- #1

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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

[tex]\frac{54^n}{100^n}[/tex]