Rules are: you pick envelopes one at a time revealing an amount. Once you reveal 5 of the same amount you win that amount.

There are:

5 of one amount

6 of one amount

9 of one amount

20 of one amount

Now to throw a twist on an alternate game play. Same concept, match 5 and win the amount but now you have 2 "free play" envelopes. If you choose a "Free play" you hold on to that and can choose to discard an envelope of a smaller amount. The denomination of envelopes would look like this:

5 of one amount

6 of one amount

8 of one amount

19 of one amount

2 of the free play

Thanks in advance.

Quote:AceHoundDon't know if anyone can figure this out? The game is "match 5".

Rules are: you pick envelopes one at a time revealing an amount. Once you reveal 5 of the same amount you win that amount.

There are:

5 of one amount

6 of one amount

9 of one amount

20 of one amount

Thanks in advance.

Odds of getting five in a random draw:

5 - 658008 to one

6 - 109668 to one

9 - 5222.3 to one

20 - 42.4 to one

Odds that you will pick five from these before you pick five from the 20:

5 - 15504 to one

6 - 2584 to one

9 - 123 to one

Quote:AyecarumbaOdds that you will pick five from these before you pick five from the 20:

5 - 15504 to one

6 - 2584 to one

9 - 123 to one

I don't immediately know how to do the math, but intuitively those numbers look awfully high to me. Especially the 123 to 1 number. I would've expected a number more in the range of 5 or maybe 10 to 1.

Quote:AyecarumbaOdds of getting five in a random draw:

5 - 658008 to one

6 - 109668 to one

9 - 5222.3 to one

20 - 42.4 to one

Odds that you will pick five from these before you pick five from the 20:

5 - 15504 to one

6 - 2584 to one

9 - 123 to one

I may have mislead the concept. You continue to pick envelopes until you match 5 of the same amount. So you could end up picking a max of 17 envelopes. I find it hard to believe the odds of winning were as high as 658,008 to one.

Quote:PapaChubbyI don't immediately know how to do the math, but intuitively those numbers look awfully high to me. Especially the 123 to 1 number. I would've expected a number more in the range of 5 or maybe 10 to 1.

I just divided the odds of drawing each in five pulls by the odds of getting the common prize. It is very difficult to pull anything other than the common prize in this game as they make up half the possible draws. Imagine that you had a deck of cards with the diamond suit removed. What are the odds that you will pull the cards required to make a Royal Flush in hearts before you draw any five non-picture black cards? That is what you are up against.

Quote:AyecarumbaI just divided the odds of drawing each in five pulls by the odds of getting the common prize. It is very difficult to pull anything other than the common prize in this game as they make up half the possible draws. Imagine that you had a deck of cards with the diamond suit removed. What are the odds that you will pull the cards required to make a Royal Flush in hearts before you draw any five non-picture black cards? That is what you are up against.

That sounds like a much more daunting task than that proposed by the original poster. There is only one way to make a royal flush in hearts. I think the problem is more like "what is the probability that you will draw any five red cards before you draw any five black cards (with the diamond suit removed)". Just to make sure we're both evaluating the same problem, if I draw RBBRRBRBR then this is a win for red, as I drew the fifth red before I got to a 5th black.

It appears that you are calculating the probability of getting 5 in a row of each category, then comparing these probabilities to determine the relative likelihood of getting 5 of any category before 5 of another category shows up. It is possible that this yields the correct solution, but I don't see it. I'd think the correct solution would require combinations and permutations out the wazoo.

Trial 1 resulted in 12 times out of 290 that I would have gotten 5 reds first. About 1 in 24.

Trial 2 resulted in 10 times out of 290. 1 in 29. The results appear fairly consistent.

So I think my intuitive guess of 1 in 5 or 10 was too low. I'll meet you half way?

[(5/40)*(4/39)*(3/38)*(2/37)*(1/36)] divided by [(20/40)*(19/39)*(18/38)*(17/37)*(16/36)] = 1/15504.

In other words, if you did this test 15,504 times you might pull the five card prize once.

5: 1887

6: 8001

9: 69043

20: 916213

Quote:mipletI did a quick sim of 1,000,000 trials for 1st one:

5: 1887

6: 8001

9: 69043

20: 916213

Hmm, I would have expected a figure closer to 64 or 65 for the "perfect 5". Your sim demonstrate a 485.5 - 1 ratio for "Common20" vs. "P5". Granted, my figures are the odds of drawing five of each, in five tries, from a full deck of 40, but wouldn't the ratio of outcomes with more attempts still be proportional to those of a "perfect pull"?