Two dice problem, a lot of questions

[flyingpig]

Suppose you are going to a night festival and you see a vendor who is willing to double your money. So he explains the game.

He has two dice, they are numbered from 0 to 9. He will roll them together in a slot such that they will make a two or one digit number. If he happens to roll a number over 55 (but not equal to), he returns your money and doubled it.

Here are some specific rules.

1. If he rolls a 00, you win.
2. If he rolls a 55, he wins
3. The dice are rolled in a slot, so 35 and 53 are distinguishable.

Now here are my questions.

1. What is the probability of the same number being rolled again? (i.e. rolling two number less than or equal to 55 consecutively)

2. What is the probablity of rolling 55 or less consecutively?

3. Can you come up with a way to cheat the vendor?

4. Suppose the two dice now disappear and now it is a computer generating random numbers from 1 to 100, how would that change? The specific rules still apply (other than 1 of course)Attempt

I think it is possible to cheat the vendor, I think it's disguising to think your chances of winning is 45/100, but really it is possible combinations

So let _ _ be the two digit places possible.

So the tenth place for the vendor to win could be

6 * 10 = 60

Because 0 to 5 have 6 digits and 10 for the other possible digits. Then subtract 1 in case we get 100

60 - 1 = 59

Then subtract 5 from 59, (to elimiante 56,57,58,59,60)

59 - 5 = 54

So the total combinations is still 100, but the vendor still has 54/100 chances of winning?

Would it mean that you only have 46/100 chances of winning? I think the best way to win is to double your loss.

Recovering from loss

Suppose you betted 100 tokens, you lose. So you bet again, 200 tokens. You then win, so the vendor gives back 400 tokens.

Now 400tokens you won - 100tokens you loss = 300 tokens

300tokens - 200tokens you gave the second time = 100 tokens.

So really you won 100 tokens? This is just an example.

Conditonal probablity?

Does that mean the vendor's chances of winning n times is

$$\frac{54^n}{100^n}$$

 

[MisterX]

Assuming a uniform distribution for the number that rolls up on each die, the probability, of each two-digit number showing up would be equal. The probability of winning would be 45/100.

1. seems like two different things
2. for n consecutive times in n games:
$$\frac{56^n}{100^n}$$
Solving for the probability of n consecutive times in more than n games is more difficult. I found http://mathforum.org/library/drmath/view/56637.html", which is relevant.
4. Win and lose probabilities wouldn't change.

As for doubling the bet after a loss, the strategy is known as "Martingale." There is http://en.wikipedia.org/wiki/Martingale_%28betting_system%29" .

 

[flyingpig]

How did you get 45/100?

56 + 45 = 101?

 

[MisterX]

For question 2 you asked "What is the probablity of rolling 55 or less consecutively?"

if you had written "What is the probability of rolling 55 or less, excluding 00, consecutively?" then I would have written 55 in my answer to question 2.

 

[CellCoree]

?I would like to clarify a few things about this game and address the questions raised.

First, the probability of rolling the same number again is not affected by the previous roll. Each roll is an independent event, so the probability remains the same for each subsequent roll.

Second, the probability of rolling 55 or less consecutively can be calculated by considering all possible combinations of two dice and counting the number of combinations that result in a number less than or equal to 55. This can be done by listing out all possible combinations or by using a mathematical formula such as the binomial coefficient.

Third, while it may be possible to come up with strategies to increase your chances of winning, it is important to remember that gambling is a game of chance and there is no guaranteed way to beat the odds. Additionally, cheating is not a ethically or morally acceptable solution.

Finally, if the game changes to a computer generating random numbers from 1 to 100, the probability calculations would change accordingly. The specific rules would still apply, but the number of possible outcomes and the probability of each outcome would be different. It is important to consider all possible outcomes and their associated probabilities in order to make informed decisions when playing this game.