHH and you choose THH ?

I can only win if the first three throws are HHH.
This occurs with probability 1/8 , so the probability that you win is 7/8.

I have eight possible choices and, for each one, you can find a triple that gives you a better than even chance of winning (ie a triple that makes your probability of winning more that 1/2).

We have seen just above that it is possible if I choose HHH. We now consider all the other possibilities in turn.

1. If I choose TTT then by symmetry you should choose HTT so that your probability of winning is 7/8.
2. I choose HTH . Suppose you try HHT. You win if the starting triple is HHH , HHT or THH. I win if the starting triple is HTH. Let p , q , r and s be the probability that you win if the starting triple is HTT , THT , TTH and TTT respectively.

By considering what happens at the fourth throw, we find

p = (r x 1/2) + (s x 1/2)
q = (p x 1/2)
r = (qx1/2) + 1/2
s = (r x 1/2) + (s x 1/2).

We find r = s = p = 2/3 and q = 1/3.
This means that
P(you win) = (1 + 1 + 0 + 2/3 + 1 + 1/3 + 2/3 + 2/3)/8 = 2/3.

3. By symmetry, if I choose THT and you choose TTH,then P(you win) = 2/3.
4. I choose HHT. Suppose you try choosing THH. I win if the starting triple is HHH or HHT , otherwise you win, so that {P(you win) = 3/4.
5. By symmetry, if I choose TTH and you choose HTT , then P(you win) = 3/4.
6. I choose HTT. Suppose you choose HHT . Then you win if the starting triple is HHH , HHT or THH , but I win if it is HTT . Carrying out calculations similar to those above, we find that if the starting triple is HTH , TTH or TTT then your winning probability is 2/3, but if it is THT your winning probability is 1/3. Thus P(you win) = 2/3.
7. By symmetry, if I choose THH and you choose TTH, then P(you win) = 2/3.

For each of my eight possible choices, there is a triple that gives you a better than even chance of winning.